Problems and solutions to _Abstract Algebra_ by John Beachy and William Blair. I will start at chapter 5 and work to the
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Problems and Solutions for Abstract Algebra John A. Beachy & William D. Blair April 24, 2012
Contents 5 Commutative Rings
3
5.1
Commutative Rings; Integral Domains . . . . . . . . . . . . . . .
5.2
Ring Homomorphisms
5.3
Ideals and Factor Rings
5.4
3
. . . . . . . . . . . . . . . . . . . . . . . .
11
. . . . . . . . . . . . . . . . . . . . . . .
17
Quotient Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27
6 Fields
32
6.1
Algebraic Elements . . . . . . . . . . . . . . . . . . . . . . . . . .
32
6.2
Finite and Algebraic Extensions . . . . . . . . . . . . . . . . . . .
36
6.3
Geometric Constructions . . . . . . . . . . . . . . . . . . . . . . .
38
6.4
Splitting Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . .
40
6.5
Finite Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45
6.6
Irreducible Polynomials over Finite Fields
6.7
Quadratic Reciprocity
. . . . . . . . . . . . .
48
. . . . . . . . . . . . . . . . . . . . . . . .
53
7 Structure of Groups
56
7.1
Isomorphism Theorems: Automorphisms . . . . . . . . . . . . . .
56
7.2
Conjugacy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
56
1
2
CONTENTS
Here's some random crap from chapter 4 that I did by accident.
9. Let f (x) = an xn + an−1 xn−1 + · · · a1 x + a0 be a polynomial with rational coecients. Show that if c is a root of f (x) and c 6= 0, then 1/c is a root of g(x) = a0 xn + a1 xn−1 + · · · + an−1 x + an . Pn a g(1/c) = j=0 n−j if this sum is equal to zero, then so cj . Now, P Pn an−j Pn an−j n n n must the sum c . Now c j=0 cj j=0 cj = j=0 an−j cn−j = f (c) = 0. n We now see that g(1/c) = f (c)/c . Since f (c) = 0 and c 6= 0, we conclude that g(1/c) = 0. Examine
10. Let f (x) = an xn + an−1 xn−1 + · · · a1 x + a0 be a polynomial with rational coecients. Show that if c is a root of f (x), and k is a nonzero constant, then kc is a root of g(x) = bn xn + bn−1 xn−1 + · · · + b1 x + b0 where bi = k n−i ai . g(kc) = Pn j j=0 aj c = 0.
Examine that
g(x).
Pn
Pn k n−j k j cj = k n j=0 aj cj . Since f (c) = 0, we know Pn n j Thus, k j=0 aj c = g(kc) = 0. Thus, kc is a root of j=0
Chapter 5
Commutative Rings 5.1
Commutative Rings; Integral Domains
1. Which of the following sets are subrings of the eld Q of rational numbers? Assume that m , n are integers with n 6= 0 and (m, n) = 1. m n |n is odd m (b) mn |n is even (c) nm|4 - n (d) n |(n, k) = 1 where k is a xed positive integer (a)
Q, we know that + is abelian and associa· distributes over +. Thus, we only need to show closure. Observe that for a, b, c, d ∈ Z, a c ad+bc b + d = bd . Now, since b and d are both odd (they do not have a factor of 2), there is no way that bd can be even. (Even if there is some cancellation that (a) Since we are very familiar with
tive and we know inverses and identity for +. We also know that
occurs, a factor of 2 can never magically appear in the denominator.)
1 1 1 / m 2 + 2 = 1 = 1 ∈ n |n is even . Thus, the group property is not satised, so (b) is not a subring. 1 2 13 (c) Examine / m 3 + 7 = 28 ∈ n |4 - n . Thus, the group property is not satised, so (b) is not a subring. a c ad+bc (d) Again we only need to show closure. since b + d := bd , and neither b nor d share any factor in common with k , it is clear that (bd, k) = 1. Thus, (d) (b) Examine
is a subring.
2. Which of the√following sets are subrings of the eld R of real numbers?
A = {m + n √2|m, n ∈ Z and n is even} B = {m + n 2|m, n ∈ Z and m is odd} √ 3 (c) C = {a + b 2|a, b ∈ Q} √ √ 3 3 (d) D = {a + b 2 + c 9|a, b, c ∈ Q} √ (e) E = {m + nu|m, n ∈ Z},where u = (1 + 3)/2 (a)
(b)
3
CHAPTER 5.
(f )
4
COMMUTATIVE RINGS
F = {m + nv|m, n ∈ Z},
where
v = (1 +
√
5)/2
(a) As in Exercise 1, we will only need to show closure of + (or lack thereof ).
√ √ √ (a + b 2) + (c + d 2) = (a + c) + (b + d) 2. The sum of two even numbers is even, so A is a subring. √ √ √ (b) Examine (3 + 2) + (3 + 2) = 6 + 2 2∈ / B . Thus, B is not a subring. √ √ √ 3 3 3 (c) Similar to (a), (a + b 2) + (c + d 2) = (a + c) + (b + d) 2. The sum of two rationals is rational, so C is a subring. (d) Similar to (c), D is a subring. √ √ 1+ 3 = 3+ 3 ∈ (e) Examine (1 + u) + (1 + u) = 2 + 2u = 2 + 2 · / E . Thus, 2 E is not a subring. (f ) Similarly, F is not a subring. (Note: Sets like A, B, C, D, E, and F can almost be thought of as vector spaces. For example, D can be thought of as a vector space over Q with basis √ √ {1, 3 2, 3 9}. Then, one only needs to check that the set from which the coWe see that
ecients come is closed, since one can always write vectors as a unique linear combination. Sets like
A
don't technically form a vector space since
Z
is a ring
and not a eld. However, there is still a concept of basis associated with this set. Such a set with a basis is called a free module.)
3. Consider which of the following conditions on the set of all 2 × 2 matrices a b with rational entries. Which conditions below dene a commutative c d
ring? If the set is a ring, nd all units. (a) all matrices with d = a, c = 0 (b) all matrices with d = a, c = b (c) all matrices with d = a, c = −2b (d) all matrices with d = a, c = −b (e) all matrices with c = 0 (f ) all matrices with a = 0 and d = 0.
(a) We know that matrix addition is commutative and we know inverses and identity. Furthermore, matrix multiplication distributes over matrix addi-
·. Exam d a+c b+d a b c d = . Also, · = c 0 a+c 0 a 0 c c d a b · . Thus, (a) denes a commutative ring. 0 c 0 a −1 1 −b a b a a2 From elementary linear algebra, we see that = which 0 a 0 a1 is dened so long as a 6= 0. So, the set of units is the set dened by (a) minus any matrix where a = 0. a b c d a+c b+d a b (b) Examine + = . Also, · b a d c b+d a+c b a tion.
Thus, we only need to check for closure and commutivity of
a b c ine + 0 a 0 ac ad + bc = 0 ac
CHAPTER 5.
5
COMMUTATIVE RINGS
a b · . Thus, (b) denes a b a −1 a −b a b 2 −b2 2 −b2 a a commutative ring. From elementary linear algebra, = −b a b a a2 −b2 a2 −b2 2 2 which is dened as long as a − b 6= 0. Thus, the set of units is the set dened 2 2 by (b) minus those elements where a − b = 0. a b c d a+c b+d (c) Examine + = . Also, −2b a −2d c −2(b + d) a + c a b c d ac − 2bd ad + bc c d a b · = = · . −2b a −2d c −2ad − 2bc ac − 2bd −2d c −2b a −1 a −b a b 1 Thus, (c) denes a commutative ring. Now, = a2 +2b 2 2b a −2b a which is dened as long as a and b are not both equal to 0. a b c d a+c b+d (d) Examine + = . Multiplica−b a −d c −(b + d) a + c c d
d c
=
ac + bd ad + bc ad + bc ac + bd
c d
=
d c
1 in (c). Thus, 2 andobtain elements −1 a b a −b 1 = a2 +b2 which −b a b a
tion is commutative since we can factor out
(d) denes a commutative ring. We see again is dened as long as
a
and
b
(e) Closure is clear. However,
3 0
2 1
=
12 0
0 2
1 0
=
6 0
0 4
2 1
4 · 0
5 6
=
12 0
27 6
4 0
6=
5 6
·
. Thus, (e) does not dene a commutative ring.
(f ) Closure is clear. However,
3 0
13 6
are not both 0.
0 2
1 0
0 · 4
3 0
=
4 0
0 6
6=
0 4
3 0
·
. Thus, (f ) does not dene a commutative ring.
√
4. Let R = {m + n √2|m, n ∈ Z}. 2 2 (a) Show that m + n 2 is a unit in R if and only if m − 2n = ±1. √ × (b) Show that 1 + 2 2 has innite order in R . × (c) Show that 1 and −1 are the only units that have nite order in R . √ m + n 2√is a unit if and only if there is an x ∈ R such that x(m + n 2) = (m + n 2) = 1. By commutivity of ·, we only need to √ √ 1√ . consider x(m + n 2). Now, x(m + n 2) = 1 if and only if x = m+n 2 Multiplying numerator and denominator by the conjugate, this implies x = √ 1 m n m2 −2n2 (m − n 2). For x to be a member of R, we need m2 −2n2 , m2 −2n2 ∈ Z. 2 2 Thus, x ∈ R if and only if m − 2n = ±1. √ r Pr √ s r (b) By the binomial theorem, we have that (1 + 2 2) = s=0 s (2 2) √ which clearly can become arbitrarily large. Thus, o(1 + 2 2) = ∞. √ √ (c) First of all, by (a), any unit must have the form ± ±1 + 2n2 + n 2. √ √ √ r−s √ r Pr s Then, by the binomial theorem, ± ±1 + 2n2 + n 2 = s=0 ± ±1 + 2n2 nr−s 2 . (a) We see that
√
CHAPTER 5.
6
COMMUTATIVE RINGS
This can become arbitrarily large unless
√
n=0
in which case
√ ± ±1 + 2n2 +
n 2 = ±1.
5. Let R be a subset of an integral domain D. Prove that if R is a ring under the operations of D, then R is a subring of D. · is assumed since R is a ring under the operations of D. (R, +) is an abelian group, we must have inverses. Thus, if a ∈ R, we have −a ∈ R. Finally, suppose that x ∈ R. Then, since 1 · x = x ∈ R, we have 1 ∈ R by closure. Thus, R is a subring of D .
Closure for + and Since must must
6. Let D be a nite integral domain. Give another proof of Theorem 5.1.9 by showing that if d is a nonzero element of D, then d−1 = dk for some positive integer k. [Theorem 5.1.9 states that any nite integral domain is a eld.] d ∈ D. By closure, d · d ∈ D. Now, D is assumed to be nite, so for m and n, we must have dm = dn . (Otherwise D would have to be innite.) m Then, d = dn ⇔ 1 = dn−m by the cancellation property for integral domains. n−m−1 −1 Thus, 1 = d · d . Set k = n − m − 1. Then clearly d = dk . Thus, every element has an inverse, so D is a eld. Let
some
7. An element a of a commutative ring R is called nilpotent if an = 0 for some positive integer n. Prove that if u is a unit in R and a is nilpotent, then u − a is a unit in R. an − 1 = 1. Now, 1 is clearly a root of an − 1, so we can write a − 1 = (a − 1) · p(a) where p(a) is a polynomial in the indeterminant a. Thus, (a − 1) · p(a) = 1, so (a − 1) is a unit. (We see that p(a) ∈ R since it is just linear combos of powers ofa.) × −1 Now, let u ∈ R . Then, (a − u) = u(au − 1). Since a is nilpotent, so is −1 au . Thus, by the previous paragraph, (au−1 − 1) ∈ R× . Since u ∈ R× , we −1 must have u(au − 1) ∈ R× . Consider the element
n
8. Let R be a commutative ring such that a2 = a for all a ∈ R. Show that a + a = 0 for all a ∈ R. (a+1)2 = a2 +2a+1. But, a2 = a for all a ∈ R, so (a+1)2 = a+1 a + 1 = a + 2a + 1. Thus, a + 1 = 3a + 1 implying 0 = 2a, as desired.
Notice that and
9. Let I be any set and let R be the collection of all subsets of I . Dene addition and multiplication of subsets A, B ⊆ I as follows: A + B = (A ∪ B) ∩ A ∩ B and A · B = A ∩ B.
Show that R is a commutative ring under this addition and multiplication.
CHAPTER 5.
7
COMMUTATIVE RINGS
We simply verify the axioms. (Closure of +) Since
A, B ⊆ I , A ∪ B ⊆ I by denition of union. I is completely contained in I .
the intersection of any set with a set in
Then Thus,
A + B ⊆ I. (Associativity of +) Omitted. (Zero Element)
∅
A + ∅ = (A − ∅) ∪ (∅ − A) = A ∪ ∅. A + A = (A − A) ∪ (A − A) = ∅. Thus,
is the zero element:
(Additive Inverses) If
A ∈ R,
then
every element is its own additive inverse. (Multiplicative identity)
I
is the multiplicative inverse:
A·I = A∩I =
I ∩ A = A. (Commutivity) Given since (Distributivity of
·
over
+)
∩
is commutative.
Omitted.
10. For the ring R dened in Exercise 9, write out addition and multiplication tables for the following cases: (a) I has two elements; (b) I has three elements. (a) Suppose
I = {a, b}.
2I = {I, {a}, {b}, ∅}.
Then
Here is the addition
table:
Here is the multiplication table:
(b) Suppose
I = {a, b, c}.
Then
2I = {I, {a, b}, {a, c}, {b, c}, {a}, {b}, {c}, ∅}.
Here is the addition table: Here is the multiplication table:
CHAPTER 5.
8
COMMUTATIVE RINGS
(b) Omitted
11. A commutative ring R is called a Boolean ring if a2 = a for all a ∈ R. Show that in a Boolean ring the commutative law follows from the other axioms. a, b ∈ R. Then ab = ba ⇔ a2 b = aba ⇔ ab = aba ⇔ ab2 = abab ⇔ ab = (ab) . Since R is boolean, the last equality is true, so all the equalities that are derived from it including ab = ba are true. This completes the proof. Let
2
12. Let I be any set and let R be the collection of all subsets of I . Dene addition and multiplication of subsets A, B ⊆ I as follows: A + B = A ∪ B and A · B = A ∩ B.
Is R a commutative ring under this addition and multiplication? Observe that
∅
is the additive identity of
R
since
A∪∅ = A
for all
However, one can never union a set to any nonempty set and obtain unless
I = ∅, R
∅.
A ⊆ I. Thus,
is not a ring.
13. Let R be the set of all continuous functions from the set of real numbers to itself. (a) Show that R is a commutative ring if the formulas (f +g)(x) = f (x)+g(x) and (f · g)(x) = f (x) · g(x) for all x ∈ R are used to dene addition and multiplication of functions. (b) Which properties in the denition of commutative ring fail if the product of two functions is dened to be (f g)(x) = f (g(x)) for all x? (a) (Associativity of +) Given since + is associative. (Additive Identity) The function for all continuous functions
f ≡0
is continuous and
(Additive Inverses) For a continuous function
f (x)
(g + f )(x) = g(x)
g. f , the function −f (x) = (−1) ·
is its additive inverse.
· is commutative. (f ·(g +h))(x) = f (x)·[g(x)+h(x)] = f (x)g(x)+f (x)h(x) =
(Commutativity) Given since (Distributivity)
(f g)(x) + (f h)(x). f (x) = x2 2 and g(x) = −x, then (f g)(x) = x , but (gf ) = −x . Also, if f (x) = x , g(x) = 3 4 3 4 2 x , and h(x) = x , then f (g + h)(x) = (x + x ) , while f (g(x)) + f (h(x)) = x6 + x8 . (b) Commutativity fails and so does distributivity. For example, if
2
2
14. Dene new operations on Q by letting a ⊕ b = a + b and a b = 2ab for all a, b ∈ Q. Show that Q is a commutative ring under these operations. (Group under +) Given.
CHAPTER 5.
COMMUTATIVE RINGS
9
) a b = 2ab = 2ba = b a. over ⊕) a (b⊕c) = 2a(b+c) = 2ab+2ac = (a b)⊕(a c)
(Commutativity of (Distributivity of
15. Dene new operations on Z by letting m ⊕ n = m + n − 1 and m n = m + n − mn, for all m, n ∈ Z. Is Z a commutative ring under these operations? (Associativity of
⊕)
Given by associativity of + and -.
m + 1 − 1 = m for all m ∈ Z. m ∈ Z, −m+2 is such that m+(−m+2)−1 = 1. (Identity element) 0 is the identity element since m 0 = m + 0 − m(0) = m. (Commutativity of ) m n = m + n − mn = n + m − nm = n m. (Distributivity of over ⊕) ` (m ⊕ n) = ` (m + n − 1) = ` + m + n − 1 − (`m + `n − `) = (` + m − `m) + (` + n − `n) − 1 = (` m) ⊕ (` n). (Zero element) 1 is the additive identity since (Additive Inverseses) For all
16. Let R and S be commutative rings. Prove that the set of all ordered pairs (r, s) such that r ∈ R and s ∈ S can be given a ring structure by dening (r1 , s1 ) + (r2 , s2 ) = (r1 + r2 , s1 + s2 ) and (r1 , s1 ) · (r2 , s2 ) = (r1 r2 , s1 s2 ).
This is called the
direct sum
of R and S , denoted by R ⊕ S .
(Associativity of +) Given by the associativity of vector addition.
(0R , 0S ) where 0R and 0S are the zero elements R ⊕ S. (Additive Inverses) If (a, b) ∈ R ⊕S , then (−a, −b) is its inverse since (a, b)+ (−a, −b) = (0R , 0S ). (Identity Element) The element (1R , 1S ) where 1R and 1S are the identity elements of R and S is the identity element since (a, b) · (1R , 1S ) = (a, b). (Commutativity) (a, b) · (c, d) = (ac, bd) = (ca, db) = (c, d) · (a, b). (Distributivity) (a, b)·[(c, d)+(e, f )] = (a, b)·(c+e, d+f ) = (ac+ae, bd+bf ) = (a, b) · (c, d) + (a, b) · (e, f ). Thus, R ⊕ S is a commutative ring. (Zero Element) The element
of
R
and
S
respectively is the zero element of
17. Give addition and multiplication tables for Z2 ⊕ Z2 . Omitted.
18. Generalizing to allow the direct sum of three commutative rings, give addition and multiplication tables for Z2 ⊕ Z2 ⊕ Z2 . Omitted.
19. Find all the units of the following rings. (a) (b)
Z⊕Z Z4 ⊕ Z9
CHAPTER 5.
10
COMMUTATIVE RINGS
(a) From Q16, the identity element is
(ac, bd) = (1, 1). Since a, b, c, d ∈ Z, a = b = c = d = 1.
(1, 1).
Now,
(a, b) · (c, d) = (1, 1) ⇔ ac = 1 and bd = 1 if
we can only have
(b) From Q16, the identity element is ([1]4 , [1]9 ). Now, ([a]4 , [b]9 )·([c]4 , [d]9 ) = ([1]4 , [1]9 ) ⇔ ([ac]4 , [bd]9 ). Now, we simply need to nd the units of Z4 and Z9 . We know Z× 4 = {[1]4 , [3]4 } and Z9 = {[1]9 , [2]9 , [4]9 , [5]9 , [7]9 , [8]9 }. Thus, (Z4 ⊕ Z9 )× = {(x, y)|x ∈ Z4 and y ∈ Z9 }.
20. An element e of a ring R is said to be idempotent elements of the following rings. (a) Z8 and Z9 (b) Z10 and Z12 (c) (d)
idempotent
if e2 = e. Find all
Z⊕Z Z10 ⊕ Z12
Z8 : [0]8 , [1]8 Z9 : [0]9 , [1]9 (b) Idempotent elements of Z10 : [0]10 , [1]10 , [5]10 , [6]10 Idempotent elements of Z12 : [0]12 , [1]12 , [4]12 , [9]12 (c) Idempotent elements of Z ⊕ Z: (0, 0), (1, 1) (d) Idempotent elements of Z10 ⊕Z12 : {(x, y)|x ∈ {idempotent {idempotent elements of Z12 }} (a) Idempotent elements of Idempotent elements of
elements of
Z10 }, y ∈
21. Let A be an abelian group, and let R = {(a, b)|a ∈ A and n ∈ Z}. Dene binary operations + and · on R by (a, n) + (b, m) = (a + b, n + m) and (a, n) · (b, m) = (am + bn, nm), for all (a, n) and (b, m) in R. Show that R is a commutative ring. (Group property) Satised since the underlying group structure is the cartesian product of two groups.
(x, y) ∈ R such that (x, y) · (a, n) = (x, y) · (a, n) = (a, n), we must have yn = n y = 1. Thus, we need xn + a = a which implies that
(Identity element) We need an element
(xn + ay, n) = (a, n).
In order for
which can only happen if
x = 0A .
Thus,
(0A , 1)
is the identity element.
(Commutativity) Given. (Distributivity)
(a, `)[(b, m)+(c, n)] = (a, `)·(b+c, m+n) = ((a(m + n) + `(b + c), `(m + n)).
(1)
(a, `) · (b, m) + (a, `)(c, n) = (am + `b, `m) + (an + `c, `n) = (am + `b + an + `c, `m + `n) = (a(m + n) + `(b + c), `(m + n)). (2) We see that (1) = (2), so distributivity has been shown.
22. Let R be a set that satises all the axioms of a commutative ring, with the exception of the existence of a multiplicative identity element. Dene binary operations + and · on R1 = {(r, n)|r ∈ R, n ∈ Z} by (r, n) + (s, m) = (r + s, n + m) and (r, n) · (s, m) = (rs + ns + mr, nm), for all (r, n) and (s, m) in R1 . Show that R1 is a commutative ring with identity (0, 1) and that {(r, 0)|r ∈ R}
CHAPTER 5.
11
COMMUTATIVE RINGS
satises all the conditions of a subring, with the exception that it does not have the multiplicative identity of R. (Group property) Satised since the underlying group structure is the cartesian product of two groups. (Identity element) Similar to Q21. (Commutativity) Given except for multiplication with the identity element. Since we showed that (0,1)
was
the identity element, we know it commutes (by
denition of identity element). (Distributivity)
(a, `)[(b, m)+(c, n)] = (a, `)·(b+c, n+m) = (a(b + c) + `(b + c) + a(n + m), `(n + m)).
(1)
(a, `) · (b, m) + (a, `) · (c, n) = (ab + `b + am, `m) + (ac + `c + an, `n) = (ab+`b+am+ac+`c+an, `m+`n) = (a(b + c) + `(b + c) + a(m + n), `(m + n)). (2) We see
5.2
(1) = (2),
so distributivity has been shown.
Ring Homomorphisms
1. Let R be a commutative ring and let D be an integral domain. Let φ : R → D be a nonzero function such that φ(a + b) = φ(a) + φ(b) and φ(ab) = φ(a)φ(b), for all a, b ∈ R. Show that φ is a ring homomorphism. We must simply show that
D is an integral domain, φ(a) = φ(1) · φ(a) ⇔ 1 = φ(1). Since
φ(1) = 1.
Now,
φ(a) = φ(1 · a) = φ(1) · φ(a).
the cancellation property can be used to obtain
2. Let F be a eld and let φ : F → R be a ring homomorphism. Show that φ is either zero or one-to-one. Clearly
φ≡0
Suppose that that
is a possible mapping.
φ(a) = φ(b)
φ(ab−1 ) = 1.
for some
a, b ∈ F . Since F is a eld, this implies ab−1 = 1 which implies that a = b.
This then implies that
3. Let F, E be elds, and let φ : F → E be a ring homomorphism. Show that if φ is onto, then φ must be an isomorphism. φ is one-to-one. Suppose that φ(x1 ) = φ(x2 ) = y . E is a eld, φ(x1 ) · φ(x2 )−1 = 1. Since F is a eld, we see that this −1 −1 that φ(x1 x2 ) = 1. This means that xx2 = 1, so x1 = x2 . Thus, φ is
We only need to show that Then, since implies
one-to-one.
4. Show that taking complex conjugates denes an automorphism of C. That is, for z ∈ C, dene φ(z) = z , and show that φ is an automorphism of C.
CHAPTER 5.
(1-1) Let
12
COMMUTATIVE RINGS
z = reiθ .
φ(reiθ ) = re−iθ .
Then
This mapping is one-to-one then
by the uniqueness of additive inverses.
w = reiθ . Then, z = re−iθ gives φ(z) = w. (Preservation) Let z = a + ib and w = c + id. Then φ(z + w) = φ[(a + c) + i(b + d)] = (a + c) − i(b + d) = (a − ib) + (c − id) = φ(z) + φ(w). iθ iω i(θ+ω) Let z = re and w = se . Then, φ(zw) = φ(rse ) = rse−i(θ+ω) = −iθ −iω re se = φ(z)φ(w). (Identity) Satised since C is a eld. (See Q1.) (Onto) Let
5. Show that the identity mapping is the only ring homomorphism from Z into Z. φ : Z → Z be a ring homomorphism. Then, we must have φ(1) = 1. n ∈ Z, n = 1 + 1 + · · · + 1 (n terms), we must have φ(n) = φ(1 + 1 + · · · + 1) = φ(1) + φ(1) + · · · + φ(1) = 1 + 1 + · · · + 1 = n. m n 6. Show that the set of all matrices over Z of the form is a ring 2n m √ isomorphic to the ring Z[ 2] dened in Example 5.1.6. Let
Since for any
Let S be the set described in the statement √ √ a b . Z[ 2] → S via φ(a + b 2) = 2b m
of the problem.
Dene
φ :
(1-1) This map is a direct substitution, so it is 1-1.
√ m n m n (Onto). Let ∈ S . Then, φ(m + n 2) = . 2n m 2n m 1 0 (Identity) φ(1) = . 0 1 √ √ √ a+c b+d (Preservation) φ[(a+b 2)+(c+d 2)] = φ[(a+c)+(b+d) 2] = = 2(b + d) a + c √ √ a b c d + = φ(a + b 2) + φ(c + d 2). 2b a 2d c √ √ √ ac + 2bd ad + bc φ[(a+b 2)(c+d 2)] = φ[(ac+2bd)+(ad+bc) 2] = . 2(ad + bc) ac + 2bd
(1)
√ √ φ(a + b 2)φ(c + d 2) =
a 2b
b a
c · 2d
d c
=
ac + 2bd ad + bc 2(ad + bc) ac + 2bd
.
(2) We see
(1) = (2),
so preservation has been shown.
√
√
√
√
7. Dene φ : Z[ 2] → Z[ 2] by φ(m √ + n 2) = m − n 2, for all m, n ∈ Z. Show that φ is an automorphism of Z[ 2]. This is similar to Q4.
We will only show preservation of products.
identity mapping is very easily veried).
(The
CHAPTER 5.
13
COMMUTATIVE RINGS
√ √ √ √ φ[(a + b 2)(c + d 2)] = φ[ac + 2bd + (ad + bc) 2] = ac + 2bd − (ad + bc) 2. (1)
√ √ √ √ √ φ(a + b 2)φ(c + d 2) = (a − b 2)(c − d 2) = ac + 2bd − (ad + bc) 2. (1) = (2), so preservation is shown.
(2)
8. Let F be a eld, and let a ∈ F . Dene φ : F [x] → F [x] by φ(f (x)) = f (x + a), for all f (x) ∈ F [x]. Show that φ is an automorphism of F . (1-1) The transformation it left
a
units. (This
does
f (x + a)
simply takes the graph of
f (x)
and shifts
make sense in an abstract setting since the denition
(x, f (x)).) Thus, this f (x) 7→ f (x + a) is a direct substitution and thus one-to-one. (Onto) Let f (x) ∈ F [x]. Then, f (x − a) 7→ f (x). (Identity) Let f ≡ 1. Then, f (x + a) ≡ 1. n n−1 (Preservation) Let f (x) = an x + an−1 x + · · · + a1 x + a0 and g(x) = m m−1 bm x + bm−1 x + · · · + b1 x + b0 . Without loss of generality, suppose deg g ≥ deg f . φ[f (x)+g(x)] = φ[bm xm +bm−1 xm−1 +· · ·+(an +bn )xn +(an−1 +bn−1 )xn−1 + · · · + (a1 + b1 )x + (a0 + b0 )]= bm (x + a)m + · · · + (an + bn )(x + a)n + · · · + a0 + b0 = (bm (x + a)m + · · · + b0 ) + (an (x + a)n + · · · + a0 ) = φ(g(x)) + φ(f (x)). Let f (x) = (x−r1 )(x−r2 ) · · · (x−rn ) and g(x) = (x−s1 )(x−s2 ) · · · (x−sm ). φ[f (x)g(x)] = φ[(x−r1 ) · · · (x−rn )(x−s1 ) · · · (x−sm )] = (x+a−r1 ) · · · (x+ a − rn )(x + a − s1 ) · · · (x + a − sm ) = φ(f (x))φ(g(x)).
of graph is simply the set of ordered pairs of the form mapping of
9. Show that the composition of two ring homomorphisms is a ring homomorphism. φ : S → T and π : R → S be ring homomorphisms. Dene φ ◦ π : R → T φ ◦ π(x) = φ(π(x)) for all x ∈ R. (Preservation of +) φ(π(x + y)) = φ(π(x) + π(y)) = φ(π(x)) + φ(π(y)). (Preservation of ·) φ(π(xy)) = φ(π(x)π(y)) = φ(π(x))φ(π(y)). (Identity) φ(π(1)) = φ(1) = 1.
Let via
10. Let R and S be rings, and let φ, θ : R → S be ring homomorphisms. Show that {r ∈ R|φ(r) = θ(r)}is a subring of R. r, s ∈ R. Then φ(r + s) = φ(r) + φ(s) = θ(r) + θ(s) = (r + s) ∈ R. (Closure under ·) φ(rs) = φ(r)φ(s) = θ(r)θ(s) = θ(rs). Thus, rs ∈ R. (Additive inverses) Let r ∈ R. Then φ(−r) = −φ(r) = −θ(r) = θ(−r). Thus, −r ∈ R. (Identity) φ(1) = 1 = θ(1). Thus, 1 ∈ R. (Closure under +) Let
θ(r + s).
Thus,
11. Show that the direct sum of two nonzero rings is never an integral domain.
CHAPTER 5.
Let
R
and
14
COMMUTATIVE RINGS
S
(1, 0) · (0, 1) = (0, 0). R ⊕ S.
be nonzero rings. Then, we always have
Neither element is zero, but the product is the zero of
12. Let R1 and R2 be commutative rings. Dene π1 : R1 ⊕ R2 → R1 by π1 ((r1 , r2 )) = r1 , for all (r1 , r2 ) ∈ R1 ⊕ R2 and dene π2 : R1 ⊕ R2 → R2 by π2 ((r1 , r2 )) = r2 for all (r1 , r2 ) ∈ R1 ⊕ R2 . Show that π1 and π2 are ring homomorphisms. (b) Let R be any ring, and let φ : R → R1 ⊕ R2 be a function. Show that φ is a ring homomorphism if and only if π1 φ and π2 φ are both ring homomorphisms. (a)
π1 [(r, s) + (v, w)] = π1 (r + v, s + w) = r + v =
(a) (Preservation of +)
π1 (r, s) + π1 (v, w). (Preservation of ·)
π1 [(r, s) + (v, w)] = π1 (rv, sw) = rv = π1 (r, s)π1 (v, w). π1 (1, 1) = 1. π1 is a ring homomorphism. Similarly, π2 is a ring homomorphism.
(Identity) Thus,
(b) (⇒)Suppose that one of
π1 φ or π2 φ is not a homomorphism. Without π1 φ is not a homomorphism. By Q9, if π1 and φ are homomorphisms, then π1 φ is a homomorphism. By contraposition, if π1 φ is not a homomorphism, then one of π1 or φ is not a homomorphism. Part (a) showed that π1 is a homomorphism, so we conclude that φ is not a loss of generality, suppose that
homomorphism. (⇐) Suppose that
φ
is not a ring homomorphism.
three things must occur: (1) (3)
φ(r + s) 6= φ(r) + φ(s),
Then, at least one of
(2)
φ(rs) 6= φ(r)φ(s),
or
φ(1) 6= 1.
φ(r + s) = (v, w) while φ(r) = (a, b) and φ(s) = (c, d). π1 (φ(v + w)) = π1 ((a + c, b + d)) = a + c. Similarly, π2 (φ(v + w)) = b + d. Now, it is possible for either a + c = v or b + d = w, but not both. For if both held, it would imply that π1 (φ(v + w)) = π1 (φ(v) + φ(w)) and π2 (φ(v + w)) = π2 (φ(v) + φ(w)), which would imply that π1 and π2 are not well dened. Contradiction. Thus, one of π1 φ or π2 φ is not a homomorphism. For (1), suppose that
Then,
(2) Follows from a similar argument. For (3), if
φ(1) 6= (1, 1),
then at least one of
to the identity. Thus, one of
π1 φ
or
π2 φ
π1 (1, 1)
or
π2 (1, 1)
is not equal
is not a homomorphism.
13. Find all ring homomorphisms from Z ⊕ Z into Z. That is, nd all possible formulas and show why no others are possible. φ : Z⊕Z → Z be a homomorphism. Then, since the identities must map (1, 1) 7→ 1. Then, since n = 1 + 1 + · · · + 1 (n terms) for any n ∈ Z, φ(n, n) = n for all n ∈ Z. Thus, we must have φ(m, n) = m or φ(m, n) = n for all (m, n) ∈ Z ⊕ Z. Let
to each other, we see that
14. Find all ring homomorphisms from Z ⊕ Z into Z ⊕ Z.
CHAPTER 5.
Thus, we must
φ
(1, 1) 7→ (1, 1). By extension, we must have (n, n) 7→ have φ(m, n) = (m, n), φ(m, n) = (m, m), or φ(m, n) =
As above, we must have
(n, n). (n, n).
15
COMMUTATIVE RINGS
15. For the rings Zn and Zk , show that if k|n, then the function φ : Zn → Zk dened by φ([x]n ) = [x]k , for all [x]n ∈ Zn , is a ring homomorphism. Show that this is the only ring homomorphism from Zn to Zk . (Preservation of +)
φ([a]n + [b]n ) = φ([a + b]n ) = [a + b]k = [a]k + [b]k =
φ([a]n ) + φ([b]n ). (Preservation of ·)
φ([a]n [b]n ) = φ([ab]n ) = [ab]k = [a]k [b]k = φ([a]n )φ([b]n ). φ : Zn → Zk is a homomorphism, we must have [1]n 7→ [1]k . Also, since for any r ∈ Z, n = 1 + 1 + · · · + 1 (r terms), we know that φ([r]n ) = φ([1]n + [1]n + · · · + [1]n ) = φ([1]n ) + φ([1]n ) + · · · + φ([1]n ) = [1]k + [1]k + · · · + [1]k = [r]k . Thus φ([x]n ) = [x]k is the only homomorphism from Zn to Zk . Now, as mentioned many times before, if
16. Are Z9 and Z3 ⊕ Z3 isomorphic as rings? Let φ : Z9 → Z3 ⊕ Z3 . If φ is ([1]3 , [1]3 ). However, 3 · [1]9 = [3]9 underlying additive groups are not
to be an isomorphism, we need
3 · ([1]3 , [1]3 ) = ([0]3 , [0]3 ), isomorphic, so Z9 Z3 ⊕ Z3 . while
[1]9 7→ so the
17. Let S be the subset of Z4 ⊕ Z4 given by {([m]4 , [n]4 )|m ≡ n(mod 2)}. (a) Show that S is a subring of Z4 ⊕ Z4 . (b) Show that S is not isomorphic (as a ring) to any ring of the form Zn , nor to any direct sum of such rings. (a) (Closure under +) Let ([a]4 , [b]4 ), ([c]4 , [d]4 ) ∈ S . Then, ([a]4 , [b]4 ) + ([c]4 , [d]4 ) = ([a + c]4 , [b + d]4 ). Now if [a]2 = [b]2 and [c]2 = [d]2 , then [a + c]2 = [b + d]2 by simple adding of two valid equations. Thus ([a + c]4 , [b + d]4 ) ∈ S . (Closure under ·) Let ([a]4 , [b]4 ), ([c]4 , [d]4 ) ∈ S . Then, ([a]4 , [b]4 )·([c]4 , [d]4 ) = ([ac]4 , [bd]4 ). Now there are two possibilities: (1) [c]2 = [d]2 = [0]2 , or (2) [c]2 = [d]2 = [1]2 . So, when we evaluate [ac]2 and [bd]2 , in (1) we get [0]2 and [0]2 and in (2) we get [a]2 and [b]2 . Thus, ([ac]4 , [bd]4 ) ∈ S . (Additive inverses) If [a]2 = [b]2 , then −[a]2 = −[b]2 . Thus, if ([a]4 , [b]4 ) ∈ S , we have −([a]4 , [b]4 ) ∈ S . (Identity Element) Clearly, [1]2 ⊂ [1]4 , so ([1]4 , [1]4 ) ∈ S . Thus, S is a subring of Z4 ⊕ Z4 . (b) We note that the element Thus, if
S
isomorphic to If
S
([1]4 , [1]4 ) ∈ S
has multiplicative order 4.
were isomorphic to a direct sum of residue rings, it would have to be
Z4 ⊕ Z4 .
However,
([1]4 , [2]4 ) ∈ Z4 ⊕ Z4 − S .
Thus
S Z4 .
were to be isomorphic to a ring of residues, it would have to be iso-
morphic to
Z4 .
However,
([n]4 , [n]4 )
are in
S
for all
n,
giving four congruence
CHAPTER 5.
16
COMMUTATIVE RINGS
classes. In addition,
([1]4 , [3]4 ) ∈ S .
Thus,
|S| > |Z4 |,
so
S Z4 .
18. Dene φ : Z → Zm ⊕ Zn by φ(x) = ([x]m , [x]n ). Find the kernel and image of φ. Show that φ is onto if and only if gcd(m, n) = 1. ker φ = {x ∈ Z : n|x and m|x}. The image is {([x]m , [x]n )|x ∈ gcd(m, n) = 1, then m and n share no factors, so no integers besides 0 and kmn where k ∈ Z can be mapped to ([0]m , [0]n ). Since |Zm ⊕ Zn | = mn, the rst mn integers will be mapped to unique elements of Zm ⊕ Zn . Thus, φ We see that
Z}.
If
is onto.
gcd(m, n) = d 6= 1, then we have m = dr and n = dq for some drq 7→ ([0]m , [0]n ). Since drq < mn, the mapping will replicate all the previous (drq−1) mappings over and over again without having generated mn unique elements (the order of Zm ⊕ Zn ). Thus, φ is not onto. Conversely, if
r, q ∈ Z.
Thus,
19. Let R be the ring given by dening new operations on Z by letting m ⊕ n = m + n − 1 and m n = m + n − mn. Dene φ : Z → R by φ(n) = 1 − n. Show that φ is an isomorphism. φ is a linear map. y ∈ R. Then φ(−y + 1) = 1 − (−y + 1) = y . Thus, φ is onto. (Preservation of sums) φ(m + n) = 1 − (m + n) = 1 − m − n (1). φ(m) ⊕ φ(n) = (1 − m) ⊕ (1 − n) = 1 − m + 1 − n − 1 = 1 − m − n (2). Now, (1) = (2), so sums are preserved. (Preservation of products) φ(mn) = 1 − mn. (1) φ(m) φ(n) = (1 − m) (1 − n) = 1 − m + 1 − n − (1 − m − n + mn) = 1 − mn (2). Again, (1) = (2), so products are preserved. (1-1) Given since
(Onto) Let
20. Let I be any set and let R be the collection of all subsets of I . Dene addition and multiplication of subsets A, B ⊆ I as follows: A + B = (A ∪ B) ∩ A ∩ B and A · B = A ∩ B.
Show that if I has two elements then R ∼ = Z2 ⊕ Z2 . (b) Show that if I has three elements then R ∼ = Z2 ⊕ Z2 ⊕ Z2 . (a)
For both of these, one can merely construct multiplication- and addition tables and note that the rings behave in the same way. The multiplication- and addition tables for
R
in these two particualar cases were already constructed in
Q10 of 5.1.
21. Let R1 , R2 , ..., Rn be commutative rings. Complete the proof of Proposition 5.2.8, to show that R = R1 ⊕ R2 ⊕ · · · ⊕ Rn is a commutative ring. Then show that R× ∼ = R1× × R2× × · · · × Rn× .
CHAPTER 5.
17
COMMUTATIVE RINGS
(Group Property) Given by Proposition 3.3.4.
R1 , ..., Rn . (1R1 , 1R2 , ..., 1Rn ) is obviously the identity element. (Distributivity) (r1 , ..., rn )[(s1 , ..., sn )+(t1 , ..., tn )] = (r1 , ..., rn )(s1 +t1 , ..., sn + tn ) = (r1 [s1 +t1 ], ..., rn [sn +tn ]) = (r1 s1 +r1 t1 , ..., rn sn +rn tn ) = (r1 , ..., rn )(s1 , ..., sn )+ (r1 , ..., rn )(t1 , ..., tn ) (Commutativity) Given by the commutativity of (Identity Element)
u = (u1 , ..., un ) ∈ R× . We must have uu−1 = 1R = (1R1 , ..., 1Rn ). Thus, (u1 , ..., un )u−1 = (1R1 , ..., 1Rn ). We see that because R1 , ..., Rn are commutative × −1 −1 × ∼ rings, this can only happen if ui ∈ Ri and u = (u−1 = 1 , ...un ). Thus, R × × × R1 × R2 × · · · × Rn . Let
22. Let R be an integral domain. Show that R contains a subring isomorphic to Zp for some prime number p if and only if char(R) = p. char(R) = p ⇔ p · 1 = 0 ⇔< 1 >∼ = Zp .
for some prime
p ⇔ 1 + 1 + · · · + 1 = 0 (p
terms)
23. Show that if R is an integral domain with characteristic p > 0, then for all a, b ∈ R we must have (a + b)p = ap + bp . Show by induction that we must n pn pn also have (a + b) = a + bp for all positive integers n. p p−k k Pp p! ap−k bk . By the binomial theorem, (a+b) = b = k=0 k!(p−k)! k=0 k a p! Clearly, k!(p−k)! will always yield a factor of p unless k = 0 or k = p. Thus, all the terms of the sum are 0 except for the terms where k = 0 or k = p. Thus, (a + b)p = ap + bp . Pp
p
We have established a basis for induction. Suppose the statement is true for N +1
(a + b)p
5.3
p N
N
= ap + bp
p N
N
p = (a + b)p p = p N. Then, p (a +Nb) N N +1 N +1 = ap + bp = ap + bp , as desired.
all positive integers less than or equal to
Ideals and Factor Rings
1. Give a multiplication table for the ring Z2 [x]/hx2 + 1i. Z2 [x]/hx2 + 1i = {ax + b|a, b ∈ Z2 } = {x + 1, x, 1, 0}. x + 1 = 0 to obtain the multiplication table.
We see that the relation
2
·
0
1
0
0
0
0
0
1
0
1
x
x x+1
0
x x+1
1
x+1 x+1
x+1
0
0
x
x+1
We use
CHAPTER 5.
18
COMMUTATIVE RINGS
2. Give a multiplication table for the ring Z2 [x]/hx3 + x2 + x + 1i. Z2 [x]/hx3 +x2 +x+1i = {ax2 +bx+c|a, b, c ∈ Z2 } = {0, 1, x, x+ 1, x , x + x, x + 1, x2 + x + 1}. We use the relation x3 + x2 + x + 1 = 0 to We see that
2
2
2
obtain the multiplication table.
x2
x2 + x
x2 + 1
x2 + x + 1
·
0
1
0
0
0
0
0
0
0
0
0
1
x x2
x+1 x2 + x x2 + 1
x2 2 x +x+1 x+1
x2 + x x+1 x2 + 1 x2 + x x2 + 1
x2 + 1 x2 + 1
x2 + x + 1
0
x2 + x x
1
x x+1 x2 2 x +x x2 + 1 2 x +x+1
x
x+1
1
x2 + 1
1
0
0
0
x2 + 1 x2
Since the multiplication is commutative, the table is symmetric about the diagonal.
3. Let R be the ring Q[x]/hx3 + 2x2 − x − 3i. Describe the elements of R and give the formulas necessary to describe the product of any two elements. Q[x]/hx3 + 2x2 − x − 3i = {rx2 + sx + t|r, s, t ∈ Q}. Using the x3 = −2x2 + x + 3 and x4 = x3 x = −2x3 + x2 + 3x = 5x2 + x − 6, nd the product of any two elements in R. (The back of the book gives
We see that relations we can
something less useful.)
4. Give a multiplication table for the ring Z3 [x]/hx2 − 1i. Z3 [x]/hx2 − 1i = {ax + b|a, b ∈ Z3 } = {0, 1, 2, x, x + 1, x + 2, 2x, 2x + 1, 2x + 2}. It is left to the reader to write out the table. Simply use 2 the relation x = 1 to simplify products. We see that
5. Show that Q[x]/hx2 − 2i ∼ = Q[x]/hx2 + 4x + 2i. φ : Q[x]/hx2 − 2i → Q[x]/hx2 + 4x + 2i via φ(f (x)) = f (x + 2). We 2 have seen from Q8 from 5.2 that φ is a homomorphism. We see that φ(x − 2) = 2 2 2 2 2 x +4x+2, so φ(hx −2i) = hx +4x+2i. Thus, Q[x]/hx −2i ∼ = Q[x]/hx +4x+2i Dene
by Example 5.3.5.
6. Let R = F [x] and let I be any ideal of R. Prove that there is a unique monic polynomial f (x) with I = hf (x)i. (b) Prove that if I is a maximal ideal of R, then I = hp(x)i for some monic irreducible polynomial p(x). (a)
CHAPTER 5.
COMMUTATIVE RINGS
19
(a) We know that since F [x] is a principal ideal domain, we have that I = hg(x)i for some g(x) ∈ F [x]. Suppose g(x) = an xn + · · · a1 x + a0 . −1 Then, f (x) = an · g(x) is a unique monic polynomial with the same roots as g(x). Let h(x) ∈ F [x]. Then, if h(x)/g(x) = q(x) + r(x), then h(x)/f (x) = h(x)/(a−1 n g(x)) = (h(x)/g(x)) · an = an q(x) + an r(x). Thus, the map φ : F [x]/hg(x)i → F [x]/hf (x)i dened by φ(h(x)) = an h(x) gives a one-to-one, onto map which gives that F [x]/hg(x)i ∼ = F [x]/hf (x)i. Thus, I = hf (x)i.
F [x], then I = hp(x)i, where I is maximal, then p(x) is irreducible. Suppose, then, that p(x) is reducible. Then p(x) = f (x)g(x) for some f (x) 6= g(x) ∈ F [x]. Clearly, hf (x)g(x)i ⊂ hg(x)i, hf (x)i. Thus, I is not (b) From (a), we know that if
p(x)
I
is an ideal of
is monic. Thus, we only need to show that if
maximal.
7. Show that the intersection of two ideals of a commutative ring is again an ideal. J are ideals of the commutative ring R. Then, let x, y ∈ I ∩ J x ± y ∈ I ∩ J since x ± y ∈ I and x ± y ∈ J (by denition of ideal). We also see that r · x ∈ I ∩ J : since x ∈ I and I is an ideal, r · x ∈ I ; also, r · x ∈ J since J is an ideal. Thus, I ∩ J is an ideal. I
Suppose
and
r ∈ R.
and
Then
8. Show that if R is a nite ring, then every prime ideal is maximal. Let I be a prime ideal of R. Suppose that J is an ideal of R such that I ⊆ J ⊆ R. Let j ∈ J . Since R is nite, there must be n > m ∈ Z+ so that j m = j n . Thus, j m − j n = 0 ∈ I . This implies that j m (1 − j n−m ) ∈ I . Now, m since I is prime, this implies that either j ∈ I or (1 − j n−m ) ∈ I . If j m ∈ I , m−1 then we have jj ∈ I , implying that j ∈ I or j m−1 ∈ I . If j ∈ I , then I = J m−1 and the problem is solved. Let us suppose that j ∈ I . Thus, jj m−2 ∈ I . m−2 Again, either j ∈ I , or j ∈ I . Continuing this factorization, we get that jj m−(m−1) = jj ∈ I which implies j ∈ I . Thus, j ∈ I and I = J . n−m If (1 − j ) ∈ I , then j m (1 − j n−m ) = 0 ∈ I (remembering that j m = j n ). m n−m Now, j (1 − j ) − (1 − j n−m ) = (j m − 1)(1 − j n−m ) = j m − j n − 1 + j n−m = n−m j − 1 ∈ I (by the closure property. Note that j m (1 − j n−m ) − (1 − j n−m ) = −(1 − j n−m ). Thus, j m − 1 = −1, showing that j m = 0 ∈ I . Then, by the argument in the previous paragraph, j ∈ I . Thus, I = J . [Somehow I think that something is wrong with this paragraph, but I can't nd any invalid statement.]
9. Find a nonzero prime ideal of Z ⊕ Z that is not maximal. I = pZ⊕Z where p is prime. Then, I is a prime ideal since (a, b)·(c, d) ∈ I a = kp or b = kp for some k ∈ Z. Thus, I is prime. We will show (Q25) that the only possible prime ideals are those of the form pZ ⊕ Z where p is prime. However, (Z ⊕ Z)/(pZ ⊕ Z) ∼ = Zp which is a eld, so the ideal Let
implies that
CHAPTER 5.
is maximum.
Z/(0 ⊕ Z) ∼ =Z
20
COMMUTATIVE RINGS
Thus, the only nonmaximal primes are
0⊕Z
or
Z⊕0
since
which is not a eld.
10. Let P be a prime ideal of the commutative ring R. Prove that I and J are ideals of R and I ∩ J ⊆ P , then either I ⊆ P or J ⊆ P . Supposee that there are
x ∈ I, y ∈ J
x, y ∈ / P . Then, since I and J xy ∈ I ∩ J . But I ∩ J ⊆ P y ∈ P . Contradiction.
such that
are ideals, they have the absorption property and so and since
P
is prime, this implies that
x∈P
or
11. Let R be a commutative ring, with a ∈ R. The dened by Ann(a) = {x ∈ R|xa = 0}. Prove that Ann(a) is an ideal of R.
annihilator
Let x, y ∈ Ann(a). Then, (x ± y) · a = xa ± ya = 0 ± 0 = 0. x ± y ∈ Ann(a). Let r ∈ R and x ∈ Ann(a). Then, (r · x) · a = r · 0 = 0. rx ∈ Ann(a). Thus, Ann(a) is an ideal.
of a is
Thus, Thus,
12. Recall that an element of a commutative ring is said to be nilpotent if
an = 0 for some positive integer n. (a) Show that the set N of all nilpotent elements of a commutative ring forms
an ideal of the ring. (b) Show that R/N has no nonzero nilpotent elements. (c) Show that N ⊆ P for each prime ideal P of R.
xm = y n = 0 for some m, n ∈ Z+ .Suppose, without Pmn mn k mn mn−k k loss of generality, that m ≥ n. Now, (x±y) = k=0 (−1) x y = k Pm(n−1) mn mn−k k Pmn mn mn−k k (−1)k x y + k=m(n−1)+1 (−1)k x y . We see k=0 k k k k that both sums are zero since x = 0 for all k > m and y = 0 for all k > n. Thus, x ± y ∈ N . m Next, suppose r ∈ R and x ∈ N . Then, (r · x) = rm xm = r · 0 = 0. Thus, rx ∈ N . Thus, N is an ideal. (a) Let
x, y ∈ N .
Then,
x + N ∈ R/N be such that (x + N )m = xm + N = N for some m m Since x + N = N if and only if x = 0, this implies that R/N has
(b) Let
+
m∈Z
.
no nonzero nilpotent elements.
x ∈ N . Then, xm = 0 for some m ∈ Z+ . P , so 0 = x · xm−1 ∈ P , implying that x ∈ P .
(c) Let ideals
Then,
0∈P
13. Let R be a commutative ring with ideals I, J . Let I + J = {x ∈ R|x = a + b for some a ∈ I, b ∈ J}.
for all prime
CHAPTER 5.
(a) (b)
21
COMMUTATIVE RINGS
Show that I + J is an ideal. Determine nZ + mZ in the ring of integers.
x, y ∈ I + J . Then, x = a + b and y = c + d for a, c ∈ I and b, d ∈ J . x ± y = (a ± c) + (b ± d). Since a ± c ∈ I and b ± d ∈ J , we have x ± y ∈ I + J. Next, let r ∈ R and x ∈ I + J where x = a + b with a ∈ I and b ∈ J . Then, r · x = r(a + b) = ra + rb. Since I and J are ideals, ra ∈ I and rb ∈ J . Thus, rx ∈ I + J . Thus, I + J is an ideal. (a) Let
Then,
(b) Suppose that gcd(n, m) = d. Then, n = dq1 and m = dq2 . Then, let rdq1 + sdq2 ∈ nZ + mZ where r, s ∈ Z. Then, rdq1 ∈ dZ and sdq2 ∈ dZ, so nZ + mZ ⊆ dZ. Since d = gcd(m, n), we have d = rm + sn for some r, s ∈ Z. Thus, d ∈ nZ + mZ. Thus, mZ + nZ = gcd(m, n)Z.
14. Let R be a commutative ring with ideals I, J . Let IJ =
( n X
) ai bi |ai ∈ I, bi ∈ J, n ∈ Z
+
.
i=1 (a) (b)
Show that IJ is an ideal contained in I ∩ J . Determine (nZ)(mZ) in the set of integers.
P P x, y ∈ IJ . Then, x ± y = ai bi ± aj bj ∈ IJ . Thus, x ± y ∈ IJ . Since ai bi ∈ I ∩ J for all i, we also have x ± y ∈ I ∩ J . P P Next, let r ∈ R and x = ai bi ∈ IJ . Then, rx = rai bi . Since I and J are ideals, we have rai ∈ I for all ai , so rx ∈ IJ . On the other hand, rbi ∈ J for all bi , so rx ∈ I ∩ J as well. (a) Let
Clearly
x∈
x=
PN
i=1 ri nsi m where mnZ. Thus, (mZ)(nZ) ⊆ mnZ.
(b) Suppose that
Next, suppose x ∈ mnZ. x = (km)(1 · n) ∈ (mZ)(nZ). Thus, (mZ)(nZ) = mnZ.
Then Thus,
ri , si ∈ Z.
Then,
x = kmn for some k ∈ Z. mnZ ⊆ (mZ)(nZ).
x = mn
P
ri si .
We could write
15. Let M = {f (x, y) ∈ F [x, y]|f (0, 0) = 0} be the maximal ideal of F [x, y] dened in Example 5.3.8. (a) Show that M = {s(x, y)x + t(x, y)y|s(x, y), t(x, y) ∈ F [x, y]}. 2 (b) Using the denition in Exercise 14, nd M . (a) Since [s(x, y)x+t(x, y)y](0,0) = 0, we clearly have {s(x, y)x+t(x, y)y|s, t ∈ F [x, y]} ⊆ M .
CHAPTER 5.
Next, if
22
COMMUTATIVE RINGS
f (0, 0) = 0,
then the constant term of
collect all the terms divisible by (which must be divisible by set. Call the rst expression
x
f
must be zero.
Hence,
into one set of brackets and everything else
y since the constant term is zero) into another s(x, y)x and the second expression t(x, y)y . Thus,
f (x, y) = s(x, y)x + t(x, y)y . P P M 2 = (s1,i (x, y)x+t1,i (x, y)y)(s2,i (x, y)x+t2,i (x, y)y) = [s1,i (x, y)s2,i (x, y)x2 + (s1,i (x, y)t2,i (x, y) + t1,i (x, y)s2,i (x, y))xy + t1,i (x, y)t2,i (x, y)y 2 ]. Not that there 2 is no constant term. Thus, M = M . (b)
√
√
16. Let R = {m+n 2|m, n ∈ Z} and let I = {m+n 2|m, n ∈ Z and m is even}. (a) Show that I is an ideal of R. (b) Find the well-known commutative ring to which R/I is isomorphic. √ √ √ x = 2a + b 2, y = 2c + d √2 ∈ I . Then x ± y =√2(a ± c) + (b ± d) 2. Thus, x ± y ∈ I . Then, if x = 2a + b 2 ∈ I and r = c + d 2 ∈ R, we have rx = √ √ √ √ (2a+b 2)(c+d 2) = 2ac+4bd+(2ad+bc) 2 = 2(ac+2bd)+(2ad+bc) 2 ∈ I . Thus, I is an ideal. (a) Let
(b) Let us nd the congruence classes. in
I,
√
Anything of the form
√ 2a + b 2
is
(2a + 1) + b 2 is not. These are clearly the only √ R. Dene φ : R → Z2 via φ(2a + b 2) = [0]2 and
so anything of the form
two classes that partition √ φ(2a + 1 + b 2) = [1]2 . Then, this is√clearly one-to-one √ and onto. We only √ really need to show that [(2a + 1) + b 2][(2c + 1) + d 2] = (2e + 1) + f 2,
which will be left to the reader (we have shown everything else). Then we will see that the multiplication- and addition tables for Thus,
R/I ∼ = Z2 .
a c
b d
R
are the same as for
Z2 .
17. Let R be the set of all matrices over Q such that a = d and c = 0. (a) Verify that R is a commutative ring. (b) Let I be the set of all matrices for which a = d = 0. Show that I is an ideal of R. (c) Use the fundamental homomorphism theorem for rings to show that R/I ∼ = Q. (a) (Closure of +) Given since
Q
is closed under +.
(Associativity) Given since + is associative in
(Zero element)
a c
b d
0 0
0 0
Q.
is the zero element since
a c
b d
+
0 0
0 0
.
(Additive Inverses)
−a −c
−b −d
is the additive inverse of
(Commutativity of +) Given since + is commutative in
Q.
a c
b d
.
=
CHAPTER 5.
COMMUTATIVE RINGS
a 0
0 1
(Commutativity of ·)
a b 0 b
b a
23
c d ac ad + bc c d · = = · 0 c 0 ac 0 c
.
(Unity element)
1 0
is the unity element.
(Distributivity) Given because matrix multiplication is distributive.
1 0 R is a commutative ring with 1 = . 0 1 0 a 0 b 0 a 0 b 0 a±b (b) Let , ∈ I . Then, ± = ∈ 0 0 0 0 0 0 0 0 0 0 c d c d 0 a 0 ac I . Now, let ∈ R. Then, · = ∈ R. Thus, 0 c 0 c 0 0 0 0 I is an ideal of R. a b (c) Dene φ : R → Q by φ = a. Clearly, φ(R) = Q. Also, 0 a a b ker φ = ∈ R a = 0 . Clearly, ker φ = I . By the fundamental ho0 a momorphism theorem, R/ ker φ ∼ = φ(R), so R/I ∼ = Q. Thus,
18. Let R be a commutative ring with ideals I, J such that I ⊆ J ⊆ R. (a) Show that J/I is an ideal of R/I , (b) Show that the factor ring (R/I)/(J/I) is isomorphic to R/J . (c) Show that J/I is a prime (or maximal) ideal of R/I if and only if J is a prime (or maximal) ideal of R. a + I, b + I ∈ J/I . Then, (a + I) ± (b + I) = (a ± b) + I = c + I ∈ J/I . R/I = {x + I|x ∈ R}, we know (a + I) ± (b + I) ∈ R/I . Next, suppose r ∈ R. Then, r(a + I) = ra + I ∈ R/I . (a) Let
Since
π : R/I → R/J
π(r + I) = r + J . (This is the natural I ⊆ J ⊆ R, we know that the map is onto, so π(R/I) = R/J . Also, the elements that are mapped to zero (aka J ) are the elements of the form j + I , where j ∈ J . These are precisely the elements from J/I . Thus, ker π = J/I . Then, by the fundamental theorem of homomorphisms, we have (R/I)/(J/I) ∼ = R/J . (b) Let
be dened by
projection mapping.) Then, since
(c) (Prime) (⇐) Suppose that J is a prime ideal of R and that we have (a + I)(b + I) = ab + I ∈ J/I . Since ab + I ∈ J/I , we have ab ∈ J and we know that a ∈ J or b ∈ J . Thus, a + I ∈ J/I or b + I ∈ J/I . Thus, J/I is a prime ideal of R/I . (⇒) Suppose that J is not a prime ideal of R. Then, there are x, y ∈ R such that xy ∈ J , yet x ∈ / J and y ∈ / J . Then, xy + I = (x + I)(y + I) ∈ J/I , yet since x, y ∈ / J , we know that (x + I), (y + I) ∈ / J/I .
CHAPTER 5.
COMMUTATIVE RINGS
24
(Maximal) By the correspondence theorem, we know that there is a one-toone correspondence between the ideals of
I.
We showed in (a) that
J/I
R/I and the ideals of R that contain R/I , so then this ideal corresponds
is an ideal of
π −1 (J/I) = J . (⇐) Suppose that J is maximal and that K is an ideal of R such that K ⊇ J/I . Then, we know that π −1 (J/I) = J , and since K ⊇ J/I , π −1 (K) ⊇ J . −1 Since J is maximal, we have π (K) = J , and so K = J/I . (⇒) Suppose that J is not maximal and that K ⊇ J/I is an ideal of R/I with k ∈ K , but k ∈ / J/I . Then, π −1 (k) ∈ / J and since π −1 (K) ⊇ J (since −1 K ⊇ J/I ), we conclude that π (K) ) J . Thus, J is not maximal. to
19. Use Exercise 18 together with Proposition 5.3.9 to determine all prime ideals and all maximal ideals of Zn . I/nZ is a prime ideal of Z/nZ if and only if I is a prime ideal of Z. Z/nZ = Zn are the ideals of the form hpi/nZ where p number. Since in Z, prime = maximal, we have also found all of the
By Q18,
Thus, the prime ideals of is a prime
maximal ideals.
20. In the ring Z[i] of Gaussian integers, let hpi be the ideal generated by a prime number. Show that Z[i]/hpi has p2 elements, and has characteristic p. Z[i]/hpi has the form [a]p + i[b]p . If |Z[i]/hpi| = p2 and char(Z[i]/hpi) = p. Now, all multiples of p are in hpi. We know that any number without an imaginary part which is not divisible by p is of the form [a]p , which are not in hpi, and so form separate equivalence classes. Similarly, pure imaginary numbers form classes of the form i[b]p . Finally, combinations of two classes of these types are also not in hpi and so form even more congruence classes. This combination completely classies all of Z[i] into congruence classes. Thus, any number in Z[i]/hpi has the form [a]p + i[b]p , so our conclusion follows. We want to show that any element in
we can show this, then clearly
21. In the ring Z[i] of Gaussian integers nd necessary and sucient conditions on integers m and n for the element m + ni to belong to the ideal h1 + 2ii. Use these conditions to determine the ideal h1 + 2ii ∩ Z of Z. h1 + 2ii = Z[i] · (1 + 2i). Thus, any element in h1 + 2ii must have (a + ib)(1 + 2i) = a − 2b + i(2a + b), where a, b ∈ Z. Clearly, h1 + 2ii ∩ Z contains precisely the elements of the form a − 2b where 2a + b = 0. This implies that b = −2a, so elements must have the form a − 2(−2a) = a + 4a = 5a. Thus, h1 + 2ii ∩ Z = 5Z. Recall that
form
22. In the ring Z[i] of Gaussian integers, show that the ideal h5 − ii is not a prime ideal.
CHAPTER 5.
25
COMMUTATIVE RINGS
5−i = (1−i)(3+2i). We need to check that 1−i, 3+2i ∈ / h5−ii. h5 − ii has elements of the form (a + ib)(5 − i) = (5a + b) + i(5b − a). If 1 − i ∈ h5 − ii, we would have to have a, b ∈ Z such that 5a + b = 1 and 2 3 and b = − 5b − a = −1. This implies that a = 13 13 . Since neither is in Z, we conclude that 1 − i ∈ / h5 − ii. Now, if 3 + 2i ∈ h5 − ii, we would have to have a, b ∈ Z such that 5a + b = 3 1 and 5b − a = 2. This implies that a = b = 2 . Again, we conclude that 3 + 2i ∈ / h5 − ii. Now, we have found that (1 − i)(3 + 2i) = (5 − i) ∈ h5 − ii, but 1 − i, 3 + 2i ∈ / h5 − ii. Thus, h5 − ii is not a prime ideal. We see that
We see that
23. Let R be the set of all continuous functions from the set of real numbers into itself. In Exercise 13 of Section 5.1, we have shown that R is a commutative ring if the following formulas (f + g)(x) = f (x) + g(x) and (f · g)(x) = f (x)g(x)
for all x, are used to dene addition and multiplication of functions. Let a be a xed real number, and let I be the set of all functions f (x) ∈ R such that f (a) = 0. Show that I is a maximal ideal of R. f (x), g(x) ∈ I . Then, (f ± g)(x) = f (x) ± g(x). We (f ± g)(0) = 0 ± 0 = 0. Thus, (f ± g)(x) ∈ I . Let h(x) ∈ R. Then, (h · f )(x) = h(x)f (x). We see that (h · f )(0) = h(0) · 0 = 0. Thus, (h · f )(x) ∈ I . (I is maximal) We see that I is the set of polynomials with constant term equal to 0. Thus, if we take the factor ring R/I we obtain the polynomials over R which only have constant terms. To see this, note that we obtain an element of R by taking an element of I and adding a real number. Thus, any element of R is of the form a + I , where a ∈ R. This ring is isomorphic to R (by the natural mapping φ(a + I) = a), which is a eld. Thus, since R/I ∼ = R, by Proposition 5.3.9, I is maximal. (I is an ideal) Let
see
24. Let I be the smallest ideal of Z[x] that contains both 2 and x. Show that I is not a principal ideal. The smallest ideal that contains 2 and is the smallest ideal that contains
a.
x
is
h2i ∩ hxi.
(We know that
hai
We have also shown that the intersection
of two ideals is an ideal, and we know that the intersection will give us the
h2i = {f (x) ∈ Z[x]|all coecients are even} hxi = {f (x) ∈ Z[x]|f (x) has no constant term}. Thus, h2i ∩ hxi = {f (x) ∈ Z[x]|f (x) has even coecients with no constant term}. Now, 2 ∈ h2i ∩ hxi, so if h2i ∩ hxi is a principal ideal, we must have f (x) ∈ Z[x] such that 2 ∈ hf (x)i. This would imply that deg f (x) = 0. But then, such an f (x) would be unable to generate x ∈ h2i ∩ hxi. The reason this fails is because Z does not have multiplicative inverses, so ax · b 6= x for any a, b ∈ Z. fewest number of elements.) Now,
and
CHAPTER 5.
26
COMMUTATIVE RINGS
25. Let R and S be commutative rings, let I be an ideal of R, and let J be an ideal of S . (a) Show that I ⊕ J as an ideal of R ⊕ S . (b) Show that (R ⊕ S)/(I ⊕ J) ∼ = (R/I) ⊕ (S/J). (c) Show that I ⊕ J is a prime ideal of R ⊕ S if and only if either I = R and J is a prime ideal of S , or else I is a prime ideal of R and J = S . (d) Show that if K is any ideal of R ⊕ S , then there exists an ideal I of R and an ideal J of S such that K = I ⊕ J . (a, b), (c, d) ∈ I ⊕ J . Then, (a, b) ± (c, d) = (a ± c, b ± d) ∈ I ⊕ J , since J are ideals. Next, let (r, s) ∈ R ⊕ S . Then, (r, s)(a, b) = (ra, sb) ∈ I ⊕ J I, J are ideals. Thus, I ⊕ J is an ideal of R ⊕ S .
(a) Let
I
and
since
(b) Dene φ : R ⊕ S → (R/I) ⊕ (S/J) via φ(r, s) = (r + I, s + J). Clearly, φ(R ⊕ S) = (R/I) ⊕ (S/J). The elements that are mapped to 0 (aka (I, J)) are any element of the form (i, j) where i ∈ I and j ∈ J . Thus, ker φ = I ⊕ J . By the fundamental theorem for homomorphisms, we have (R ⊕ S)/(I ⊕ J) ∼ = (R/I) ⊕ (S/J). (c) By Proposition 5.3.9, I ⊕ J is a prime ideal of R ⊕ S if and only if (R ⊕ S)/(I ⊕ J) is an integral domain. By (b), (R ⊕ S)/(I ⊕ J) ∼ = (R/I) ⊕ (S/J). Thus, I ⊕ J is prime if and only if (R/I) ⊕ (S/J) is an integral domain. Now, for any I, J , we can always form the product (i, 1)(1, j) = (i, j) where i ∈ I , j ∈ J . This product will always be such that (i, 1)(1, j) ∈ I ⊕J , but (i, 1), (1, j) ∈ / I ⊕J unless 1 ∈ I (in which case I = R) or 1 ∈ J (in which case J = S ). Without loss of generality, suppose J = S . Then, (R ⊕ S)/(I ⊕ S) ∼ = R/I ⊕ S/S ∼ = R/I . Thus, (R⊕S)/(I ⊕J) is an integral domain if and only if J = S and R/I is an integral domain. This happens if and only if I is prime (Proposition 5.3.9) and J = S . (d) Suppose that
(r, s) ∈ R ⊕ S ,
K
R ⊕ S . Then, if (a, b), (c, d) ∈ K , and (a ± c, b ± d) ∈ K and (ra, sb) ∈ K . This implies
is an ideal of
we must have
that the coordinates are closed under addition and enjoy the absorption property separately. Thus,
K =I ⊕J
for ideals
IER
and
J E S.
26. Let R be the set of all rational numbers m/n such that n is odd. (a) Show that R is a subring of Q. k k (b) Let 2 R = {m/n ∈ R|m is a multiple of 2 and n is odd}, for any posik tive integer k. Show that 2 R is an ideal of R. k (c) Show that each proper nonzero ideal of R has the form 2 R, for some positive integer k. k (d) Show that R/2 R is isomorphic to Z2k . (e) Show that 2R is the unique maximal ideal of R.
CHAPTER 5.
(a) Let
a R. If 2m+1
27
COMMUTATIVE RINGS
a(2n+1)+b(2m+1) b a b a = a(2n+1)+b(2m+1) 2m+1 , 2n+1 ∈ R. Then, 2m+1 + 2n+1 = (2m+1)(2n+1) 4mn+2(m+n)+1 a ∈ R, then clearly − 2m+1 ∈ R. Thus, R is a subring of Q.
m2k (2j+1)+`2k (2n+1) `2k m2k `2k m2k Then, = 2n+1 , 2j+1 ∈ I . 2n+1 + 2j+1 = (2n+1)(2j+1) (2mj+m+2`n+`)2k 2mj2k +m2k +2`n2k +`2k a a = ∈ 2k R. If 2b+1 ∈ R, then 2b+1 · (2n+1)(2j+1) (2n+1)(2j+1) k k m2 am2 k k 2n+1 = 4bn+2(b+n)+1 ∈ 2 R. Thus, 2 R is an ideal of R. (b) Let
m d where m and d are odd. Then, by the absorption property, we should be able to obtain any a element of R. For instance, if we wish to obtain the element b ∈ R where b is ad m a odd, we have the product · = . Thus, I = R. mb d b (c) Suppose
I
is an ideal of
2k R
(d) We see that
R
that contains the element
is a principal ideal (as its notation suggests).
h2k i.
Now,
r ∈ R ∩ h2k i, then r = s2k , where k s ∈ h2 i. Otherwise, by the division algorithm, r = qs2k + r0 , where 1 ≤ r < 2k . 0 Based on the value for r , we nd the coset for any r ∈ R. Use the mapping k φ(r + 2 R) = [r]2k to see that R/2k R ∼ = Z2k .
let us determine the cosets of
If
(e) We have found that the only ideals of
R
k 2R ⊇ 4R ⊇ · · · ⊇ 2 R ⊇ · · · . (For instance m2 n Thus, 2R is the unique maximal ideal.
k
5.4
2k R for some k . Clearly, k 2 m2k−1 ∈ 2R since m2 .) n = 1 · n are
Quotient Fields
Notational convention:
The notation
a
b ∼ is equivalent to [a, b]∼ . They Q(D) for an integral domain
both refer to the equivalence relation that builds
D.
1. Complete the proof of Lemma 5.4.3, to show that multiplication of equivalence classes in Q(D) is well-dened. p1 p2 ···pk m a b = p1 p2 ···pk n ∈ mr (p1 ···pk )(q1 ···qn )mr (p1 ···pk )(q1 ···qn )ns ∈ ns ∼ Let
m
c n ∼, d m = n ∼
·
`r ∈ = qq11 qq22 ···q r ···q` s
r
s ∼.
a b
Then,
·
c d
=
ac bd
=
s ∼.
2. Show that the associative and commutative laws hold for addition in
Q(D).
(Associative)
+ ∼
c
h i
=
a b
d
+ ∼
h i e f
=
a b
+ ∼
h
cf +ed df
i ∼
=
h
adf +b(cf +ed) bdf
i ∼
.
(1)
a b ∼
+
c d ∼
+
e f
ad+bc bd
∼
+
h i e f
∼ ∼ We see that (1) = (2), so + is associative. a c ad+bc (Commutative) = b ∼+ d ∼ = bd ∼ + is commutative.
=
h
f (ad+bc)+ebd bdf
bc+ad db
∼
=
i ∼
c d ∼
. (2)
+
a
b ∼ . Thus,
∈
CHAPTER 5.
28
COMMUTATIVE RINGS
3. Show that the associative and commutative laws hold in Q(D). h i h i h i a e ce ace c · · = · = = ac b ∼ d ∼ f b ∼ df bdf bd ∼ · ∼ ∼ ∼ h i h i e = ab ∼ · dc ∼ · fe . f ∼ a c∼ ac ca c a (Commutative) b ∼ · d ∼ = bd ∼ = db ∼ = d ∼ · b ∼ . (Associative)
a
4. Let φ : D → Q(D) be the mapping φ(D) = [d, 1] dened in Theorem 5.4.4. Show that φ is an isomorphism if and only if D is a eld. (⇒)
If
φ
is an isomorphism, then every element in
Q(D)
can be written as
h 0i Q(D) is a eld, so for any 1 ∼ ∈ Q(D), we must have d1 ∈ ∼ d h d0 i h dd0 i 1 d Q(D) such that 1 · 1 = 1 = 1 . Thus, for any 1 ∈ Q(D), we have h 0 i h 0i 1 = φ−1 11 = φ−1 d1 · d1 = φ−1 d1 · φ−1 d1 = d · d0 . Thus, any d ∈ D has an inverse, so D is a eld. (⇐) If D is a eld, then D contains all inverses. (1-1) φ is a direct map and is thus one-to-one. h i x (Onto) Let y ∈ D . Since the numerators and denominators come from D , h i h −1 i x a eld, this expression is equivalent to = x1 · y 1 = φ(xy −1 ). xy x yy (Preservation) φ(xy) = = · 1 = φ(x)φ(y). x 1y 1 φ(x + y) = x+y = + = φ(x) + φ(y). 1 1 1 d
d
1 ∼ . Now,
5. In Theorem 5.4.6, verify that θˆ is a one-to-one ring homomorphism. = θˆ dc . This means that θ(a)θ(b)−1 = θ(c)θ(d)−1 . b −1 Now, F is a eld, so the previous equality is the same as θ(c) θ(b)−1 = −1 −1 −1 θ(a)−1θ(d) ⇔ θ (cb) = θ (ad) . Since θ is one-to-one, this implies cb = ad. c ad−bc 0 a c a Thus, . Thus, = d . b − d = bd = bd = [0] ad+bc b a c ˆ ˆ (Preservation of +) θ = θ(ad + bc)θ(bd)−1 = (θ(ad) + b + d = θ bd −1 −1 −1 −1 −1 (1-1) Suppose
θˆ
a
θ(bc))θ(bd) + θ(b)θ(c)) θ(b) θ(d) = θ(a)θ(b) + θ(c)θ(d) = (θ(a)θ(d) = θˆ ab + θˆ dc . ˆ a · c = θˆ ac = θ(ac)θ(bd)−1 = θ(a)θ(c)θ(b)−1 θ(d)−1 = (Preservation of ·) θ b d bd θ(a)θ(b)−1 θ(c)θ(d)−1 = θˆ a θˆ c . b
d
6. Let D1 and D2 be integral domains, with quotient elds Q(D1 ) and Q(D2 ), respectively. Let θ : D1 → D2 be a ring homomorphism. (a) Prove that if θ is one-to-one, then there exists a ring homomorphism ˆ θˆ : Q(D1 ) → Q(D2 ) such that θ([d, 1]) = [θ(d), 1] for all d ∈ D1 . (b) Prove that if θ is not one-to-one, then it is impossible to nd a ring homomorphism θˆ : Q(D1 ) → Q(D2 ) that satises the conditions of part (a). i h ˆ 1 = θ(1) = 1 since θ is a homomorphism. (a) (Identity) θ 1 1 1
CHAPTER 5.
(Preservation of +)
h
θ(a) 1
i
h
θ(c) 1
29
COMMUTATIVE RINGS
θˆ
a 1
+
c 1
= θˆ
a+c 1
=
h
θ(a+c) 1
i
=
h
θ(a)+θ(c) 1
i
=
i
ˆ + θ(c) ˆ . = θ(a) h i h i h i ˆ a · b ˆ ab = θ(ab) = θ(a) · θ(b) = (Preservation of ·) θ = θ 1 1 1 1 1 1 ˆ θ(b) ˆ . θ(a) +
θ is not one-to h one,i then hthereiare x 6= y ∈ D1 such that θ(x) = θ(y). x y ˆ Then, θ = θ(x+y) = 2 θ(x) , but since x 6= y , x + y 6= 2x and 1 + 1 1 h1 i x+y h θ(x+y) i θ(x) 6= 2 1 . θˆ 1 = 1 (b) If
√
7. Determine Q(D) for D = {m + n 2|m, n ∈ Z}. For every
h
i
√ a + b 2 ∈ D, we h
1√ . For two elements a+b 2 h h √ i √ i a−b 2 c−d 2 pression + 2 2 2 a −2b c −2d2
must introduce the elements
1√ a+b 2
i
√ ∈ Q[ 2].
+
h
Thus,
√ i a+b 2 and 1
i
1√ , we have the equivalent exc+d 2
Thus,
√ r + s ∈ Q[ 2]
Also, doing the same sort of trickery, we easily see that
rs ∈ Q(D).
h
√ Q(D) = Q[ 2].
r, s ∈ Q(D). √ rs ∈ Q[ 2] for all for all
8. Let p be a prime number and let D = {m/n|m, n ∈ Z and p - n}. Verify that D is an integral domain and nd Q(D). n mn m q1 p+r1 · q2 p+r2 = (q1 p+r1 )(q2 p+r2 ) m only way this can happen is if one of m or n is 0, in which case q p+r (D is an integral domain) Let
would be 0. Thus,
D
1
= 0. 1
or
The
n q2 p+r2
is an integral domain. β
β
p` p1 1 ···pkk For the element α n , where p1 1 ···pα n
pi 6= p for all i, the element p1` must be introduced to Q(D). This means that Q(D) must have elements whose denominator can contain any power of p. Since D already has elements that can contain other primes, we simply have Q(D) = Q.
9. Determine Q(D) for D = {m + ni|n, m ∈ Z} ⊆ C. Very similar to Q7,
Q(D) = Q[i].
10. Considering Z[x] as a subring of Q[x], show that both rings have the same quotient elds. 1 k nx is a eld, this implies that
nxk ∈ Z[x],
∈ for all n, k ∈ Z. Then, since PQ(Z[x]) m k x ∈ Q(Z[x]) for all m, n, k ∈ Z. We n now see that Q[x] ⊆ Q(Z[x]). Since Q(Z[x]) is a eld, it must contain all of the inverses for Q[x]. Thus, Q(Q[x]) ⊆ Q(Z[x]). On the other hand, we have that Z[x] ⊆ Q[x], so by the same argument about Q(Q[x]) being a eld, we have Q(Z[x]) ⊆ Q(Q[x]). Thus, Q(Z[x]) = Q(Q[x]). Since
Q(Z[x])
surely
CHAPTER 5.
30
COMMUTATIVE RINGS
11. Show that if P is a prime ideal of D, then DP = {a/b ∈ Q(D)|b ∈/ P } is an integral domain with D ⊆ DP ⊆ Q(D). a c ac b · d = bd = [0]. Now, since P is an ideal, 0 ∈ P , so neither b nor d can equal 0. Also, since D is an integral domain, we cannot have bd = 0 since b, d 6= 0. Thus, ac = 0 and since D is an a integral domain, either a = 0 (in which case b = [0]) or c = 0 (in which case c = [0] ). Thus, DP is an integral domain. d Clearly, DP ⊆ Q(D), since the elements come from Q(D). Let a ∈ D . Then, by denition of DP , a ∈ DP . Thus, D ⊆ DP . (It might be worth pointing out that 1 ∈ / P since P is an ideal.) (DP is an integral domain) Suppose that
12. In the ring DP dened in Exercise 11, let M = {a/b ∈ DP |a ∈ P }. (a) Show that M is an ideal of DP . (b) Show that DP /M ∼ = Q(R/P ), and conclude that M is a maximal ideal of DP . (a) Let
h
p1 q1
i h i , pq22 ∈ M .
h i h i 2 q1 ± pq22 = p1 qq21±p . Since P is q2 h i h i h i p2 p1 ∈ M . If qd0 ∈ DP , then so q1 ± q2 h i h i p d ideal, dp ∈ P , so q 0 · q ∈ M . Thus, M is
Then,
p1 q2 ± p2 q1 ∈ P , and h i h i h i p dp d q 0 · q = q 0 q . Since P is an an ideal of DP . an ideal,
h
p1 q1
i
(b) I can only assume that there was a misprint in this question as dened. My best guess is that If that is the case, dene
R shouldhbe Di. φ ab = a+P b+P .
R
We need to show that
h
is not
φ
is a
i ∈ DP , then φ + =φ = h a+P i h c+P i h (d+P )(a+P )+(b+P )(c+P ) i h ad+bc+P i c a + φ Also, φ = . b d = b+P + d+P = (b+P )(d+P ) bd+P a c homomorphism. First, if b, d
a b
c d
ad+bc bd
ad+bc+P . bd+P
Thus, + is preserved. Next,
·
φ
a b
is preserved.
·
c d
=φ Thus, φ
ac bd
=
h
ac+P bd+P
i
=
h
a+P b+P
i h i c+P · d+P =φ
a b
+φ
c d . Thus,
is a homomorphism.
We see that the elements that get mapped to 0 (which is to say that the
P ) are the ones whose numerators are memM . Thus, by the fundamental theorem for DP /M ∼ = Q(D/P ).
numerator of the function value is bers of
P.
This set is precisely
homomorphisms, we see that
derivation
13. Let R be a commutative ring. A on R is a function ∂ : R → R such that (i) ∂(x+y) = ∂(x)+∂(y) and (ii) ∂(xy) = ∂(x)y+x∂(y). Show that if ∂ is a derivation on an integral domain D with quotient eld Q(D), then ∂ can be extended to a derivation ∂ of Q(D) by dening ∂(a/b) = (b∂(a) − a∂(b))/b2 for all a, b ∈ D with b 6= 0.
CHAPTER 5.
COMMUTATIVE RINGS
31
bd∂(ad+bc)−(ad+bc)∂(bd) c a ad+bc = bd[∂(ad)+∂(bc)]−(ad+bc)[d∂(b)+b∂(d)] = = b + d =∂ bd b2 d2 b2 d2 2 2 2 [bda∂(d)+bd ∂(a)+b d∂(c)+bdc∂(b)]−[ad ∂(b)+adb∂(d)+bcd∂(b)+b2 c∂(d)] bd2 ∂(a)+b2 d∂(c)−ad2 ∂(b)−b2 c∂(d) = b2 d2 b2 d2 d2 [b∂(a)−a∂(b)]+b2 [d∂(c)−c∂(d)] . (1) b2 d2 b∂(a)−a∂(b) d∂(c)−c∂(d) 2 2 [d∂(c)−c∂(d)] a c Then, ∂ + = d [b∂(a)−a∂(b)]+b b +∂ d = b2 d2 b2 d2 (2). Now, (1) = (2), so (i) is satided. bd∂(ac)−ac∂(bd) a c ac (ii) ∂ = bd[a∂(c)+c∂(a)]−ac[b∂(d)+d∂(b)] . (1) b · d = ∂ bd = b2 d2 b2 d2 b∂(a)−a∂(b) c[b∂(a)−a∂(b)] a c a c c a d∂(c)−c∂(d) Next, ∂ ·d+b· = + b · d + b ·∂ d = b2 d2 b2 d a[d∂(c)−c∂(d)] cd[b∂(a)−a∂(b)]+ab[d∂(c)−c∂(d)] bd[a∂(c)+c∂(a)]−ac[b∂(d)+d∂(b) = = . (2) bd2 b2 d2 b2 d2 Now, (1) = (2), so (ii) is satised. Thus, ∂ is a derivation on Q(D). (i)
∂
14. Show that ∂ : Q[x] → Q[x] dened by ∂(f (x)) = f 0 (x) for all f (x) ∈ Q[x] is a derivation. Describe the derivation ∂ dened on the quotient eld of Q[x]. ∂(f (x) + g(x)) = f 0 (x) + g 0 (x) and ∂(f (x)g(x)) = g(x)f (x) + f (x)g (x) from calculus. Thus, ∂ is a derivation of Q[x]. We have shown that Q(Q[x]) = Q(Z[x]), which is just the set of rational functions. We already know that
0
Thus, is,
∂
0
∂
f (x) g(x)
f (x) g(x)
=
should be dened according to the ordinary quotient rule. That
g(x)f 0 (x)−f (x)g 0 (x) (g(x))2
=
g(x)∂(f (x))−f (x)∂(g(x)) . (g(x))2
=
Chapter 6
Fields 6.1
Algebraic Elements
1. Show that the following complex numbers are algebraic over Q. √
√2 + √ n, for √n∈Z (c) p3 + √5 (d) 2 + √3 (e) (−1 + √ √ 3i)/2 3 (f ) 2+ 2
(a)
(b)
√ x2 − 2 has √2 as a root. 2 (b) x − n has as a root. √ √ √ n√ 3 + 5. Then, x − 3 = 5. Squaring (c) Let x = √ √ both sides, we get x2 − 2 3x + 3 = 5 which is the same thing as x2 − 2 = 2 3x. Again, squaring 4 2 2 4 2 both sides, we get x − 4x + 4 = 12x . Thus, the polynomial x − 16x + 4 has √ √ 3 + 5 as a root. p √ √ √ (d) Let x = 2 + 3. Then, x2 = 2p+ 3, so x2 − 2 = 3. Squaring, √ x4 − 4x2 + 4 = 3. Thus, x4 − 4x2 + 1 has 2 + 3 as a root. √ 2π i( 2π 3 ) . Thus, if x = ei( 3 ) , then x3 = e2πi = 1. (e) Note that (−1+ 3i)/2 = e 3 Thus, x − 1 has the number as a root. √ √ 3 1/3 (f ) Let x = 2√+ 2 = 21/3 + 21/2 . Then,√x − 21/2 √ √ = 2 3 . Cubing, 3 2 3 2 x √ − 3 2 2x + 6x − 2 2 = 2. Thus, x + 6x − 2 = 3 2x + 2 2, so x + 6x − 2 = 2(3x + 2). 6 4 3 2 4 2 Squaring, we get x + 12x − 4x + 36x − 24x + 4 = 2(9x + 12x + 4). 6 4 3 2 Thus, x − 6x − 4x + 12x − 24x − 4 has the number as a root. (a)
2. Let F be an extension eld of K , and let u be a nonzero element of F that is algebraic over K . Show that u−1 is also algebraic over K .
32
CHAPTER 6.
33
FIELDS
f (x) = an xn +· · ·+a1 x+a0 ∈ K[x] be such that f (u) = 0. Then, examine g(x) = a0 xn +· · ·+an−1 +a We will n . (The coecients are in the reverse P Pnorder.) n −1 −k n −k show that g(u ) = 0P . If u a = 0 , then surely u u a n−k n−k = k=0 k=0 n 0. This is the same as k=0 un−k an−k which we are given to be equal to 0. Thus, Pn −k it must be true that our rst statement, an−k = 0,P is true. (Note that k=0 u Pn n −k K[x] is an integral domain and so k=0 u an−k = 0 ⇔ un k=0 u−k an−k = 0 since u 6= 0.) Let
3. Suppose that u is algebraic over the eld K , and that a ∈ K . Show that u + a is algebraic over K , nd its minimal polynomial over K , and show that the degree of u + a over K is equal to the degree of u over K . It is pretty clear that if u ∈ F , we must have K(u) ⊆ F . So, since u + a = u + a · 1 ∈ K(u) (keeping in mind the notion of K(u) being a vector space 2 over K with basis {1, u, u , ...}), we see that u + a ∈ F . Furthermore, since K(u) = K(u + a), the degrees of the extension are also equal. However, since we have to nd the minimal polynomial anyway, we don't really need the preceeding argument. Note that
f (x)
f (x + a) gives the graph of f (x) shifted left a units. Thus, if u, then f (x + a) has a root at u − a. We need to prove this,
has a root at
though, since there is no notion of shifting left in the abstract setting. Suppose
f (x) = an xn + · · · + a1 x + a0 . Then, set f (x + a) = g(x) = an (x + a)n + · · · + a1 (x+a)+a0 . (This polynomial is still in K[x] since K[x] is a ring.) We see that g(u − a) = an (u − a + a)n + · · · + a1 (u − a + a) + a0 = an un + · · · + a1 u + a0 = 0. This proves that g(x) is the minimal polynomial, for if there were a polynomial h(x) of smaller degree for which h(u + a) = 0, we would see that h(x − a) has u as a root. Since f (x) was the minimal polynomial of u, we know that this is impossible. Thus, g(x) is the minimal polynomial and deg f = deg g . As such, [K(u) : K] = [K(u + a) : K].
4. Show that
√
√ 3∈ / Q( 2).
√ √ Q( space over Q has a basis {1, 2}. Now, √ 2) as a vector √ √ suppose 3 ∈ Q( 2). Then 3 = ab + dc 2 for a, b, c, d ∈ Z. This means 2√ √ a2 d2 √ − 2adb 3 +3d2 d( 3−a/b) b2 that = 2 . Squaring both sides, we get = 2. Since c √ √ √c2 3∈ / Q, we know that this is impossible. Thus, 3 ∈ / Q( 2). We see that
√
5.
Show that f (x) = x3 + 3x + 3 is irreducible over Q. −1 (b) Let u be a root of f (x). Express u and (1+u)−1 in the form a+bu+cu2 where a, b, c ∈ Q. (a)
(a) By the rational root theorem, the only possible rational roots are
±1. Q.
By plugging these numbers in, we quickly see that
f (x)
±3 and
is irreducible over
CHAPTER 6.
34
FIELDS
Q(u), we have u(a+bu+cu2 ) = 1. This gives au+bu2 +cu3 = 1. Since u + 3u + 3 = 0, we have u3 = −3(u + 1) and we obtain au + bu2 − 3cu − 3c = 1 1 2 or (a − 3c)u + bu − 3c = 1. Thus, −3c = 1 ⇒ c = − . Then, a − 3c = 0 ⇒ 3 1 a − 3 · − 3 = 0 ⇒ a + 1 = 0 ⇒ a = −1. Finally, b = 0. Thus, u−1 = −1 − 31 u2 . 2 3 2 In Q(u), we have (1+u)(a+bu+cu ) = 1. Thus, cu +(b+c)u +(a+b)u+a = 2 1. Exploiting the same relation as above, this gives (b + c)u + (a + b − 3c)u + a − 3c = 1. This implies that a = 4, b = −1, c = 1. Thus, (1 + u)−1 = 4 − u + u2 . (b) In
3
6. Show that the intersection of any collections of subelds is again a subeld. T x, y ∈ Fi . This means that x and y are in every Fi . Thus, x ± y ∈ Fi for all i. Next, if x ∈ Fi T −1 for all i, then since Fi are elds, we must have x ∈ Fi for all i. Thus, Fi is a subeld of E . Let
E
be a eld and
{Fi }
be a collection of subeld. Let
7. Let F = K(u), where u is transcendental over the eld K . If E is a eld such that K ⊂ E ⊆ F , then show that u is algebraic over E . To restate what is given, we have K $ E ⊆ K(u). Now, all elements in F = K(u) are of the form a0 + a1 u + a2 u2 + · · · where ai ∈ K . The elements n of K must not have u's. Since E 6= K , we must have b1 + b2 u + · · · + bn u ∈ E n where bi ∈ K . But, K(b1 + b2 u + · · · + bn u ) = K(u), so E = F , and obviously u would have to be algebraic over E .
8. Let F be an extension eld of K . (a) Show that F is a vector space over K . (b) Let u ∈ F . Show that u is algebraic over K if and only if the subspace spanned by {1, u, u2 , ...} is a eld. (a) Since
F
is a eld, it is an abelian group under addition. Thus, i-v (the
axioms for vector spaces are listed on page 458) are satised. Since
K ⊆ F,
we know that
k·f ∈ F
for all
k∈K
and
f ∈ F.
Thus, (vi)
is satised.
K ⊆ F and F is a eld (and thus · is associative), we have k1 (k2 f ) = k1 , k2 ∈ K and f ∈ F . Thus, (vii) is satised. Since K ⊆ F and F is a eld (and thus + distributes over ·), we have (k1 + k2 )f = k1 f + k2 f for k1 , k2 ∈ K and f ∈ F . Thus, (viii) is satised. Since
(k1 k2 )f
for all
For the same reasons, (ix) is satised. Since Thus,
F is a eld, we must have 1 · f = f F is a vector space over K .
for all
f ∈ F.
Thus, (x) is satised.
n Then, let f (x) = x + · · · + K[x]/hf (x)i ∼ K(u) is a eld. = n We know that K(u) is a vector space over K . It is clear that {1, u, ..., u } is a n basis for K(u). Thus, span{1, u, ..., u } = K(u). (b) (⇒) a1 x + a0 be
Suppose that
u
is algebraic over
the minimal polynomial of
u.
K.
Then,
CHAPTER 6.
35
FIELDS
(⇐) Suppose that span{1, u, u2 , ...} is 2 (1) {1, u, u , ...} is linearly independent or
a eld. (2)
There are two possibilities:
{1, u, u2 , ...}
is not linearly inde-
pendent. In the case of (1), since span{1, u, u
2
, ...} is a eld, we must have elements a0 + a1 u + · · · + an u , b0 + b1 u + · · · + bm um ∈ span{1, u, u2 , ...} such that (a0 + a1 u + · · · + an un )(b0 + b1 u + · · · + bm um ) = c0 + c1 u + · · · + c` u` = 1. This would involve coecients of u terms multiplying to 0. However, the coecients come from K which is a eld, so multiplication of nonzero elements should never n
equal 0. Contradiction. Thus, (1) is not possible.
k0 + k1 u + · · · + k` u` = 0 where k` x` + · · · + k1 x + k0 has u as a
In the case of (2), then we know that there is
ki 6= 0
for some
root. Thus,
u
i > 0.
Thus, the polynomial
is algebraic over
K.
9. Let F be an extension eld of K . If u ∈ F is transcendental over K , then show that every element of K(u) that is not in K is also transcendental over K . u = k0 + k1 u + · · · + kn un ∈ K(u) where ki 6= 0 for some i ≥ 1 (which is to say the element is from K(u) − K ). Suppose that u is algebraic over K . Then, there is some polynomial over K for which u is a root. But, if we take powers of u, we obtain an expression that has only powers of u. Thus, if we have a polynomial that has u as a root, there must be a polynomial that has u Let
as a root.
10. Let u and r be positive real numbers, with u 6= 1. It follows from a famous theorem of Gelfand and Schneider that if r is irrational and both u and r are algebraic over Q, then ur must be transcendental over Q. You may use this result √to show that the following numbers are transcendental over Q. (a) (b)
√ 3
5
7√ √ 5 3 7 +7
√ √ √5 √ √ 5 3 7 = 37 . Both 7 and 5 are both well-known √ √ 5 3 irrational numbers, so 7 is transcendental over Q. √ √ 5 3 (b) We already proved that 7 was transcendental over Q, so by Q9 √ √ √ √ 5 5 3 3 7 +7∈Q 7 − Q is also transcendental. (a) We see that
√ 3
11. Show that there exist irrational numbers a, b ∈ R such that ab is rational. √ We see that
(2
2
√
)
2
= 22 = 4 ∈ Q.
√ (Note that
2
2
is transcendental by
Q10 and is thus irrational.)
12. Assuming that π is transcendental over Q, prove that either π + e or
π · e is irrational.
CHAPTER 6.
36
FIELDS
π · e = ab for a, b ∈ Z. Then, ab ∈ Q, so a minimal polynomial c for π · e is p(x) = x − (π · e). Thus, we have πe ∈ Q(π)(e) = Q(π, e) and d πe ∈ Q. Thus, we have πe ∈ Q, but πe ∈ / Q(π). This is imposssible since Q ⊆ Q(π). Thus, e is irrational. Suppose that
6.2
Finite and Algebraic Extensions
Notational convention:
E is an extension eld of F , we say E/F is a(n V is a vector space over the eld F , we say V /F
If
extesion) eld. Similarly, if
is a vector space. This saves lots of unnecessary typing.
1. Find√ the degree and a basis for each of the given extensions. (a) Q( 3) over Q √ √ (b) Q( 3, 7) over Q √ √ (c) Q( 3 + 7) over Q √ √ 3 (d) Q( 2, 2) over Q √ √ 3 (e) Q( 2 + 2) over Q √ (f ) Q(ω) over Q, where ω = (−1 + 3i)/2
√ . Degree: 2. {1, √3}√ √ (b) Basis: {1, 3, √ √ 7, √ 21}. Degree: 4. (c) Basis: {1, 3, 7, 21}. Degree: 4. 1/2 1/3 2/3 5/6 1/6 (d) Basis: {1, 2 , 2 , 2 , 2 , 2 }. 1/2 1/3 2/3 5/6 1/6 (e) Basis: {1, 2 , 2 , 2 , 2 , 2 }. i 2π −i 2π 3 }. Degree: 3. (f ) Basis: {1, e 3 , e (a) Basis:
Degree: 6. Degree: 6.
2. Find√ the√degree and a√basis for each of the given eld extensions. (a) Q( 3, 21) √ √ over Q( √7) (b) Q( 3 + √ √ 7) over Q( √ 7)√ (c) Q( 3, 7) over Q( 3 + 7) (a) Basis: (b) Basis: (c) Basis:
√ {1, √3}. Degree: {1, 3}. Degree: {1}. Degree: 1. √
2. 2.
√
3. Find the degree of Q( 3 2, 4 5) over Q. √ √ √ √ √ √ [Q( 3 2, 4 5) : Q] = [Q( 3 2) : Q][Q( 3 2, 4 5) : Q( 3 2)] = 3 · 4 = 12.
4. Let F be a nite extension of K such that [F : K] = p, a prime number. If u ∈ F but u ∈/ K , show that F = K(u). We have that p = [F : K] = [F : K(u)][K(u) : K]. Thus, [F : K(u)] = p or [K(u) : K] = p. Since u ∈ / K , we must have [K(u) : K] = p, so [F : K(u)] = 1. Thus, F = K(u).
CHAPTER 6.
37
FIELDS
5. Let f (x) be an irreducible polynomial in K[x]. Show that if F is an extension eld of K such that deg (f (x)) is relatively prime to [F : K], then f (x) is irreducible in F [x]. f (x). Then, deg (f (x)) = q[K : K(u)]. If f (x) is reF [x], then u ∈ F , so we have to have K(u) ⊆ F and [F : K] = [F : K(u)][K(u) : K]. However, since gcd (deg (f (x)) , [F : K]) = 1, this is impossible. Thus, f (x) is not reducible over F [x]. Let
u
be a root of
ducible over
6. Let K ⊆ E ⊆ F be elds. Prove that if F is algebraic over K , then F is algebraic over E and E is algebraic over K . If
F/K
E ⊆ F.
Now,
is transcendental. In particular, suppose that there is
u∈F
is algebraic, then clearly,
suppose that
F/K
E/K
is algebraic since
E . Then, by Q9 of 6.1, we have that every element E(u)−E is transcendental over E . Let u0 ∈ E(u)−E . Then, since K ⊆ E , we 0 have u transcendental over K . But, E(u)−E ⊆ F , so we would have an element of F being transcendental over K . But F/K is algebraic. Contradiction.
which is transcendental over of
7. Let F ⊇ K be elds, and let R be a ring such that F ⊇ R ⊇ K . If F is an algebraic extension of K , show that R is a eld. What happens if we do not assume that F is algebraic over K ? Let r ∈ R − K . Since R ⊆ F and F/K is algebraic, there is an irreducible p(x) ∈ K[x] such that p(r) = 0. Then, since p(x) is irreducible, we obtain K[x]/hp(x)i ∼ = K(r) is a eld. Now, K(r)/K is a vector space, so any element n of K(r) can be written as k0 + k1 r + · · · + kn r where ki ∈ K . This expression is an element of R by the axioms for subring. (Remember that k ∈ K ⇒ k ∈ R S since K ⊆ R.) Let E = r∈R−K K(r). We have shown (Q6 from 6.1) that E is a subeld of F . We see that the addition and multiplication of these two elements of E is again in R. Thus, E ⊆ R. Obviously, we also have R ⊆ E (since we take K ⊆ R and adjoin any missing elements from R). Thus, R = E and is thus a eld. If we do not have conclude that
R
F/K
algebraic, we cannot construct
E,
so we cannot
is a eld.
√
8. Determine [Q( n) : Q] for all n ∈ Z+ . If
n
√ √ [Q( n) : Q] is {1, n}. Thus, n ∈ Q, so the degree would be 1.
is not a square, then a basis for
degree would be 2. If
n = m2 ,
then,
√
√
the
√ √
9. For any positive integers a, b, show that Q( a + b) = Q( a, b). We know that a basis for a basis for
√ Q( a +
√
b).
√ √ Q( a, b)
Any element of
√ √ √ {1, √ a,√ b, ab}. Let us now nd Q( a + b) will have the form n0 + is
CHAPTER 6.
38
FIELDS
√ √ √ √ n1 ( a+ b)+· · ·+nk ( a+ b)k .
Pn √ √ n √ k √ n−k ( a+ b)n = k=0 a b . k √ √ √ √ We see that every element in Q( a + b) can have a, b, √ √ a√term involving √ √ √ ab, or none of them at all. Thus, a basis for Q( a + b) is {1, a, b, ab}. √ √ √ √ Thus, Q( a, b) = Q( a + b). Examine
10. Let F be an extension eld of K . Let a ∈ F be algebraic over K , and let t ∈ F be transcendental over K . Show that a + t is transcendental over K . (a + t) is algebraic over K . Then, there is p(x) ∈ K[x] such p(a + t) = 0. Thus, there is q(x) = p(x + a) ∈ K(a)[x] such that t is a root of q(x). Thus, t is algebraic over K(a). Thus, K(a, t) = K(a)(t) = K(a). Thus t is algebraic over K by Proposition 6.2.4. Contradiction. Suppose that
that
11. Let F be an algebraic extension of K , and let S be a subset of F such that S ⊇ K , S is a vector space over K , and sn ∈ S for all s ∈ S and all positive integers n. Prove that if char(K) 6= 2, then S is a subeld of F . s1 , s2 ∈ S . Then, s1 + s2 ∈ S by vector addition. Since sn ∈ S for all s ∈ S and n ∈ Z+ , we know that (s1 + s2 )2 = s21 + s22 + 2s1 s2 ∈ S . Again, by 2 2 2 2 vector addition, we have s1 + s2 + 2s1 s2 − s1 − s2 = 2s1 s2 ∈ S . Thus, s1 s2 ∈ S since char(K) 6= 2. Now, distributivity in S follows from the fact that products are closed and by the fact that S inherits the distributivity of F . Thus, S is a ring. Since F/K is algebraic and K ⊆ S ⊆ F , S is a eld by Q7. Let
6.3
Geometric Constructions
1. Show that the roots of the polynomial 8x3 − 6x − 1 used in Theorem 6.3.9 7π are u1 = cos π9 , u2 = cos 5π 9 , and u3 = cos 9 . 3 If we make the substitution x = cos θ , then we obtain the expression 8 cos θ− 6 cos θ − 1 = 2(4 cos3 θ − 3 cos θ) − 1 = 2 cos(3θ) − 1. Then, the polynomial has (6k−1)π (6k+1)π the roots where 2 cos(3θ) = 1. This happens at θ = or θ = . 9 9 Pick any three values of θ to nd that cos θ is equal to u1 , u2 , or u3 .
2. Use the identity 4 cos3 θ − 3 cos θ − cos(3θ) = 0 to show that the roots of 4π 8π the polynomial x3 − 3x + 1 are u1 = 2 cos 2π 9 , u2 = 2 cos 9 , u3 = 2 cos 9 . x = 2 cos θ. The polynomial becomes (2−8(1−cos2 θ)) cos θ+1 = 2 cos θ − 8 cos θ(1 − cos2 θ) + 1 = −6 cos θ + 8 cos3 θ + 1. This equals 0 when 8 cos3 θ − 6 cos θ = −1 ⇔ 4 cos3 θ + −3 cos θ = − 12 ⇔ cos(3θ) = − 12 . This 2(3k+1)π 2(3k−1)π happens at θ = orθ = . Hence, the polynomial is equal to zero 9 9 for x values u1 , u2 , and u3 . Substitute
CHAPTER 6.
39
FIELDS
3. In this exercise we outline how to construct a regular pentagon. Let
ζ = cos(2π/5) + i sin(2π/5). (a) Show that ζ is a primitive fth root of unity. −1 2 (b) Show that (ζ + ζ ) + (ζ +√ζ −1 ) − 1 = 0. −1 (c) Show that ζ + ζ = (−1 + 5)/2. p √ √ (d) Show that cos(2π/5) = (−1+ 5)/4 and that sin(2π/5) = 10 + 2 5 /4. (e)
Conclude that a regular pentagon is constructible.
(a) By example A.5.3, this is a solution to
z 5 = 1.
ζ
Thus,
is a primitive
fth root of unity.
ζ = cos(2θ)+i sin(2θ) where θ = 2π/5, we see that (ζ +ζ −1 )2 + (ζ + ζ ) − 1 = 4 cos2 (θ) + 2 cos(θ) − 1 where θ = 2π/5. Thus, 4 cos2 θ + 2 cos θ − 1 = 0 if and only if 4 cos2 θ = 1 − 2 cos θ. Then, 4 cos2 θ = 1 − 2 cos θ ⇔ 2(cos(2θ) + 1) = 1 − 2 cos θ ⇔ 2 cos(2θ) = −2 cos θ − 1. Now, we can check √ −1+ 5 . to see that this equality is true. In (c), we shall nd that cos(2π/5) = 4 √ −1− 5 Using this, we nd that LHS= = RHS. 2 −1 (c) We see that ζ + ζ = 2 cos(2π/5). In (d), we nd that √ cos(2π/5) = √ (−1 + 5)/4, so we must have 2 cos(2π/5) = ζ + ζ −1 = (−1 + 5)/2. 2 (d) From (b), we have 4 cos θ + 2 cos θ − 1 = 0, so if x = cos θ , we have √ √ −1+ 5 2 4x + 2x − 1 = 0 which has solutions at x = −1−4 5 or x = . Hence, 4 (b) Plugging
−1
cos θ =
√ −1+ 5 . Since 4
√ − 5−5 . Thus, 8
sin2 θ + cos2 θ = 1, √ √
we know that
sin2 θ =
√ −1+ 5 4
−1 =
5 sin θ = 10+2 . 4 √ h i √ √ √ 10+2 5 −1+ 5 −1+ 5 (e) We need to nd Q , : Q . Let us rst nd Q : Q . 4 4 4 √ √ √ −1+ 5 −1+ 5 −1+ 5 In Q , we have a polynomial of the form x − which has 4 4 4 √ √ −1+ 5 −1+ 5 , we have an x such that x = as a root. Thus, in Q . Hence, 4 4 √ 2 2 2 16x + 8x − 4 4x + 1 = 5 and 16x + 8x + 1 = 5,√so 16xh +8x −√4 = 0. Thus, i −1+ 5 −1+ 5 is the minimum polynomial for , so Q : Q = 2. 4 4 √ √ √ √ √ 10+2 5 10+2 5 −1+ 5 Let us now nd Q : Q . As above, x = ⇔ 4 4 4 p √ √ √ 4x = 10 + 2 √5 ⇔ 16x2 −10−2 5 = 0. Thus, 16x2 −10−2 5 is the minimum √ √ 10+2 5 −1− 5 polynomial of over Q . 4 4 √ √ i √ √ √ √ h √ 10+2 5 10+2 5 −1+ 5 −1+ 5 −1+ 5 Now, Q , : Q = Q : Q Q : Q = 4 4 4 4 4
2 · 2 = 4 = 22 .
Thus, a regular pentagon is constructible.
4. Prove that a regular heptagon is not constructible. Needless to say, root of unity.
As
ζ = cos(2π/7) + i sin(2π/7) = ei i 2π 7 above, x = e 7 ⇔ x − 1 = 0.
2π 7
is a primitive seventh Thus,
[Q(ζ) : Q] = 7
CHAPTER 6.
40
FIELDS
which is not a power of 2.
Hence, a regular heptagon is not constructible
[Q(ζ) : Q] = [Q(cos(2π/7), sin(2π/7)) : Q(cos(2π/7))] [Q(cos(2π/7) : Q] thus [Q(cos(2π/7), sin(2π/7)) : Q(cos(2π/7))] cannot be a power of 2.
since and
6.4
Splitting Fields
1. Determine the splitting elds in C for the following polynomials (over
Q).
x2 − 2 2 (b) x + 3 4 2 (c) x + x − 6 3 (d) x − 5 (a)
√ 2 2 √x − 2 = 0 ⇔ x = 2 ⇔ x = ± 2. Thus, the polynomial has all its roots in Q( 2). √ 2 2 (b) x + 3 = 0 ⇔ x = −3 ⇔ x = ±i 3. Thus, the polynomial has all its √ roots in Q(i 3). 2 2 (c) Let u = x . Then, u + u + 6 = 0 ⇔ (u − 2)(u + 3) = 0 ⇔ u = 2 or √ √ √ √ u = −3. Thus, x = ± 2 or x = ±i 3. Thus, the splitting eld is Q( 2, i 3). 4π 3 3 1/3 1/3 1/3 i 2π (d) x − 5 = 0 ⇔ x = 5 ⇔ x = 5 . Thus, Q(5 , 5 e 3 , 51/3 ei 3 ) is (a)
the splitting eld.
2. Determine the splitting elds in C for the following polynomials (over
Q).
x3 − 1 4 (b) x − 1 3 2 (c) x + 3x + 3x − 4 (a)
x3 − 1 = 0 ⇔ x3 = 1. 4 4 (b) x − 1 = 0 ⇔ x = 1.
2π
4π
Q(ei 3 , ei 3 ) is the splitting eld. iπ i 3π Thus, Q(e 2 , e 2 ) is the splitting eld. 3 2 3 (c) By completing the cube, we nd that x +3x +3x−4 = 0 ⇔ (x+1) = 5. 4π 3 1/3 1/3 i 2π i If y = x+1, then the solutions to y = 5 are 5 , 5 e 3 , e 3 . Thus, the roots 2π 4π 1/3 of the original polynomial are ω1 = 5 −1, ω2 = 51/3 ei 3 −1, ω3 = 51/3 ei 3 −1. Thus, a splitting eld for the polynomial would be Q(ω1 , ω2 , ω3 ). (a)
Thus,
3. Determine the splitting elds over Z2 for the follwing polynomials.
x2 + x + 1 2 (b) x + 1 3 (c) x + x + 1 3 2 (d) x + x + 1 (a)
(a) Following Example 6.4.4, identify
0 1
1 1
is such that
A2 + A + I = 0.
Z2
with GL2 (Z2 ). We see that
Hence,
A
A=
is the companion matrix of
CHAPTER 6.
x2 + x + 1 .
41
FIELDS
a0 I + a1 A where ai ∈ Z2 form 2 Z2 in which x + x + 1 has a root. 0 1 2 (b) Similarly, let A = . We see that A +I = 0, so A is the companion 1 0 matrix. Thus, {a0 I + a1 A ∈ GL2 (Z2 )|a0 , a1 ∈ Z2 } is a splitting eld. 0 0 1 0 1. We see that A3 + A + I = 0, so A is the (c) Similarly, let A = 1 0 1 0 2 companion matrix. Thus, {a0 I + a1 A + a2 A ∈ GL2 (Z2 )|a0 , a1 , a2 ∈ Z2 } is a Thus, the set of matrices of the form
an extension eld of
splitting eld.
0 0 1 0 0. We see that A3 +A2 +I = 0, so A is the companion (d) Let A = 1 0 1 1 2 matrix. Thus, {a0 I + a1 A + a2 A ∈ GL2 (Z2 )|a0 , a1 , a2 ∈ Z2 } is a splitting eld.
4. Let p be a prime number. Determine the splitting eld for xp − 1 over Q. 2π
4π
2(p−1)π
xp −1 = 0 ⇔ xp = 1. Thus, the roots are {1, ei p , ei p , ..., ei p }. If p = 2, then the solutions are ±1, so the splitting eld would be Q. If p 6= 2, 2(p−1)π i 2π i p then the splitting eld would be Q(e p , ..., e ). (None of the roots except for 1 are from Q because p is prime and so the fraction in the exponent will never simplify to an expression in Q). We see that
5. Determine the splitting eld for xp − x over Zp . Since p must be prime for Zp to be a eld, by Fermat's little theorem, xp ≡ x(mod p), so xp − x ≡ 0(mod p) and Zp would be the splitting eld.
6. Determine the splitting eld for x9 − x over Z3 . splitting eld is
Q5
x9 − x = (x3 )3 − x ≡ x3 − x(mod 3) ≡ 0(mod 3). Z3 .
We have that
Thus, the
7. Prove that if F is an extension eld of K of degree 2, then F is the splitting eld over K of some polynomial. [F : K] = 2, then F is a vector space over K with some basis {1, u} u ∈ F − K . Since [F : K] = 2, we know that u2 can be written as k0 + k1 u for some k0 , k1 ∈ K . Hence, u2 − k1 u − k0 = 0. Thus, the polynomial x2 − k1 x − k0 has u as a root and u ∈ F − K . (The other root must also be in F , since we can factor out the root to obtain x2 − k1 x − k0 = (x − u)(x − r). If r ∈ K , when we foil the expression on the RHS out, we get a polynomial whose coecients do not come from K . Contradiction.) Thus, F is the splitting eld 2 for x − k1 x − k0 . If
for some
CHAPTER 6.
42
FIELDS
8. Let K be a eld. For a monic polynomial f (x) = a0 + a1 x + · · · + an−1 xn−1 +xn in K[x], the following matrix C is called the of f (x): 0 1 0 ··· 0 0 0 1 ··· 0 0 0 0 ··· 0 . .. .. .. .. .. . . . . . 0 0 0 ··· 1 −a0 −a1 −a2 · · · −an−1
companion matrix
This exercise outlines the proof that f (C) = 0. (That is, a0 I + a1 C + · · · + an−1 C n−1 + C n = 0, where I is the n × n identity matrix.) Let v1 = (1, 0, ..., 0), n v2 = (0, 1, ..., 0), ... ,vn = (0, 0, ..., 1) be the standard basis vectors Pn for K . (a) Show that vi C = vi+1 for i = 1, ..., n − 1 and vn C = j=1 −aj−1 vj . 2 n−1 (b) Find similar expressions for v1 C , ..., v1 C , v1 C n , and show that v1 f (C) = 0. (c) Show that vi f (C) = 0, for i = 2, ..., n and conclude that f (C) = 0.
(a) We see that
0
···
1
···
0 0 0
0 · . .. 0 −a0
1 0 0
0 1 0
··· ··· ···
0 0 0
. . .
. . .
..
. . .
.
will
··· 1 · · · −an−1 give a 1 × n vector. By matrix multiplication, to obtain the (1, k)th entry of the ··· 1 ··· 0 vector, we take the dot product of the rst (and only) row of 0 with the k th column of C . Thus, the (1, k)th entry will be 0 if the (k, i)th entry of C is 0. The only nonzero entries of C are the superdiagonal and the bottom row. So the (1, k)th entry of the product will be 1 if k − 1 = i and will be −ak Pn if i = n. Hence, vi C = vi+1 for i = 1, ..., n − 1 and vn C = j=1 −aj−1 vj . 0 −a1
0 −a2
v1 C 2 = v1 CC = v2 C = v3 . Arguing similarly, we see n for 1 ≤ m ≤ n − 1. Then, v1 C = v1 C n−1 C = vn C =
(b) We can write
m
Pn v1 C = vm+1 j=1 −aj−1 vj . n−1 Next, v1 f (C) = a0 v1 I + a1 v1 C + · · · + an−1 v1 C + v1 C n = a0 v1 + a1 v2 + Pn · · · + an−1 vn + j=1 −aj−1 vj . Note that the sum at the end cancels every term before it. Thus, v1 f (C) = 0.
that
vi f (C) = v1 C i−1 f (C) = v1 a0 C i−1 + a1 C i + · · · + an−1 C n−1+i + C n+i = v1 a0 C i−1 +· · · v1 an−1 C n−1+i +v1 C n+i = (v1 a0 I+· · ·+v1 an−1 C n−1 +v1 C n )C i = 0 · C i = 0. Now vi f (C) = 0 for all i ≥ 2, so by the axioms for eld, f (C) = 0. (c) Examine
9. Let K be a eld, let f (x) = a0 + a1 x + · · · + an−1 xn−1 + xn ∈ F [x], and let C be the companion matrix of f (x), as dened in Exercise 8. Show that the set
CHAPTER 6.
43
FIELDS
R = {b0 I + b1 C + · · · + bn−1 C n−1 |bi ∈ F for i = 0, ..., n − 1} is a commutative ring isomorphic to the ring F [x]/hf (x)i. K F.
First of all, I believe that there is a typo in this question, as but not used in any way. I assume that
K
should have been
is introduced
Example 6.4.4 practically establishes the ring properties, so let us prove the isomorphic claim.
F [x]/hf (x)i can only be a ring if f (x) is irreducible. Dene φ : φ(b0 + b1 x + · · · + bn−1 xn−1 ) = b0 I + b1 C + · · · + bn−1 C n−1 . n−1 (1-1)Recall that by how C was dened, b0 I + b1 C + · · · + bn−1 C = 0 if and only if the coecients of C terms correspond to the coecients of x terms in f (x). Thus, ker φ = 0. Thus, φ is one-to-one. n−1 (Onto)If we wish to obtain a0 I + a1 C + · · · + an−1 C , we simply take the n−1 image φ(a0 I + a1 x + · · · + an−1 x ). Thus, φ is onto. (Preservation of +)Addition in R is clearly the same as for polynomials, so Note that
F [x]/hf (x)i
by
addition is preserved.
F [x]/hf (x)i, we have the relation xn = −b−1 n−1 x(b0 + b1 x + · · · + bn−2 x ). Since b0 I + b1 C + · · · + bn−1 C n−1 = 0, we have the −1 n n−2 relation C = −bn−1 C(b0 + b1 C + · · · + bn−2 C ). We have similar relations n+1 n+2 for x ,x , .... Thus, multiplication in R corresponds to multiplication in F [x]/hf (x)i. Thus, F [x]/hf (x)i ∼ = R. (Preservation of ·) In
n−2
10. Strengthen Theorem 6.4.2 by proving under the conditions of the theorem there exists a splitting eld F for f (x) over K for which [F : K] is a divisor of n!. We need to show that by induction on
n.
[F : K] = n − k for some 0 ≤ k ≤ 1. We procede n = 1. Then, F = K is a splitting eld and
Suppose that
[F : K] = 1|n!. F/K such that [F : K]|n! for all n ≤ N . deg f (x) = N + 1. By Kronecker's Theorem, there is E/K where r ∈ E is a root of f (x). Hence, f (x) = g(x)(x − r) over K(r) where deg g(x) = N . Now, by the inductive hypothesis, there is F/K(r) such that g(x) splits over F and [F : K(r)] = N − k where 0 ≤ k < n. Since the minimum polynomial of r is of degree N +1 at most, we have [K(r) : K] ≤ N +1, so we see that [F : K] = [F : K(r)][K(r) : K] ≤ (N − k)(N + 1). Thus, [F : K]|(N + 1)!, as desired. (Recall that if x, y ≤ z , we have xy|z!.) Suppose that there exists an Now, suppose that
11. Let K be a eld, and let F be an extension eld of K . Let φ : F → F be an automorphism of F such that φ(a) = a, for all a ∈ K . Show that for any polynomial f (x) ∈ K[x], and any root u ∈ F of f (x), the image φ(u) must be a root of f (x). Suppose that
f (x) = an xn +· · ·+a1 x+a0 has a root at u ∈ F . Then, f (u) = 0 an un + · · · + a1 u + a0 = 0. Then, since φ is a homomorphism,
or in other words
CHAPTER 6.
44
FIELDS
φ(0) = 0 and we get 0 = φ(an un +· · ·+a1 u+a0 ) = an φ(un )+· · ·+a1 φ(u)+a0 = an φ(u)n + · · · + a1 φ(u) + a0 (since φ xes all elements of K ). Hence, φ(u) is a root.
12. Use Exercise 11 to show that there are only two automorphisms of the eld Q(i): the identity automorphism and, and the one dened by φ(a + bi) = a − bi, for all a, b ∈ Q. Q(i) must x Q in order to use φ : Q(i) → Q(i) is an Then, φ(m) = automorphism. 1 −1 −1 = φ(n ) = [φ(n)] = n−1 = n1 . φ(1 + 1 + · · ·+ 1) = m · φ(1) = m . Also, φ n m m 1 Hence φ n = φ(m)φ m = n . The element a + ib 7→ a + φ(i)b, so let us simply consider what the possible mappings of i are. We know that a minimal polynomial for i is found by x = i ⇒ x2 = −1. Thus x2 + 1 is the minimal polynomial of i over Q. We see that ±i are the roots to this polynomial. Thus, there are two possible mappings of i: the identity mapping or mapping to its additive inverse. Thus, a + ib 7→ a + ib or a + ib 7→ a − ib. We must show that any automorphism of
Q11.
Suppose that
13. Use Exercise √ 11 to√ show that there are at most four distinct automorphisms of the eld Q( 2, 3). be xed by any automorphism √Q must √ √ √ φ : √ Q( 2,√ 3) → is an automor√ Q( 2, 3) √ √ phism. Then, we must have φ(a + b 2 + c 3 + d 6) = a + bφ( 2) + cφ( 3) + √ √ √ √ dφ( 2)φ( 3). As in Q12,√we only√need to consider the mappings of 2 and 3. 2 2 Minimal polynomials for 2 and 3 over √ Q are√x − 2 and x − 3, respectively. The roots of these polynomials are ± 2 and ± 3, respectively. Thus, by Q11, √ √ √ √ 2 7→ ± 2 and 3 7→ ± 3. Thus, there are four possible mappings. As proved in Q12, every element of
containing
14.
Q.
Suppose that
√
Show that√the splitting eld of x4 − 2 over Q is Q( 4 2, i). 4 4 (b) Show that Q( 2, i) is also the splitting eld of x + 2 over Q. (a)
x4 − 2 = (x + 21/4 i)(x − 21/4 i)(x − 21/4 )(x + 21/4 ). (These 4 1/4 solving x = 2 as discussed in A.5.) Hence, Q(2 , i) is the
(a) We see that were obtained by splitting eld.
−1/4 x4 + 2 are 2−1/4 + 2−1/4 i, 2−1/4 − 2−1/4 i, −2 + 2−1/4 i, √ 4 and −2 −2 i. All of these elements are contained in Q( 2, i) and since √ 4 each element needs both i and 2, it is the splitting eld. (b) The roots of
−1/4
−1/4
15. Use Exercise 11 to show √ that there are at most eight distinct automorphisms of the splitting eld Q( 4 2, i) of x4 − 2 over Q. √ √ φ : Q( 4 2, i) → Q( 4 2, i) be an automorphism. Then, φ(a + b(21/4 ) + c(21/2 ) + di + e(21/4 i) + f (21/2 i)) = a + bφ(21/4 ) + cφ(21/4 )φ(21/4 ) + dφ(i) + Let
CHAPTER 6.
45
FIELDS
eφ(21/4 )φ(i) + f φ(21/4 )φ(21/4 )φ(i). A minimal polynomial for 21/4 over Q is x4 − 2. By Q11, this means that φ(21/4 ) can be mapped to any of the four 2 solutions. Now, x + 1 is the minimal polynomial for i and so i 7→ ±1. Thus, there are 4 · 2 = 8 possible automorphisms. 6.5
Finite Fields
1. Give addition and multiplication tables for the nite eld GF(23 ), as described in Example 6.5.3. Use the relation
x8 = x (among others) to construct the multiplication table.
Addition is the same as polynomial addition.
3 ). We see that x3 + 1 is 3 ∼ irreducible over Z[x], so Z[x]/hx +x+1i = GF(2 ). Additionally, we previously 3 found that Z[x]/hx +1i ∼ = {a0 I +a1 C +a2 C 2 ∈ GL3 (Z2 )|a0 , a1 , a2 ∈ Z2 } where 0 1 0 C = 0 0 1 is the companion matrix for x3 + 1. Thus, we can use ordinary 1 0 0 Let us examine another way of constructing GF(2
3
matrix multiplication and addition to nd the desired tables. This may be an easier option if a computer is available.
2. Give addition and multiplication tables for the nite eld GF(32 ), and nd a generator for the gyclic group of nonzero elements under multiplication. The tables are left to the reader. After constructing the multiplication table, the reader should be able to readily identify the generator. Simply start with the rst nonzero element (call it
x),
take the product and multiply it by
look up the product of it with itself. Then
x
again. Continue the process until (1) all
elements are accounted for or (2) a duplicate is generated. In the case of (2), move on to the next nonzero element and start the process over.
3. Find a generator for the cyclic group of nonzero elements of GF(24 ). We see that GF(2
4
).
x4 + x + 1
is irreducible over
Z2 [x]
and so
Z2 [x]/hx4 + x + 1i ∼ =
One can construct the multiplication table of the nonzero elements of
this eld and identify the generator. The generator happens to be
x+hx4 +x+1i.
4. Find the splitting elds over GF(3) for the following polynomials.
x4 + 2 4 (b) x − 2. (a)
x = 1 is a root, so x4 + 2 = (x − 1)(x3 + x2 + x + 1). Then, x + x + x + 1 has x = 2 as a root, so x4 + 2 = (x − 1)(x − 2)(x2 + 1). This is 2 completely factored over Z3 , so the splitting eld is Z3 [x]/hx + 1i ∼ = GF(32 ). (a) We see that
3
2
CHAPTER 6.
46
FIELDS
(b) We see that
2i ∼ = GF(34 ).
x4 −2 is irreducible over Z3 , so the splitting eld is Z3 [x]/hx4 −
5. Show that x3 − x − 1 and x3 − x + 1 are irreducible over GF(3). Construct their splitting elds and explicitly exhibit the isomorphism between these elds. Just plug in
x = 0, 1, 2
to both polynomials and observe that the output is
Z3 [x]/hx3 − x − 1i
never zero. Thus, both polynomials are irreducible. Thus, is the splitting eld of of
x3 − x + 1.
3
x −x−1
and
3
Z3 [x]/hx − x + 1i
is the splitting eld
Since both elds have the same number of elements (namely 27),
they must be isomorphic.
φ : Z3 [x]/hx3 − x − 1i → Z3 [x]/hx3 − x + 1i dened by φ(f (x) + hx3 − x − 1i) = (f (x) − 2) + hx3 − x + 1i. (1-1) We have just subtracted 2 from every element, so φ is one-to-one. 3 (Onto) If we wish to obtain g(x) + hx − x + 1i, we use the input g(x) + 2 + 3 hx − x − 1i. Thus, φ is onto. Dene
(Preservation of +) Addition is corresponds to polynomial addition, so + is clearly preserved.
Z[x]/hx3 −x−1i, we have the relation x3 +hx3 −x−1i = x + 1 + hx − x − 1i. In Z[x]/hx3 − x + 1i, we have the relation x3 + hx3 − x + 1i = x − 1 + hx3 − x + 1i. Since φ(x + 1 + hx3 − x − 1i) = x − 1 + hx3 − x + 1i, (Preservation of ·) In
3
we see that multiplication will correspond.
Thus, the two splitting elds are
isomorphic.
6. Show that x3 − x2 + 1 is irreducible over GF(3). Construct its splitting eld and explicitly exhibit the isomorphism between this eld and the splitting eld of x3 − x + 1 over GF(3). Just plug in
x = 0, 1, 2
and observe that the output is never 0. Thus, the
polynomial is irreducible. The splitting eld for the polynomial would then be
Z3 [x]/hx3 − x2 + 1i.
One can construct multiplication tables for both elds,
identify the generator and map them to each other.
7. Show that if g(x) is irreducible over GF(p) and g(x)|(xp − x), then deg(g(x)) is a divisor of m. m
We know that then
g(x)
m
xp − x
m
has splitting eld GF(p
must split over GF(p
m
)
since
xp − x
).
Now, if
splits over it.
m
is the splitting eld of
a subeld of GF(p
over it.
since
Also,
g(x). Then, F is [F : GF(p)] = deg(g(x))
m
g(x)
)
is the minimal polynomial
m
over its splitting eld. By Lemma 6.5.5 and its proof, [GF(p
m = [GF(pm ) : GF(p)] = [GF(pm ) : F ][F : F ] · deg(g(x)). Thus, deg(g(x)) is a divisor of m.
Furthermore,
m
g(x)|(xp − x), Suppose that F since g(x) splits ):
GF(p)]
GF(p)]
= m. = [GF(pm ) :
8. Let m, n be positive integers with gcd(m, n) = d. Show that, over any eld, the greatest common divisor of xm − 1 and xn − 1 is xd − 1.
CHAPTER 6.
47
FIELDS
Let m = qd and n = rd. We see by simple polynomial long division that xdq − 1 = (xd − 1)(xd(q−1) + xd(q−2) + · · · + x + 1) and similarly, xrd − 1 = (xd − 1)(xd(r−1) + xd(r−2) + · · · + x + 1). We now examine p1 (x) = xd(q−1) + · · · + x + 1 d(r−1) and p2 (x) = x + · · · + x + 1 to show that these two factors share no divisors. Assume without loss of generality that q ≥ r . Then, we can write p1 (x) = p2 (x) + xd(q−1) + xd(q−2) + · · · + xd(r+1) + xdr . We see clearly that p2 (x) + xd(q−1) + · · · + xr p2 (x) xd(q−1) + · · · + xr p1 (x) = = + . Clearly, the p2 (x) p2 (x) p2 (x) p2 (x) second term is not perfectly divisible. (If it were, we'd have p2 (x) as a factor of xd(q−1) +· · ·+xr which would imply that p1 (x) = p2 (x)[1+p3 (x)] for some p3 (x) of degree less than p2 (x). But since p1 (x) is of degree q and p2 (x) is of degree r such that (q, r) = 1, this is impossible.) Thus, gcd (xm − 1, xn − 1) = xd − 1.
9. If E and F are subelds of GF(pn ) with pe and pf elements respectively, how many elements does E ∩ F contain? Prove your claim. F are subelds, they have pe and pf elements where e|n and f |n. Then, E ∩ F is a subeld of both E and F , so the number of elements e f gcd(e,f ) must divide both p and p . Thus, E ∩ F contains p elements. Since
E
and
10. Let p be an odd prime. n n (a) Show that the set S of squares in GF(p ) contains (p + 1)/2 elements. n (b) Given a ∈ GF(p ), let T = {a − x|x ∈ S}. Show that T ∩ S 6= ∅. n (c) Show that every element of GF(p ) is a sum of two squares. n (d) What can be said about GF(2 )? (a) We know that multiplication in GF(p
n ×
) is cyclic with some generator 0 ≤ k < p . This element generates (pn − 1)/2 2 n elements. Let us not forget that 0 = 0 so 0 ∈ S . Thus, |S| = (p + 1)/2. (b) Suppose a = 0. Then, 0 ∈ S , so 0 − 0 = 0 ∈ T and so T ∩ S 6= ∅. Also, if a = 1, then 1 − 1 = 0 ∈ T ∩ S . k k 2` Otherwise, we have a = x 0 for some xed k0 ∈ Z. Then, x 0 − x = k0 2`−k0 2` 2`−k n × x (1 − x ). Thus, choose x such that x = 1. (Since GF(p ) is x.
Then
x
2
x
generates
2k
n
for all
cyclic, we can always do this.)
0 = 02 + 02 and 1 = 12 + 02 are sums of squares. Now, k 2` let x ∈ GF(p ). Then, by (b), x − x = x2j for some x2` ∈ S and j ∈ Z+ . k 2j 2` j 2 ` 2 = (x ) + (x ) . Thus, any element of GF(pn ) is the sum Hence, x = x + x (c) It is clear that
k
n
of two squares.
y = a2 + b2 = (a + b)2 . Since n GF(2 ) is closed under addition, we have that a + b = c ∈ GF(2 ) and that 2 y = c . Hence, y is a sum of squares if and only if y is itself a square. (For 2 2 2 instance, if GF(2 ) = {0, 1, a, b}, then a = b and b = a, and so neither a nor b can be a sum of squares.) (d) Suppose that
n
y ∈
GF(2
n
)
is such that
CHAPTER 6.
48
FIELDS
11. Show that xp − x + a is irreducible over GF(p) for all nonzero elements a ∈ GF(p). We know that
xp = x for any x ∈ GF(p).
xp − x + a = x − x + a = a. x − x + a is irreducible over
Hence,
There are no roots to a constant polynomial, so
p
GF(p).
12. Dene the function φ : GF(23 ) → GF(23 ) by φ(x) = x2 , for all x ∈ GF(23 ). (a) Show that φ is an isomorphism. 3 (b) Choose an irreducible polynomial p(x) to represent GF(2 ) as Z2 /hp(x)i, 2 3 and give an explicit computation of φ,φ , and φ . 3
(a) (1-1) Here is the multiplication table for GF(2
):
Note that every element is a square. Thus, the map is one-to-one. (Onto) Any nite map that is one-to-one is also onto. (Preservation of +)
φ(a + b) = (a + b)2 = a2 + b2 = φ(a) + φ(b).
(See Lemma
6.5.4.) (Preservation of ·) (b) Let
φ(ab) = (ab)2 = a2 b2 = φ(a)φ(b).
p(x) = x3 + 1.
(Check that this has no roots by using the multipli-
Z2 /hp(x)i has 8 elements and is thus isomorphic to ). We must have φ(ax2 + bx + c + p(x))2 = a2 x4 + b2 x2 + c2 + p(x). Since x3 = 1, we know x4 = x, so φ(ax2 + bx + c) = bx2 + a2 x + c2 . The rest of the
cation table above.) Then, GF(2
3
maps are similar.
6.6
Irreducible Polynomials over Finite Fields
1. Verify Theorem 6.6.1 in the special case of x16 − x over GF(2), by multiplying out the appropriate irreducible polynomials from the list given in the answer to Exercise 12 of section 4.2.
CHAPTER 6.
49
FIELDS
x, x + 1, x2 + x + 1, x3 + x2 + 1, x + x + 1, x + x + x + 1, x + x + 1, and x4 + x + 1. Now, by Theorem 24 6.6.1, the monic irreducible factors of x − x should be the monic irreducible 2 polynomials of degree 4, 2, and 1. Thus, we must show that x(x + 1)(x + x + 4 3 2 4 3 4 16 1)(x + x + x + x + 1)(x + x + 1)(x + x + 1) = x − x. 2 2 2 4 3 2 4 First, x(x+1) = x +x. Next, (x +x)(x +x+1) = x +2x +2x +x = x +x. 4 4 3 2 8 7 6 3 2 Then, (x + x)(x + x + x + x + 1) = x + x + x + x + x + x. Next, (x8 + x7 + x6 + x3 + x2 + x)(x4 + x3 + 1) = x12 + x9 + x8 + x6 + x4 + x3 + x. 12 Finally, (x + x9 + x8 + x6 + x4 + x3 + x)(x4 + x + 1) = x16 + x = x16 − x. The list of irreducible polynomials is:
3
4
3
2
4
3
2. Use Theorem 6.6.1 to show that over GF(2) the polynomial x32 +x factors as a product of the terms x, x + 1, x5 + x2 + 1, x5 + x3 + 1, x5 + x4 + x3 + x + 1, x5 + x4 + x2 + x + 1, x5 + x3 + x2 + x + 1, and x5 + x4 + x3 + x2 + 1. We have that
5
x32 + x = x2 − x,
so by Theorem 6.6.1, the irreducible factors
must be the irreducible polynomials of degree 5 and 1. It is left to the reader to show that the list of factors is complete. To do this, realize that in order for the polynomial to be irreducible, it must be linear or have an odd number of terms. Since 5 has no other divisors than 5 and 1, the non-linear polynomials must be of degree 5 and have an odd number of terms.
3. Let F be a eld of characteristic p, with prime subeld K = GF(p). Show that if u ∈ F is a root of the polynomial f (x) ∈ K[x], then up is also a root of f (x). f (x) = an xn + · · · + a1 x + a0 . If f (u) = 0, then 0 = f (u)p = (an un + · · · + a1 u + a0 )p = apn (up )n + · · · + ap1 up + ap0 . (The last equality holds since p char(F ) = p.) Now, by Fermat's Little Theorem, ai ≡ ai (mod p), so the last p n p p expression in the equality becomes an (u ) + · · · + a1 (u ) + a0 = f (u ). Thus, p f (u ) = 0. Let
4. Let u be a primitive element of GF(pm ) and let M (i) (x) be the minimum k polynomial of ui over GF(p). Show that every element of the form uip is also a root of M (i) (x). (i) u is primitive, then hui = GF(pm )• . By denition of M (x), we have (i) ip i (i) i p M (u ) = 0. By Q3, we know that M (u ) = M (u ) = 0. Similarly, 2 (i) ip since M (u ) = 0, we know that M (i) (uip )p = M (i) (uip ) = 0. Continuing (i) ipk this argument, we see that M (u ) = 0. If (i)
5. Let GF(26 ) be represented by Z2 [x]/hx6 +x+1i, and let u be any primitive element of GF(26 ). Show that GF(23 ) = {0, 1, u9 , u18 , u27 , u36 , u45 , u54 }. I'm not sure why GF(2 above. If
u
6
)
is primitive, then
would have to have the representation discussed
hui = GF(26 )×
(which has a multiplicative group
CHAPTER 6.
structure). GF(2
3 ×
)
≤
50
FIELDS
Since GF(2 GF(2
6 ×
)
have order dividing be a generator is
.
64.
u9 .
3
6
)
is a subeld of GF(2
)
(since
3|6),
we know that
By Cauchy's theorem, any generator of GF(2
) must )| = 8, the only possible element that can 3 × 9 3 9 54 GF(2 ) = hu i. Thus, GF(2 ) = {0, 1, u , ..., u }.
Since |GF(2
Hence
3
3
6. Let F be a eld, and let n be a positive integer. An element ζ ∈ F is called a primitive nth root of unity if it has order n in the multiplicative group F × . Show that no eld of characteristic p > 0 contains a primitive pth root of unity. If
p
F
has characteristic
is not a divisor of
k
p,
p − 1.
then
|F × | = pk − 1
for some
k.
Now, by clearly,
(The division algorithm can be used to show this.)
Thus, by Cauchy's Theorem, there cannot exist an element of Hence, there can be no primitive
pth
F×
of order
p.
root of unity.
7. Let n ∈ Z+ , and dene τ (n) to be the number of divisors of n. (a) Show that τ is a multiplicative function. αk α1 α2 (b) Show that if n = p1 p2 · · · pk , then τ (n) = (α1 + 1)(α2 + 1) · · · (αk + 1). (c) Show that τ (n) is odd if and only if n is a square. P (d) Show that d|n τ (d)µ(n/d) = 1. n has divisors r and s such that (r, s) = 1. Then, r and s τ (rs) = τ (r)τ (s). αk α1 α2 αi (b) By (a), we have τ (n) = τ (p1 )τ (p2 ) · · · τ (pk ). Then, each pi has αi αi 2 1, pi , pi , ..., pi as divisors, so τ (pi ) = αi + 1. Hence, τ (n) = (α1 + 1)(α2 + 1) · · · (αk + 1), as desired. (c) If n is a square, then all of its prime factors are raised to an even power, αk α1 so by (b), τ (n) is odd. If τ (n) = τ (p1 · · · pk ) is odd, then by (b) all of the factors of (α1 + 1) · · · (αk + 1) must be odd. (For if there were so much as one αi + 1 that were even, then the expression (α1 + 1) · · · (αk + 1) would be even.) This implies that αi + 1 = 2mi for some mi for all αi . Thus, αi = 2mi + 1, so αi is odd for all i. (d) We have established that τ is a multiplicative function, so if we dene P f (n) ≡ 1, then τ (n) = d|n f (d) and so by the Mobius Inversion Formula, we P obtain 1 = f (m) = n|m µ(m/n)τ (n). (a) Suppose that
share no factors, so
8. Let n ∈ Z+ and dene σ(n) = d|n d, the sum of positive divisors of n. (a) Show that σ is a multiplicative function. Qk αk αi +1 α1 α2 (b) Show that if n = p1 p2 · · · pk , then σ(n) = − 1)/(pi − 1) . i=1 (pi (c) Show that σ(n) is odd if and only if n is a square or two times a square. P (d) Show that n|d σ(d)µ(n/d) = n. P
(a) Suppose that and
s
n
r and s such σ(rs) = σ(r)σ(s). Thus, σ
has two divisors
share no divisors, so
that
(r, s) = 1.
is multiplicative.
Then,
r
CHAPTER 6.
51
FIELDS
(b) By (a), we see that i +1 pα i
σ(n) =
Qk
i=1
i σ(pα i ).
We need to show that
i σ(pα i )=
−1 . We do this by induction on αi . For αi = 1, we get σ(pi ) = p + 1. pi − 1 2 pi − 1 (pi + 1)(pi − 1) Also, = = pi + 1. Thus, the statement is true for pi − 1 pi − 1 αi = 1. Suppose it is true for all αi ≤ N . Then, for αi = N + 1, we see N +1 N +1 N +1 that σ(pi ) = σ(pN . Thus, by the inductive hypothesis, σ(pi )= i ) + pi N +1 N +1 N +2 N +1 N +2 p − 1 + p − p p − 1 pi −1 +1 +pN = i = , as required. Hence, i pi − 1 pi −1 pi − 1 Qk i +1 σ(n) = i=1 (pα − 1)/(pi − 1) . i αi (c) For any prime number pi , we have τ (pi ) = αi + 1. Hence, if αi is even, αi αi we will have that σ(pi ) is a sum of an odd number of terms. Thus, σ(pi ) is αk α1 odd for all pi if αi is even. Hence, if n is a square, σ(n) = σ(p1 ) · · · σ(pk ) is a product of odd numbers and thus odd.
2αk 2 n is two times a square, then σ(n) is of the form σ(n) = σ(22α1 +1 )σ(p2α 2 ) · · · σ(pk ). 2αk 2α2 2α1 +1 As previously discovered, σ(p2 ) · · · σ(pk ) is odd. Then, τ (2 ) = 2α + 2. 2α +1 One of the factors is 1, and the rest are even. Thus, σ(2 1 ) is odd. αi Conversely, σ(pi ) is even if αi is odd. (See rst paragraph of (c)). Hence, α1 k σ(n) = σ(p1 ) · · · σ(pα k ) would be even since one even factor makes the entire If
product even. (d) We have shown
P
d|n f (d).
σ
to be multiplicative.
Let
Thus, by the Mobius Inversion Formula,
f (n) = n. Then, σ(n) = P n|d σ(d)µ(n/d) = n.
9. A positive integer n is called perfect if it is equal to the sum of its proper positive divisors. Thus, n is perfect if and only if σ(n) = 2n. Prove that n is an even perfect number if and only if n = 2p−1 (2p − 1), where p and 2p − 1 are prime numbers. p and 2p − 1 are prime. We show that n = 2p−1 (2p − 1) p−1 is perfect. Examine σ(n) = σ(2 )σ(2p − 1). Since 2p − 1 is prime, its only p p−1 divisors are 1 and 2 − 1. Thus, σ(n) = σ(2 ) · 2p = (2p − 1)2p from Q8. Thus, p−1 p σ(n) = 2 · 2 (2 − 1) = 2n, as needed. (⇒) Suppose that p is prime, but 2p − 1 is not. Then σ(n) = σ(2p−1 )σ(2p − 1) = (2p − 1) · (something greater than 2p ). (Since at very least 2p − 1 has three p factors:1, 2 − 1 and p ¯, for some prime p¯. If we add these three up we get p p 2 + p¯ > 2 .) Thus, σ(n) > 2n. Thus, n is not perfect. (⇐)
Suppose that
+ 10. Let D be an integral P domain. Show that Q if f : Z → D is a nonzero multiplicative function, then d|n µ(d)f (d) = p|n (1 − f (p)), for all n ∈ Z+ , where the product is taken over all prime divisors p of n. Suppose that
n
has the prime factorization
αk 1 α2 n = pα 1 p2 · · · pk .
We form di-
n by selecting prime factors of n and multiplying them together. Both f α1 α2 α1 α2 α α and µ are multiplicative, so we see that µ(d)f (d) = µ(q1 q2 · · · qmm )f (q1 q2 · · · qmm ) = α1 α1 αm αm µ(q1 )f (q1 ) · · · µ(qm )f (qm ). Now, if αi > 1 for any pi prime factor of a divisors of
CHAPTER 6.
52
FIELDS
visor, then by the denition of i µ(pα i ) = 0.
since
µ,
we have that the entire term must equal zero
Thus, the only divisors of
n
that will ever contribute to the
sum are those that are products of prime factors.
P
Q
d|n µ(d)f (d) = p|n (1 − f (p)) by induction on the number of prime factors of n. Suppose that n = p for some prime p. Then, We now show that
P
d|n µ(d)f (d) = µ(1)f (1) + µ(p)f (p) = f (1) − f (p) = 1 − f (p). (We see that f (1) = 1 since we must have f (x) = f (1 · x) = f (1)f (x).) Thus, we have established a basis for induction. Let us suppose that this formula holds for M prime factors. Now, we need to show that it works for M + 1 factors. αM +1 αM +1 αM α1 α1 Let n = p1 · · · pM +1 . Then, n = mpM +1 where m = p1 · · · pM . Then, P P PM P d|n µ(d)f (d) = d|m µ(d)f (d)−f (pM +1 )+ i=1 f (pi pM +1 )− i6=j f (pi pj pM +1 )+ P · · · + (−1)M i1 6=i2 6=···6=iM f (pi1 pi2 ·P · · piM pM +1 ). Since f is multiplicative, we P can factor out f (pM +1 ) and obtain µ(d)f (d) = d|m µ(d)f (d) − f (pM +1 ) d|n PM P P 1 − i=1 f (pi ) + i6=j f (pi pj ) − · · · + (−1)M −1 i1 6=i2 6=···iM f (pi1 pi2 · · · piM ) .
−f (pM +1 ) P on the RHS of the equation is equal P µ(d)f (d) = µ(d)f (d)−f (pM +1 ) d|m µ(d)f (d). d|n d|m Q Q By the inductive hypothesis, this is equal to p|m (1−f (p))−f (pM +1 ) p|m (1− Q Q f (p)) = p|m (1 − f (p)) (1−f (pM +1 )) = p|n (1−f (p)), as desired. (Whew!)
We notice that the coecient of to
P
d|m µ(d)f (d).
Hence,
P
11. Let R be a commutative ring. Let R be the set of all functions f : Z+ → R. For f, g ∈ R, dene f + g by ordinary addition of functions: (f + g)(n) = f (n) + g(n), for all n ∈ Z+ . Dene a product * on R as follows: (f ∗ g)(n) =
X
f (d)g(n/d), for all n ∈ Z+ .
d|n
The product * is called the convolution product of the functions f and g . Dene : Z+ → R by (1) = 1 and (n) = 0, for all n > 1. (a) Show that R is a commutative ring under the operations + and *, with identity . (b) Show that f ∈ R has a multiplicative inverse if and only if f (1) is invertible in R. (c) Show that if f, g ∈ R are multiplicative functions, then so is f ∗ g . (d) Show that if R is an integral domain, then the set of nonzero multiplicative functions in R is a subgroup of R× , the group of units of R. (e) Dene µR ∈ R as in Denition 6.6.3, with the understanding that 0 and 1 are (respectively) the additive and multiplicative identities of R. Let ν ∈ R be dened by ν(n) = 1, for all n ∈ Z+ . Show that µR ∗ν = , and that µR ∗ν ∗f = f , for all f ∈ R. (a) (Group property of +) Already shown in Ch. 5.
P f ∗(g+h)(n) = f ∗(g(n)+h(n)) = d|n f (d)(g(n/d)+ P P P h(n/d)) = d|n [f (d)g(n/d)+f (d)h(n/d)]= d|n f (d)g(n/d)+ d|n f (d)h(d/n)= (f ∗ g)(n) + (f ∗ h)(n). (Distributivity) Examine
CHAPTER 6.
53
FIELDS
The other distributive property is shown in a similar way.
P P (f ∗g)(n) = d|n f (d)g(n/d) = d|n f (n/d)g(d) = (g ∗ f )(n). (Whatever factors aren't in n/d are in d and vice-versa.) (b) Suppose that there is g ∈ R such that g ∗ f = f ∗ g = . This happens if and only if n = 1 implies that 1 = (1) = f (1)g(1) = g(1)f (1). This happens if and only if f (1) is invertible. (c) Suppose that f and g are multiplicative functions. Suppose that n = pq where p and q are prime. Then, (f ∗ g)(pq) = f (1)g(pq) + f (p)g(q) + f (q)g(p) + f (pq)g(1). (1) Next, (f ∗ g)(p)(f ∗ g)(q) = [f (1)g(p) + f (p)g(1)] · [f (1)g(q) + f (q)g(1)] = g(p)g(q) + g(p)f (q) + f (p)g(q) + f (p)f (q). (2) We see that (1) = (2) (keeping in mind that for multiplicative functions f and g , we must have f (1) = g(1) = 1). Hence, f ∗ g is multiplicative. × (d) Suppose that f is multiplicative. Then, as always, f (1) = 1 ∈ R . × Hence, f ∈ R . By (c), the set of multiplicative functions is closed. Thus, the × set of multiplicative functions are a subgroup of R . αk α1 α1 (e) Suppose that n = p1 · · · pk . Then, (µR ∗ ν)(n) = (µR ∗ ν)(p1 ) · · · (µR ∗ Qk Pαk αk αi ν)(pk ) = j=1 i=0 µR (pj ) . Now, clearly (µR ∗ ν)(1) = 1. Also, it is clear that if αi > 1 for some i, then (µR ∗ ν)(n) = 0. So, suppose that αi = 1 for all Qk i. Then, we have (µR ∗ ν)(n) = j=1 (1 − 1) = 0. Hence, (µR ∗ ν) = . Now that this has been established, we have that µR ∗ ν ∗ f = ∗ f = f (keeping in mind that the ring R is commutative). (Commutativity) We see that
+ 12. Let R be a commutative ring. function, and let P Let f : Z → R be any + F : Z → R be dened by F (n) = d|n f (d), for all n ∈ Z+ . Show that if F is a multiplicative function, then so is f . Suppose we have (r, s) = 1. Now, by the Mobius Inversion formula, we have P f (rs) = n|rs µ(rs/n)F (n) = 1 − F (r) − F (s) + F (rs) = 1 − F (r) − F (s) + F (r)F (s). (1) Next, f (r)f (s) = (−F (r) + 1)(−F (s) + 1) = F (r)F (s) − F (r) − F (s) + 1. (2) We see that (1) = (2), so
6.7
f
is multiplicative.
Quadratic Reciprocity
1. Prove that
ab p
=
a p
b p
for all a, b ∈ Z such that p - a and p - b.
On the other hand,
ab p
≡ (ab)(p−1)/2 ≡ a(p−1)/2 b(p−1)/2 (mod p). b a (p−1)/2 (mod p) and pb ≡ b(p−1)/2 , so ap p ≡a p ≡ b p). Thus, ab = ap p p .
We see that by Euler's Criterion
a(p−1)/2 b(p−1)/2 (mod
2. Compute the following values of the Legendre symbol.
CHAPTER 6.
(a) (b)
54
FIELDS
231 997 783 997
231 3·7·11 3 7 11 = 997 997 = 997 997 997 . Following Example 997 3 (997−1)(3−1)/4 498 ≡ 9972/2 (−1) (mod 3) ≡ 6.7.1, we nd that 997 = 3 (−1) 3 7 11 1(mod 3). Thus, = 1 . Similarly, = −1 and = −1 . Hence, 997 997 997 231 = (1)(−1)(−1) = 1 . 997
(a) From Q1,
(b) Again, Also,
29 997
=
33 29 997 997 . We compute, as above, that 783 −1. Hence, 997 = (1)3 (−1) = −1. 783 997
=
3 997
= 1.
3. Is the congruence x2 ≡ 180873(mod 997) solvable? 180873 ≡ 416(mod 997). To determine if the congruence is 416 2 5 13 2 = 997 = 997 997 . Now, obviously 997 13 = −1. Also, 997 = 1. Hence 416 997 = −1. Thus, the congruence is
First of all,
solvable, we must compute
(−1)
9972 −1 8
not solvable.
*4. Determine the value of pr for the indicated values of r, where p is an odd prime subject to the indicated conditions. r = 5, p 6= 5 r = 6, p 6= 3 (c) r = 7, p 6= 7 (d) r = 11, p 6= 11 (e) r = 13, p 6= 13 (a)
(b)
5 p
(a) We see that
=
p 5
(−1)4(p−1)/4 =
p 5 . Thus,
5 p
≡ p2 (mod 5).
Thus, we have Value of
(b) We see that possible values of
6 p
=
p
modulo 5
5 p
1
1
2
-1
3
-1
4
1
3 p
2 p
=
p 3
(−1)
p2 −1 8
. Example 6.7.3 gives the
p 3 . Using these, we construct the following table.
Value of
p
modulo 12
1
6 p
1
5
1
7
-1
11
-1
CHAPTER 6.
55
FIELDS
p 7
(−1)6(p−1)/4 . Now, p7 ≡ p3 (mod 7). If p7 = 1, then p ≡ 1, 2, or 4(mod 7). If p7 = −1, then p ≡ 3, 5, or 6(mod 7). In addition, we have that p ≡ 1 or 3(mod 4). Thus, we create the following table: 7 Value of p modulo 28 p (c) We see that
7 p
=
1
-1
9
-1
25
-1
15
1
23
1
11
1
17
-1
5
-1
13
-1
3
1
19
1
27
1
[FINISH]
5. If a is a quadratic nonresidue of each of the odd primes p and q , is the congruence x2 ≡ a(mod pq) sovlable? Suppose that
k.
x2 ≡ a(mod pq) has a solution. Then, x2 − a = kpq for some x2 = a + p(kq) = a + q(kp) imlying that x2 ≡ a(mod p) q) have solutions. Contradiction.
This implies that
and
x2 ≡ a(mod
6-8.
[FINISH]
Chapter 7
Structure of Groups 7.1
Isomorphism Theorems: Automorphisms
See other Scribd document.
7.2
Conjugacy
1. Let H be a subgroup of G. Prove that N (H) is a subgroup of G. a, b ∈ N (H). Then, we have aHa = H and b−1 Hb = H . Then, H(ab−1 )−1 = ab−1 Hba−1 = a(b−1 Hb)a−1 = aHa−1 = H . Thus, N (H) ≤
Let −1
ab G.
2. Let H be a subgroup of the group G. Prove that the subgroups of G that are conjugate to H are in one-to-one correspondence with the left cosets of N (H) in G. K ∼ H . Then K = aHa−1 for some a ∈ G. Dene the prescription φ aHa 7→ aN (H). Now, if φ(aHa−1 ) = φ(bHb−1 ), then we have aN (H) = bN (H) which implies that ab−1 ∈ N (H). Hence ab−1 Hba−1 = H . This is −1 equivalent to aHa = bHb−1 . Thus, members of G/N (H) are in a one-to-one correspondence with subgroups of G conjugate to H . Let
−1
3. Let G be a group with subgroups H and K such that H ⊆ K . Show that H is a normal subgroup of K if and only if K ⊆ N (H). H E K if and only if aHa−1 = H for all a ∈ K . Since any −1 element a such that aHa = H is contained in N (H) by denition, we see that H E K if and only if K ⊆ N (H). We see that
56
CHAPTER 7.
57
STRUCTURE OF GROUPS
4. Let p be a prime number, and let C be a cyclic subgroup of order p in Sp . Compute the order of N (C). The only element of order element of
K
p
since the only element of order
C E Sp .
in
Sp
is
(1, 2, ..., p).
has the same structure as an element in
Thus,
p
is
(1, 2, ..., p),
Let
C
K ∼ C.
Then, every
by Example 7.2.3. But
it must be that
K = C.
Thus,
|C| = p.
5. Let G be a group, let H be a subgroup of G, and let a ∈ G. Show that there is a subgroup K of G such that K is conjugate to H and aH = Ka. aHa−1 ≤ G. Let ah1 a−1 , ah2 a−1 ∈ aHa−1 . −1 = (ah1 a−1 )(ah−1 ) = ah1 h2 a−1 ∈ aHa−1 . Thus, 2 a
We must simply show that
−1
−1 −1
(ah1 a )(ah2 a ) aHa−1 ≤ G, as desired.
Then,
6. Let G be a group, let x, y ∈ G and let n ∈ Z. Show that y is a conjugate of xn if and only if y is the nth power of a conjugate of x. y = (gxg −1 )n ⇔ y = (gxg −1 )(gxg −1 ) · · · (gxg −1 ) ⇔ y = gxn g −1 ⇔ y ∼ xn .
7. Find the conjugate subgroups of D4 . First, all normal subgroups are self-conjugate, and since each subgroup of order 4 has index 2, the subgroups
{e, a2 , b, a2 b}, {e, a, a2 , a3 },
and
{e, a2 , ab, a3 b}
are all self conjugate.
{e, b} ∼ {e, a2 } ∼ {e, a2 b} (from simple verication) and (ab)a3 b(ba3 ) = 3 so {e, ab} and {e, a b} are self-conjugate.
Next
2
3
a ba
,
8. Find the conjugacy classes of D5 . First,
D5 = {e, a, a2 , a3 , a4 , b, ab, a2 b, a3 b, a4 b}.
The cyclic subgroup
{e, a, a2 , a3 , a4 }
has order 5 and is of index 2 and thus normal. Thus, it is its own conjugacy class. There can be no more subgroups of order 5 (for if there were, there would be more than ten elements). The only possible orders of subgroups would be order 2. There are ve such subgroups:
{e, b}.
We multiply
{e, ab}
{e, ab}, {e, a2 b}, {e, a3 b}, {e, a4 b},
and
on the left and right by the non-identity elements in
the subgroups above and see that all ve subgroups are conjugate.
9. Describe the conjugacy classes of S5 by listing the types of elements and the number of each type in each class.
CHAPTER 7.
58
STRUCTURE OF GROUPS
Class
Calculation of Cardinality
Cardinality
[(1)]
The identity mapping is unique
1
[(1,2)]
[(1,2)(3,4)]
5 2
5 2
[(1,2,3)] [(1,2,3)(4,5)]
5 3
10
3 2
·
1 2
15
· 2!
20
Same
[(1,2,3,4)] [(1,2,3,4,5)]
5 4
20
· 3!
30
4!
24
120
Total:
10. Find the conjugacy classes of A4 . A4 consists of the identity element (1), all 3-cycles (of which there are 8), and three single products of disjoint transpositions (those being (1,2)(3,4), (1,3)(2,4), and (1,4)(3,2)) .
Since disjoint transpositions commute, we know that the
products of disjoint transpositions cannot be conjugate to any 3-cycle or the identity. We now nd the conjugacy classes: (1,3)(2,4)[(1,2)(3,4)](1,3)(2,4) = (1,3)(4,2)(2,1)(4,3)(3,1)(2,4) = (1,2,4)(2,4)(1,3)(1,3,4) = (2,1)(4,3) (1,4)(3,2)[(1,2)(3,4)](1,4)(3,2) = (3,2)(4,1)(1,2)(3,4)(4,1)(3,2) = (3,2)(2,1,4)(1,4,3)(3,2) = (2,4)(1,3) Thus, (1,2)(3,4)
∼
(1,3)(2,4)
(1,2,3)
∼
∼
(1,4)(3,2). Doing a similar analysis on the 3-cycles, we get that
(4,3,2)
∼
(3,4,1)
∼
(2,1,4) and (1,3,2)
∼
(4,2,3)
∼
(3,1,4)
∼
(2,4,1).
11. Find the conjugacy classes for the quaternion group Q dened in Example 3.3.7. iji−1 = j; kjk−1 = j. Thus, j is its own conjugacy class. jij−1 = −i = kik−1 . Thus, i ∼ −i. iki−1 = −k = jkj−1 . Thus, k ∼ −k. Then, ±1 are
We see Then, Then,
their own
conjugacy classes.
12. Write out the conjugacy class equations for the following groups. (a) (b)
A4 S5 P |A4 | = |Z(A4 )| + P[A4 : C(x)] = 1 + (3 + 4 + 4) = 12. clearly |S4 | = |Z(S5 )| + [S5 : C(x)] = 1 + (10 + 15 + 20 +
(a) Using Q4, clearly (b) Using Q9,
20 + 30) = 120.
13. Let the dihedral group Dn be given by elements a of order n and b of order 2, where ba = a−1 b. Show that am is conjugate to only itself and a−m , and that am b is conjugate to am+2k b for any integer k.
CHAPTER 7.
59
STRUCTURE OF GROUPS
x ∼ am in Dn . We see that x cannot have a factor of b in it since if a b = ya y , then whether y has a factor of b in it or not, the b's in y and y −1 will cancel. Thus, x must be a power of a. Thus, a` = ak bam ban−k −1 for some k . By the fact that ba = a b, we see that this implies that a` = k n−m n−k k+n−m+n−k 2n−m a a a = a = a = a−m . Of course, every element is Suppose that
m −1
`
conjugate to itself as well.
ak bam bban−k = ak bam+n−k = ak+n−m+k = a2k+m b.
We see that
14. Show that the Frobenius group F20 (dened in Exercise 12 of Section 7.1) is isomorphic to the subgroup of S5 generated by the permutations (1, 2, 3, 4, 5) and (2, 3, 5, 4). Use this fact to help in nding the conjugacy classes of F20 , and its conjugacy class equation. h(1, 2, 3, 4, 5), (2, 3, 5, 4)icomes
The fact that
from the work done in Q13 of
Section 7.1, where we found elements a of order 5 and b of order 4 and that any element of
F20
could be uniquely expressed as a product of these two elements
F20 ). Z(F20 ) = 4. Now, h(1, 2, 3, 4, 5)i = {(1), (1, 2, 3, 4, 5)), (1, 3, 5, 2, 4), (1, 4, 2, 5, 3), (1, 5, 4, 3, 2)} and h(2, 3, 5, 4)i = {(1), (2, 3, 5, 4), (2, 5)(3, 4), (2, 4, 5, 3)}. So, it is easliy seen that h(1, 2, 3, 4, 5), (2, 3, 5, 4)i = {(1), (1, 2, 3, 4, 5), (1, 3, 5, 2, 4), (1, 4, 2, 5, 3), (1, 5, 4, 3, 2),(2, 3, 5, 4), (2, 5)(3, 4), (2, 4, 5, 2 Thus, the conjugacy classes of F20 are [a], [b], and [b ]. The class equation would then be |F20 | = 4 + 4 + 10 + 2 = 20. (i.e., they generate
From Q12 of 7.1,
15. Show that if a group G has an element a which has precisely two conjugates, then G has a nontrivial proper normal subgroup. Suppose that
C(a) E G p3 .
[a]∼
has precisely two elements.
[G : C(a)] = 2
and
*16. Show that for each prime p, there exists a nonabelian group of order Let us examine the possible centers of
side's theorem, to
Then,
(since subgroups of index two are normal.)
Zp ),
or
p2 .
Z
p2
and
Zp × Zp .
G.
By Lagrange's theorem and Burn-
Any group of order
and the abelian groups of order
are abelian) are
G/Z(G)
|Z(G)| = p,
p2
Z p2
is cyclic (isomorphic
p2
The following conclusions are made from the
theorem.
Z(G) Zp
p
(recall that all groups of order
G/Z(G) Zp2 Zp × Zp Zp
Conclusion
G
is abelian
G
is abelian
???
CHAPTER 7.
60
STRUCTURE OF GROUPS
Thus, we seek to nd a nonabelian group of order and
G/Z(G) ∼ = Zp × Zp .
p3
with center
Z(G) ∼ = Zp
[FINISH]
17. Let G be a nonabelian group of order p3 , for some prime number p. Show that Z(G) must have order p. p- group is nontrivial by Burnsides Theorem, so the pos|Z(G)| are p, p2 , or p3 . However, if |G| = p2 , then G would be 3 abelian. Since G is given to be nonabelian (so its center cannot have order p ), we must have |Z(G)| = p. The center of any
sibilities for
18. Determine the conjugacy classes of the alternating group A5 and use this information to show that A5 is a simple group. We have already found the conjugacy classes for Class
in Q9.
Order
[(1)]
1
[(1,2)(3,4)]
15
[(1,2,3)]
20
[(1,2,3,4,5)]
24
We know that a normal subgroup of
A5
A5
(or any group, for that matter) is
the union of conjugacy classes. In the following table, for every row, construct a union of conjugacy classes corresponding to the column headers who have X's in their cell. The order of this union is given in the right-most column. [(1)]
[(1,2)(3,4)]
[(1,2,3)]
[(1,2,3,4,5)]
X X
1 X
X
16 X
X
21 X
X
X
X
X
X X
Order of union
X
X
25 36
X
40
X
X
45
X
X
60
Now, by Lagrange's theorem, the order of any group must divide |A5 | = 60. The only possible unions that have orders dividing 60 are [(1)] = (1) and [(1)] ∪ [(1, 2)(3, 4)] ∪ [(1, 2, 3)] ∪ [(1, 2, 3, 4, 5)] = A5 . Hence, A5 has no proper nontrivial normal subgroups. Thus, A5 is simple.