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√ MATH 371 - THE RING Z[ D] DR. ZACHARY SCHERR

1. Background Throughout this note, D will denote a squarefree integer different from 1. This means that n2 does not divide √ D for any integer n > 1. The purpose of many of the ideas we’ve studied this note is to study the ring Z[ D] using √ in chapter 8 of DF. As a ring, we define Z[ D] to be √ Z[ D] := Z[x]/(x2 − D). √ More concretely, we can view Z[ D] as the set √ √ Z[ D] = {a + b D|a, b ∈ Z} with addition and multiplication given by √ √ √ (a + b D) + (c + d D) = (a + c) + (b + d) D √ √ √ (a + b D) × (c + d D) = (ac + bdD) + (ad + bc) D. √ √ For each α = a + b D ∈ Z[ D] we define the conjugate of α, denoted α, to be √ α = a − b D. The first observation we can make about the conjugate is √ Lemma 1.1. αβ = αβ for any α, β ∈ Z[ D]. Proof. This is just a direct calculation. We have √ √ αβ = (a − b D)(c − d D)

√ = (ac + bdD) − (ad + bc) D = αβ.

√ Define the function N : Z[ D] → Z given by √ √ √ N (a + b D) = a2 − b2 D = (a + b D)(a − b D).



For reasons we will discover later, N (α) = αα is known as the norm of α. We will soon see that many properties of α can be deduced from its norm. Our first observation about the norm is that it is multiplicative. Lemma 1.2. N (αβ) = N (α)N (β) 1

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Proof. N (αβ) = αβ · αβ = αβαβ (Lemma 1.1) = ααββ = N (α)N (β).  As an immediate consequence of Lemma 1.2 we have √ Proposition 1.3. α ∈ Z[ D] is a unit if and only if N (α) = ±1. √ Proof. If α is a unit then we can find β ∈ Z[ D] so that αβ = 1. Taking norms gives 1 = N (1) = N (αβ) = N (α)N (β) showing that N (α) = ±1. Conversely, if N (α) = ±1 then αα = ±1 so (α)(±α) = 1 proving that α is a unit.



This gives us a useful criterion for testing whether or not an element in √ √ Z[ D] is a unit. In Z[i] = Z[ −1] for example, the element a + bi is a unit if and only if N (a + bi) = a2 + b2 = 1 √ proving that √ the only units in Z[ D] are ±1, ±i. For D < −1 it is easy to check that Z[ D] only has as units ±1. For √ D > 1, on the other hand, it is a consequence of number theory that Z[ √ D] has infinitely many units. In √ Z[ 2], for example, one can see that 3 + 2 2 is a unit since √ N (3 + 2 2) = 32 − 22 · 2 = 1. √ This immediately that Z[ 2] has infinitely many units since for √ implies example (3 + 2 2)n √ is a unit for every n ≥ 1. Of course one would have to know that (3 + 2 2) does not have finite multiplicative order, but it shouldn’t of √ √ nbe too hard to convince yourself that the coefficients of powers (3 + 2 2) increase as n increases. Notice that in these rings a + b D is a unit if and only if a2 − b2 D = ±1. This is the subject of the very famous Pell equation from number theory. From Proposition 1.3 one can deduce Proposition√1.4. If N (α) = ±p, where p is a prime in Z, then α is irreducible in Z[ D].

√ MATH 371 - THE RING Z[ D]

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√ Proof. Suppose α = βγ for some β, γ ∈ Z[ D]. Taking norms gives ±p = N (α) = N (βγ) = N (β)N (γ). As this is an equation in Z, √ one of N (β), N (γ) must equal ±1 showing that either β or γ is a unit in Z[ D].  Proposition 1.4 doesn’t have a converse per se, but one can prove √ Proposition 1.5. Suppose π ∈ Z[ D] is a prime element. Then either π is associate to some prime p ∈ Z or N (π) = ±p. √ Proof. If π is prime then ideal it generates, (π), is a prime ideal of Z[ D]. Let I = (π) ∩ Z. Then I is non-empty since N√(π) = ππ ∈ I and so I is an ideal in Z. Moreover, since (π) is prime in Z[ D] we have that I is prime in Z. Thus √ I = (p) for some prime in Z. Since p ∈ (π) we know that there is α ∈ Z[ D] so that πα = p and so taking norms gives N (π)N (α) = p2 . Thus either N (π) = ±p or N (α) = ±1 in which case π and p are associates since α is a unit.  Remark. Be careful when comparing the last two propositions. In every ring primes are irreducible, but in some rings irreducible elements aren’t necessarily prime. We can √ use these preliminaries to say some very interesting things about specific Z[ D]. √ 2. Z[ −5] is not a UFD √ In Z[ −5] we have √ √ (1 + −5)(1 − −5) = 6 = 2 · 3. √ I claim that this proves that Z[ −5] is not a UFD. To do that, we check that each of the terms showing up in the above √ factorization are irreducible. Consider 2 for example. If 2 = αβ in Z[ −5] then taking norms gives 4 = N (α)N (β). We cannot have N (α) = 2 since one can easily check that the equation a2 + 5b2 = 2 has no integer solutions. This forces one of N (α), N (β) to equal 1 in which case the corresponding element is a unit. This proves that 2 is irreducible. The same argument works for 3. One only needs to show that there is no integer solution to a2 + 5b2 = 3 √ in. √ We can use our calculations to verify that 1 + −5 is irreducible. If 1 + −5 = αβ then taking norms gives 6 = N (α)N (β). We already verified

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√ that no element be 1 and √ of Z[ −5] has norm 2 so one of N (α), N (β) must√ indeed 1 + −5 is irreducible. The same argument works for 1 − −5 and indeed √ √ (1 + −5)(1 − −5) = 6 = 2 · 3 √ proves that Z[ −5] does not have unique factorization. In fact, the element 2 is irreducible but not prime. It cannot√be prime since the above equation √ tells us that 2 divides the product (1 + −5)(1 − −5) yet doesn’t divide either √ term. This is because 2α would have even components for any α ∈ Z[ −5]. It is a very deep √ and profound idea that one can √ “fix” lack of unique factorization in Z[ −5] by proving that ideals in Z[ −5] have unique factorization into prime ideals. This is one of the key subjects of the field algebraic number theory. One shows, for example that the ideals: √ √ √ I2 = (1 − −5, 3), I3 = (1 + −5, 2) I1 = (1 + −5, 3), √ are prime ideals in Z[ −5]. Then on the level of ideals we have

(1 + (1 −

√ √

(2) = I32 (3) = I1 I2 −5) = I1 I3 −5) = I2 I3

so that the two factorizations of 6 in ideals corresponds to I1 I3 I2 I3 = (6) = I32 I1 I2 and we see that, at least for ideals, we haven’t broken unique factorization. Let’s just do one example calculation to check why the ideal equations are correct. We have, for example: √ √ I32 = (1 + −5, 2)(1 + −5, 2) √ √ = ((1 + −5)2 , 22 , (1 + −5)(2)) √ √ = (−4 + 2 −5, 4, 2 + 2 −5). This last ideal is certainly contained in (2) since every element generating it has even components. Conversely, √ √ 2 = (2 + 2 −5) − ((−4 + 2 −5) + 4) showing that I32 contains (2). This proves the equality. 3. The ring Z[i] √ √ We can contrast Z[ −5] with the ring Z[i] = Z[ −1]. This ring will turn out to be a Euclidean domain under the norm function N (a + bi) = a2 + b2 . We prove

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Theorem 3.1. Let α, β ∈ Z[i] with β 6= 0. Then there exist θ, γ ∈ Z[i] so that α = βθ + γ with γ = 0 or N (γ) < N (β). Proof. Let α = a + bi and β = c + di. Simplify the complex number a + bi = r + si c + di where r and s are rational numbers. Let m and n be r and s rounded to the nearest integer. Then |m − r| ≤ 12 and |n − s| ≤ 12 as well. Take θ = m + ni and γ = α − βθ. Then α γ = − θ = (r − m) + (s − n)i. β β This shows that   1 1 1 γ = N ((r − m) + (s − n)i) = (r − m)2 + (s − n)2 ≤ + = . N β 4 4 2 But then 1 N (γ) ≤ N (β) 2 and so we have our desired quotient and remainder.  Theorem 3.1 shows that Z[i] is a Euclidean domain, so in particular it is a principal ideal domain and in fact a unique factorization domain. In particular, in Z[i] we see that prime elements and irreducible elements are the same. We can use the norm to classify which elements of Z[i] are prime. Immediately we see 1 + i is prime since N (1 + i) = 2 is prime in Z. We also have: Proposition 3.2. Let p ∈ Z be a prime for which p ≡ 3 mod 4. Then p remains prime in Z[i]. Proof. Assume for the sake of contradiction that p = ππ 0 where π and π 0 are non-units in Z[i]. Then taking norms gives p2 = N (π)N (π 0 ) and so N (π) = N (π 0 ) = p. This means that there is an integral solution to the equation a2 + b2 = p and upon reducing modulo 4 we get a2 + b2 ≡ 3 02

mod 4.

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Notice that ≡ ≡ 0 mod 4 and 12 ≡ 32 ≡ 1 mod 4 so the above equation is absurd and we have our desired contradiction.  So far we have identified that ±1 ± i is prime in Z[i] and that every p ≡ 3 mod 4 is prime in Z[i]. The other prime elements are precisely those πs for which N (π) = p where p is a prime in Z congruent to 1 modulo 4. To see this, we we must use some basic number theory. We will set this up in the next section.

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4. Number Theory The basic tool from number theory that makes everything work in Z[i] is the following proposition. Proposition 4.1. Let p be an odd prime. Then there exists n ∈ Z with n2 ≡ −1 mod p if and only if p ≡ 1 mod 4. Proof. Suppose first that there exists n ∈ Z so that n2 ≡ −1 mod p. Then the order of n in the group of units Z/pZ× is necessarily 4 and so Lagrange tells us that 4 = o(n) divides p − 1 = o(Z/pZ). That is, p ≡ 1 mod 4. The converse is harder to prove. Later in the course we will see that Z/pZ× is always a cyclic group. If g generates this group then o(g) = p − 1 p−1 and if p ≡ 1 mod 4 then x = g 4 will have the property that x2 ≡ −1 mod p. We don’t need such a strong result to see that Z/pZ has a square root of −1. For example, let G denote the group Z/pZ× . Then the subgroup p−1 H = {±1} has index p−1 2 in G. Then if p ≡ 1 mod 4 we have that 2 is even, so the group G/H has an element of order 2 by Cauchy’s theorem. Let x ∈ G be an element such that o(x) = 2 in G/H. This says that x2 ∈ H so x2 = 1 or x2 = −1. Of course if x2 = 1 then necessarily x = ±1 (since Z/pZ is an integral domain) and in this case o(x) = 1. Thus x2 = −1 and we have our desired element. If you don’t like group theory, you can be clever and prove this more directly. First we show that (p − 1)! ≡ −1 mod p for any prime p. This is because in Z/pZ× , i 6= i−1 unless i = −1. The reason is that i = i−1 if and only if i2 = 1 and in an integral domain this occurs if and only if i = ±1. But then every element in the product (p − 1)! cancels out with its inverse except −1 and so indeed (p − 1)! ≡ −1 mod p. Let y = p−1 2 . Then I claim that if p ≡ 1 mod 4 we have (y!)2 ≡ −1 mod p. To see this, note that p−1 p−1 (y!)2 ≡ (1 · 2 · · · )( · · · 2 · 1) 2 2 p−1 p+1 ≡ (1 · 2 · · · )(− · · · − (p − 2) · −(p − 1)) 2 2 p−1 ≡ (−1) 2 (p − 1)! = −1 mod p when p ≡ 1 mod 4.



Using Proposition 4.1 we can prove Proposition 4.2. Let p ≡ 1 mod 4 be a prime in Z. Then there exists a prime π ∈ Z[i] such that N (π) = p. In particular, p is not prime in Z[i] and p factors as p = ππ. Proof. By the above proposition we know the existence of n ∈ Z such that n2 ≡ −1 mod p. This tells us that p|n2 + 1, but of course in Z[i] n2 + 1 = (n + i)(n − i).

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Thus p|(n + i)(n − i), but notice that p does not divide neither n + i nor n − i. This shows that p is not prime in Z[i] so it is not irreducible. Let π be an irreducible factor of p. Then ππ 0 = p for some π 0 ∈ Z[i]. Thus N (π)N (π 0 ) = p2 so N (π) = p or N (π) = p2 . If N (π) = p2 then π 0 is a unit, but then (π) = (p) saying that (p) is a prime ideal, which we’ve shown is absurd. Thus N (π) = p and indeed p = ππ.  Putting everything we’ve proven together yields: Theorem 4.3. There are exactly three types of primes in Z[i]. They are given by • ±1 ± i. • p ∈ Z so that p ≡ 3 mod 4 • π ∈ Z[i] so that N (π) = p where p ≡ 1 mod 4. As a final application, we prove the celebrated two squares theorem of Fermat. Theorem 4.4. Let p ≡ 1 mod 4 be a prime. Then there exist integers a, b so that a2 + b2 = p. Proof. If p ≡ 1 mod 4 then we’ve just shown that there exists π ∈ Z[i] with N (π) = p. But then if π = a + bi we have N (π) = a2 + b2 = p, proving the claim.