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Notes on Algebraic

Robin Chapman

January 20, 1995 (corrected November 3, 2002)

1 Introduction

This is a summary of my 1994–1995 course on Algebraic Numbers. (Revised and improved on 1993– 1994!) The background assumed is standard elementary theory—as found in my Level III course—and a little (Abelian) group theory. Corrections and suggestions for improvement are wel- come, and will be credited in future editions! I first learned algebraic from Stewart & Tall’s book ([3]) and this is an excellent account. However it’s more abstract than the approach of this course and deals with general theory while I deal mainly with the theory of quadratic fields. A book dealing mainly with quadratic fields is Cohn ([1]); I have incorporated many of the ideas in this book into this course, but this is a rather difficult book to read. I am grateful to Jeremy Bygott for corrections to and suggestions on a previous version, and to Paul Epstein for a further correction.

2 Algebraic Numbers and

An algebraic number is a root of a with rational coefficients, i.e., α is an algebraic number if and only if there exist a1, a2, . . . , an ∈ Q with

n n−1 n−2 α + a1α + a2α + ··· + an−1α + an = 0. Similarly an algebraic is a root of a equation with integer coefficients, i.e., β is an algebraic number if and only if there exist b1, b2, . . . , bm ∈ Z with

m m−1 m−2 β + b1β + b2β + ··· + bm−1β + bm = 0. Clearly every is also an algebraic number. (As a convention I shall use Greek letters for algebraic numbers, and Roman letters for rational numbers.) We think of algebraic integers and algebraic numbers as generalizations of integers and rationals. All integers are algebraic integers, and all rationals are algebraic numbers. Some examples of algebraic √ √ √ q integers are i, 2, 3 10, (1 + 5)/2 and 2 cos 2π/9. The algebraic numbers include (1 + i)/2, 1/5 and sin 2π/7. The numbers e and π are not algebraic—they are transcendental. Transcendence proofs are very difficult, and will not be covered in this course. (See [2] for proofs that e and π are transcendental.) A basic fact about algebraic integers is that a which is also an algebraic integer is an (ordinary) integer:

Proposition 1 If α is an algebraic integer and α ∈ Q, then α ∈ Z.

1 Proof Write α = a/b in lowest terms. Suppose that α isn’t an integer so b 6= ±1. As α is an algebraic integer, there exist c1, c2, . . . cn ∈ Z with

n n−1 α + c1α + ··· + cn−1α + cn = 0.

Writing α = a/b and clearing denominators gives

n n−1 n−1 n a + c1a b + ··· + cn−1ab + cnb = 0, which implies that an ≡ 0 (mod b). As b 6= ±1 then b has a prime factor, p, say. Hence p | an which implies that p | a. This means that p is a factor of both a and b contradicting the assumption that a/b is in its lowest terms. This contradiction establishes the result. 2 This result is useful as it shows that some algebraic numbers aren’t algebraic integers. The most important result in this section is that sums and product of algebraic numbers (resp. algebraic integers) are also algebraic numbers (resp. integers). Before we can prove this we need an alternative description of algebraic numbers and integers by means of matrix theory.

Lemma 1 A number α is an algebraic number if and only if it is an eigenvalue of a () matrix with rational entries, and it is an algebraic integer if and only if it is an eigenvalue of a (square) matrix with integer entries.

Proof Suppose first that A is a square matrix with integer entries. The characteristic polynomial of A is χA(λ) = det(λI − A). As the entries in λI − A involve only integer coefficients, it follows that its determinant χA(λ) is a monic polynomial with integer coefficients. Now if α is an eigenvalue of A, then χA(α) = 0 and so α is an algebraic integer. If A has rational entries then an identical argument shows that its eigenvalues are algebraic numbers. For the converse suppose that α is an algebraic integer, so that there exist b1, b2 . . . , bn ∈ Z with

n n−1 α + b1α + ··· + bn−1α + bn = 0.

We can rewrite this as n n−1 α = −b1α − · · · − bn−1α − bn. (∗) Let v be the (column) vector (1 α α2 ··· αn−1)T , so

 α   0 1 0 ··· 0   1   2       α   0 0 1 ··· 0   α         .   ......   .  αv =  .  =  . . . . .   .  = Av        αn−1   0 0 0 ··· 1   αn−2        n n−1 α −bn −bn−1 −bn−2 · · · −b1 α where A is a matrix with integer entries. (NB we have used (∗) here.) Hence α is an eigenvalue of A, with eigenvector v. Similarly if α is an algebraic number then we can find a matrix A with rational entries having α as an eigenvalue. 2 A slightly more elaborate argument proves that sums and products of algebraic numbers or integers are also algebraic numbers or integers.

Theorem 1 If β and γ are algebraic numbers (resp. algebraic integers), then β + γ and βγ are also algebraic numbers (resp. algebraic integers).

2 Proof I’ll just give the proof for algebraic integers, as that for algebraic numbers follows mutatis mutandis as in Lemma 1. Suppose β and γ are algebraic integers. The idea of the proof is to find a non-zero vector w and two integer matrices B and C with Bw = βw and Cw = γw. This will imply that β + γ and βγ are eigenvalues of B + C and BC respectively. Suppose that n n−1 β + b1β + ··· + bn−1β + bn = 0 and m m−1 γ + c1γ + ··· + cm−1γ + cm = 0 where the bjs and cks are integers. Let w be the mn by 1 vector (1 β β2 ··· βn−1 γ βγ β2γ ··· βn−1γ γ2 βγ2 ··· βn−1γm−1)T . Consider βw. The entries in this vector all have the form βj+1γk where 0 ≤ j < n and 0 ≤ k < m. If j + 1 < n this is already an entry in w; if j + 1 = n it equals

n k n−1 k k k β γ = −b1β γ − · · · − bn−1βγ − bnγ . In all cases βj+1γk is a linear combination, with integer coefficients, of entries in w. Hence there is an integer matrix B with Bw = βw. Similarly there is an integer matrix C with Cw = γw. (Writing down general expressions for B and C is fairly horrible so I won’t do it, but from the numerical examples I’ll give in lectures I hope it’s plain that this construction is a lot easier than it looks.) Now (B + C)w = Bw + Cw = βw + γw = (β + γ)w and (BC)w = B(Cw) = B(γw) = γBw = (βγ)w so β + γ and βγ are eigenvalues of B + C and BC respectively. As B + C and BC have integer coefficients, then by Lemma 1 β + γ and βγ are algebraic integers. 2

3 Quadratic Fields

From now on we abandon the general study of algebraic numbers to concentrate on a specific class of such numbers, quadratic numbers, i.e., those which are solutions of quadratic with√ rational coefficients. By the solution formula for quadratics each such number has the form a + b d where a, b and d are rational numbers. We won’t even look at all quadratic numbers at once;√ if we fix a non-square rational number d we shall consider the of all numbers of√ the form a + b d with a√and b ∈ Q. This set is called a quadratic field and is denoted by Q(√d). (NB if d >√0 then d is by convention the positive√ , and if d = −D < 0 then d will denote i D.)√ If d > 0 then every element of Q( d) is real, and we say is a real quadratic field; if d < 0 then Q( d) contains non-real numbers—it is an imaginary quadratic field. The structure of real quadratic fields is radically different from that of imaginary quadratic fields, as we shall see on many subsequent occasions. We should perhaps pause to justify our terminology; a field is by definition an algebraic structure which admits , , and (except by 0) where these have all the familiar properties. This is indeed the case. √ Lemma 2 Let K = Q( d) be a quadratic field, where d is a non-square rational. Then if α, β ∈ K then α + β, α − β, αβ and, provided β 6= 0, α/β ∈ K.

3 Proof The proofs for addition, subtraction and multiplication are easy so I’ll skip them. For division it’s OK to treat the special case√ of α = 1, for if 1/β ∈ K then by the multiplication rule α/β = α(1/β) ∈ K. Now if β = a + b d then we might try to write √ √ 1 1 a − b d a − b d = √ = √ √ = . β a + b d (a + b d)(a − b d) a2 − b2d

But remembering the commandment, “thou shalt not divide by zero”, we must check that the denominator a2 − b2d is non-zero. But if a2 − b2d = 0 then either b 6= 0 and then d = (a/b)2 is a square contrary to hypothesis, or b = 0 and then a = 0 implying that β = 0, again contrary to hypothesis. Hence 1 a −b √ = + d ∈ K β a2 − b2d a2 − b2d as desired. 2 In this proof we’ve employed the useful trick of rationalizing a denominator√ by multiplying√ by its conjugate. Formally speaking we say that the conjugate of β = a + b d is β∗ = a − b d. The operation of conjugation reflects the idea that both square roots of a number d are “equally good”. (For a much deeper application of this principle see√ the course or books on , e.g. [2].) Conjugation respects the algebraic properties of Q( d). √ Lemma 3 Suppose that α, β ∈ Q( d). Then (α∗)∗ = α, (α + β)∗ = α∗ + β∗, (α − β)∗ = α∗ − β∗, (αβ)∗ = α∗β∗ and, provided β 6= 0, (α/β)∗ = α∗/β∗.

Proof These are all straightforward apart, perhaps, for division, which can be deduced from multiplication by the following artifice. Put γ = α/β so α = βγ. Then α∗ = β∗γ∗ and so ∗ ∗ ∗ ∗ α /β = γ = (α/β) . 2 √ In showing that 1/β was in Q( d) I multiplied β by β∗, obtaining a non-zero rational. Hence we define the norm of β to be N(β) = ββ∗. Similarly we define the trace of β to be T (β) = β + β∗. The norm and the trace of β are both rational numbers. Hence β and β∗ are the roots of

(x − β)(x − β∗) = x2 − T (β)x + N(β) = 0, an equation with rational coefficients which is called the characteristic equation of β. The trace respects addition and subtraction, and the norm respects multiplication and division. √ Lemma 4 If α, β ∈ Q( d) then T (α + β) = T (α) + T (β), T (α − β) = T (α) − T (β), N(αβ) = N(α)N(β) and, provided β 6= 0, N(β) 6= 0 and N(α/β) = N(α)/N(β).

Proof First we note that T (α ± β) = (α ± β) + (α ± β)∗ = α ± β + α∗ ± β∗ = (α + α∗) ± (β + β∗) = T (α) ± T (β). Similarly N(αβ) = (αβ)(αβ)∗ = αβα∗β∗ = (αα∗)(ββ∗) = N(α)N(β). Finally if β 6= 0 then N(β) 6= 0 by the proof of Lemma 2, and so N(α/β) = (α/β)(α/β)∗ = (α/β)(α∗/β∗) = (αα∗)/(ββ∗) = N(α)/N(β). 2 ∗ 2 In imaginary quadratic fields β = β, the complex conjugate, and so√N(β) = ββ = |β| ≥ 0; but in real quadratic fields β = β 6= β∗ in general, and as N(1) = 1 and N( d) = −d the norm takes on both positive and negative values. √ √ How many quadratic fields are there? If we have two such fields Q( d1) and Q( d2) how can we tell if they’re the same? The following result tells us. √ √ Proposition 2 If K1 = Q( d1) and K2 = Q( d2), then K1 = K2 if and only if d1/d2 is a square of a rational.

4 √ √ √ √ √ Proof If d /d = r2 where r is rational, then d = r d and d = 1 d . Hence a + b d = √ 1 2 √ √ 1 2 2 r 1 1 a + br d2 and u + v d2 = u + (v/r) d1. It follows that√ K1 ⊆ K2 and K2 ⊆ K1, i.e.,√ that K1 =√K2. The converse is slightly trickier. If K1 = K2 then d1 ∈ K2 and so we can write d1 = a + b d2 with a and b rational. Squaring gives

2 2 q d1 = (a + b d2) + 2ab d2.

2 2 This means that d1 = a + b d2 and 2ab = 0. Hence either a = 0 or b =√ 0; a = 0 means that 2 2 d1/d2 = b and we’re OK, while b = 0 means that d1 = a contrary to Q( d1) being a quadratic field. 2 We can now distinguish between quadratic fields. Can we now classify them? We say that an integer d is squarefree if it’s not divisible by any square bigger than 1. This means that d = ±1 or ±p1p2 ··· pk where the pj are distinct primes. √ Proposition 3 If K is a quadratic field, then K = Q( d) where d is a uniquely determined square- free integer and d 6= 1. q q √ Proof If K = ( a/b) where a and b are integers then K = ( b2(a/b)) = ( ab). Hence √ Q Q Q K = Q( r) where r is an integer. If r isn’t squarefree then c2 | r where c is an integer bigger than 1. q √ Then K = ( r/c2) and we can keep removing square factors until we reach K = ( d) with d Q √ Q squarefree. Note that d 6= 1 as Q( 1) isn’t a quadratic√ field. √ Now we address the question of uniqueness. If Q( d1) = Q( d2) with d1 and d2 squarefree then 2 d1/d2 = s where s is a rational number. It’s now clear that d1 and d2 must have the same sign. 2 2 Putting s = u/v where u and v are integers we get that d1v = d2u . If p is a prime dividing d1 then 2 2 p occurs to an odd power in the prime of d1v and so of d2u . But this means that p | d2 also. Similarly every prime factor of d2 is also a prime factor of d1. As d1 and d2 are squarefree, have the same prime factors and the same sign,√ they are equal. 2 Hence from now on when we write Q( d) we tacitly assume that d is a squarefree integer.

4 Rings of Integers √ To do number theory we need the idea of integer. We have been considering quadratic fields Q( d) as generalizations of the rationals Q. What should we consider instead of the integers Z? Our first section gives us the answer. As the intersection of the set of algebraic integers and the set of rationals is the set of integers, we shall try to identify which elements of each quadratic field are algebraic integers. The norm and trace provide a useful characterization. √ Proposition 4 If α ∈ Q( d) then α is an algebraic integer if and only if N(α) and T (α) are (ordinary) integers.

Proof Suppose first that N(α) and T (α) are integers. Then

α2 − T (α)α + N(α) = 0 and so α is an algebraic integer. Now suppose that α is an algebraic integer. Then there exist integers bj with

n n−1 α + b1α + ··· + bn−1α + bn = 0.

5 Taking conjugates gives ∗n ∗n−1 ∗ α + b1α + ··· + bn−1α + bn = 0 which means that α∗ is also an algebraic integer. As sums and products of algebraic integers are algebraic integers, then T (α) = α + α∗ and N(α) = αα∗ are algebraic integers. But T (α) and N(α) are rational numbers, and so T (α) and N(α) must be integers by Proposition 1. 2 In a given quadratic field we can now identify the algebraic integers.√ If d is a squarefree integer other than 1 we√ denote by Rd the set of all algebraic integers in Q( d). We call Rd the of ∗ integers of Q( d). Incidentally note that the above Proposition shows that if α ∈ Rd then α ∈ Rd. Theorem 2 If d 6≡ 1 (mod 4) then √ Rd = {a + b d : a, b ∈ Z} while if d ≡ 1 (mod 4) then √ (r + s d ) R = : r, s ∈ , r ≡ s (mod 2) . d 2 Z √ 2 2 Proof If α = a + b d then T (α)√ = 2a ∈ Z and N(α) = a − b d ∈ Z and α is an algebraic integer. 1 1 2 2 If d ≡ 1 (mod 4) and α = 2 (r + s d), then T (α) = r ∈ Z and N(α) = 4 (r − s d). If r and s are both even then r2 ≡ s2 ≡ 0 (mod 4), while if r and s are both odd then r2 ≡ s2 ≡ 1 (mod 4). In 2 2 any case then r − s d ≡ 0 (mod 4) and so N(α) ∈ Z and α is an√ algebraic integer. Conversely suppose that α ∈ Rd. We can write α = a + b d with a and b rational. Now T (α) = 2a is an integer, so we’ll split the argument into two cases: 2a even or 2a odd. If 2a is even, then a is an integer. As N(α) = a2 − b2d is an integer it follows that so is b2d. Now b is rational and d is squarefree. I claim that this means that b must be an integer. Put b = u/v in lowest terms with u and v integers. Then du2/v2 is an integer. But if p is a prime factor of v we must have p2 | du2, and p - u. This means that p2 | d which is squarefree—a contradiction! So v = ±1 and b is an integer. We now turn to the case where 2a is odd. Put r = 2a and s = 2b. The number r2−ds2 = 4(a2−b2d) is an integer which is divisible by 4. Hence ds2 is an integer, and repeating the above argument we see that s is an integer also. As r2 − ds2 is even then s can’t be even. As r and s are both odd then 2 2 2 2 r ≡ s ≡ 1 (mod 4) and so 0 ≡ r − ds ≡ 1 − d (mod 4), i.e.,√ d ≡ 1 (mod 4). 2 √ 1 We can rephrase this result as follows. If we define τd = d if d 6≡ 1 (mod 4) and τd = 2 (1 + d) if d ≡ 1 (mod 4) then R = {a + bτ : a, b ∈ }. d d Z √ 1 It may seem odd at first to be considering numbers like 2 (1 + 5) as elements of Rd as these appear to have “denominators”, but to exclude them would make the theory a lot harder—the crucial Corollary 3 would be false if we excluded these numbers. (In future I’ll usually write τ instead of τd when it’s clear what d is.) One of the most important things about Rd is that it’s a ring, i.e., it’s closed under addition, subtraction and multiplication. Proposition 5 If α, β ∈ R , then α + β, α − β and αβ ∈ R . d d √ Proof Recall that α ∈ Rd if and only if α is an algebraic integer and α ∈ Q( d). Now the√ algebraic integers are closed under addition, subtraction and multiplication (Theorem 1), and Q( d) is also closed under these operations. The result is now immediate. Alternatively you can use the description

Rd = {a + bτ : a, b ∈ Z} to give a direct proof, putting α = a + bτ, β = a0 + b0τ and computing α ± β and αβ explicitly. (Details left as an exercise!) 2

6 5 Divisibility and Factorization

We now have the basic materials we need to start to do number theory in quadratic fields. Classical number theory deals with divisibility, congruences and primes, all in the integers. Can we define similar notions inside Rd? Divisibility and congruences are easy: if α, β ∈ Rd with α 6= 0 then we say that α divides β (and write α | β) if β/α ∈ Rd. Similarly we say that α is congruent to β modulo γ (and write α ≡ β (mod γ)) if γ | (α − β). All the familiar properties of divisibility and congruences are true in Rd also, so I won’t repeat them. Note though that if α | β then N(α) | N(β).√ √ Some new things do happen in the various Rd. Consider ε = 1 + 2 ∈ R2. As 1/ε = −1 + 2 it follows that α/ε ∈ R2 for all α ∈ R2, and so ε divides every element of R2. An element of Rd with this property is called a ; formally we define a unit to be an element ε ∈ Rd whose inverse 1/ε is also an element of Rd. We denote the set of units of Rd by Ud. The existence of units complicates slightly the question of how to generalize the notion of prime. We digress slightly to study the theory of units. There’s a simple characterization of units.

Proposition 6 An element ε of Rd is a unit if and only if N(ε) = ±1.

∗ ∗ Proof Suppose first that N(ε) = ±1. This means that εε = ±1 and so 1/ε = ±ε ∈ Rd, and ε is a unit. Conversely suppose that ε ∈ Ud. Then 1/ε ∈ Rd and so

1 = N(1) = N(ε(1/ε)) = N(ε)N(1/ε).

But as N(ε) and N(1/ε) are integers this implies that N(ε) = ±1. 2 Another important fact is that Ud is a group under multiplication.

Lemma 5 If ε and η ∈ Ud, then εη ∈ Ud and 1/ε ∈ Ud. Also 1 ∈ Ud. Proof An easy exercise. 2 We now determine the size of Ud. This result shows a radical difference between the of real and imaginary quadratic fields.

Theorem 3 (a) If d < 0 then Ud = {±1} unless d = −1 when U−1 = {±1, ±i}, or d = −3 when 1 √ U−3 = {±1, 2 (±1 ± −3)}. (b) If d > 0 then Ud is an infinite group.

Proof (a) It’s easy to check that the numbers claimed√ to be units really are units. Put D = −d > 0. 1 Let ε ∈ Ud. We can certainly write ε = 2 (r + s d) with r and s integers. As N(ε) = 1 (remember that as d < 0, N(ε) ≥ 0) we get 4 = r2 − s2d = r2 + s2D. 2 2 As s D ≥ 0 we must have |r| ≤ 2. If r =√±2 then s = 0 and ε = ±1. If r = ±1 then s D = 3, and 1 2 so D = 3 and s = ±1; hence ε = 2 (±1 ± −3). If r = 0 then s D = 4, and as D is squarefree D = 1 and s = ±2; hence ε = ±i. 2 2 (b) By the theory of Pell’s equation,√ there is a solution of the equation x − dy = 1 with x and 2 3 n y positive integers. If ε = x + y d then ε ∈ Ud and ε > 1. Now ε , ε , . . . , ε ,... are all units and 2 3 n ε < ε < ε < ··· < ε < ··· so they are all distinct. Hence Ud forms an infinite group. 2 In the language of group theory we see easily that for negative d, Ud is a cyclic group (a generator 1 √ is −1, i or 2 (1 + −3) as appropriate). What is the structure of Ud when d > 0?

7 Theorem 4 If d > 0 there exists a unique unit ε ∈ Ud such that ε > 1 and every ξ ∈ Ud has the form ±εn for some n ∈ Z. Proof We try to take ε to be the smallest unit that’s bigger than 1. But how do we know that this description makes sense? By Theorem 3 there is a unit bigger than 1, but maybe the set of all such units doesn’t have a least element? (Cf. the set of all positive rationals.) Let η be any unit bigger than 1, and let

U = {ξ ∈ Ud : 1 < ξ ≤ η}.

If there is a least unit bigger than 1, it must be the least element of U. I claim that U is a finite non-empty set. It’s non-empty as η ∈ U. If ξ ∈ U then ξξ∗ = ±1 and so |ξ∗| = 1/ξ < 1. As T (β) = ξ + ξ∗ = ξ ± |ξ∗| it follows that 0 < T (ξ) < η + 1. As ξ2 − T (ξ)ξ + N(ξ) = 0 and N(ξ) = ±1 then q √ T (ξ) ± T (ξ)2 − 4N(ξ) t ± t2 ∓ 4 ξ = = 2 2 where t is an integer between 0 and η + 1. Hence ξ is confined to a finite set, and so the set U is finite. As U is finite and non-empty it has a least element ε; ε is a unit, ε > 1 and there are no other units between 1 and ε. Now consider any ξ ∈ Ud. First assume that ξ > 0. Consider the (log ξ)/(log ε). There is an integer n with log ξ n ≤ < n + 1, log ε and so log ξ log(ξε−n) 0 ≤ − n = < 1 log ε log ε which implies that 1 ≤ ξε−n < ε. Now ξε−n is a unit and so can’t be bigger than 1 and less than ε. Hence ξε−n = 1 and ξ = εn. If ξ < 0 then −ξ is a positive unit and so −ξ = εn for some integer n, and ξ = −εn. I claim that ε is uniquely determined, for if ε0 also satisfied the same conditions then ε0 = εn and ε = ε0m for some integers n and m and so ε = εmn meaning that mn = 1. Hence m = n = ±1 and we must have n = +1 as ε0 = εn > 1. 2 The unit ε in the above theorem is called the fundamental unit of Rd. In terms of group theory the theorem states that the group Ud is isomorphic to the direct product Z × Z2. If d 6≡ 1 (mod 4) the fundamental unit can be found by applying the continued fraction algorithm for Pell’s equation. If d ≡ 1 (mod 4) this algorithm√ may or may√ not find the fundamental unit, but a variant method 1 where you start with 2 (1 + d) instead of d always does work. The above proof illustrates a useful technique. If α ∈ Rd and we know N(α) and have inequalities bounding the size of α, then we can bound the size of T (α) and can restrict α to lie in a explicitly specified finite set. We now try to generalize the notion of prime. As every element of Rd is divisible by every unit, we make the following definition. We say that α ∈ Rd is irreducible if it’s non-zero, not a unit, and whenever α = βγ with β, γ ∈ Rd then either β or γ is a unit. (We don’t call such elements primes, the reason being is that we reserve this term for elements satisfying a more stringent condition which we’ll meet later.) It’s easy to show (by induction on |N(α)|) that every α ∈ Rd is either 0 or a unit or irreducible or a product of irreducibles. It’s now natural to ask how unique this factorization into irreducibles is. For instance in R−1 we have 14 − 5i = (4 + i)(3 − 2i) = (1 − 4i)(2 + 3i), and all of these factors are irreducible. But these two are really the same in disguise. For i is a

8 unit and 4+i = i(1−4i) and (3−2i) = −i(2+3i). We can go to and fro between these factorizations by juggling with units. We make the definition that non-zero elements α and β ∈ Rd are associates if α/β is a unit. An easy exercise shows that being associates is an equivalence relation. We now say that two factorizations

α = π1π2 ··· πm = ρ1ρ2 ··· ρn of α into irreducibles are equivalent if m = n and we can re- the πjs so that πj is an associate of ρj for each j. We might hope that factorization into irreducibles is unique up to equivalence. Alas this is not so. In R−5 consider √ √ 6 = 2 · 3 = (1 + −5)(1 − −5).

These are two non-equivalent factorizations into irreducibles. We thus say that Rd has unique fac- torization if every factorization into irreducibles is unique up to equivalence. It follows that R−5 doesn’t have unique factorization, but we shall see that there do exist some Rd which do have unique factorization. (NB to show that some Rd has unique factorization one needs a proof not just an example!) If we go back to standard number theory and study the proof of unique factorization of natural numbers, we find that the proof ultimately depends on the . If we could find a generalized version of this algorithm inside a particular Rd then perhaps we could show that this Rd has unique factorization. This is indeed so. We say that Rd is Euclidean if whenever α, β ∈ Rd and β 6= 0, then there exists√ a λ ∈ Rd with |N(α − βλ)| < |N(β)|. Equivalently Rd is Euclidean if and only if for all ξ ∈ Q( d) there exists γ ∈ Rd with |N(ξ − γ)| < 1. (To see this put ξ = α/β.) Several examples of Euclidean Rd are known, including R−1, R−2, R−3, R−7, R−11, R2, R3, R5 and R13.I shall prove some of these in lectures, and set others as exercises. By means of the classical Euclidean algorithm one can find the greatest common divisor g of two numbers a and b, and write g = ra + sb. We can also do this in Rd provided that Rd is Euclidean. I’ll give an “abstract” proof of this result, but it can also be proved algorithmically, and I’ll illustrate this in lectures.

Proposition 7 Let Rd be Euclidean. If α and β are non-zero elements of Rd then there exists γ ∈ Rd with (i) γ | α and γ | β, and (ii) if δ | α and δ | β then δ | γ. Also there exist η, ξ ∈ Rd with γ = ηα + ξβ. Proof Let I be the set {ηα + ξβ : η, ξ ∈ Rd}. We look for a non-zero element γ in this set as “small” as possible in an appropriate sense. It’s easy to check that I contains α and β, it’s closed under addition and subtraction and if δ ∈ I and ζ ∈ Rd then ζδ ∈ I. If γ is a non-zero element of I, then |N(γ)| is a positive integer, so we can choose such a γ with |N(γ)| as small as possible. Certainly there exist η, ξ ∈ Rd with γ = ηα + ξβ. If δ | α and δ | β then δ | ηα and δ | ξβ and so δ | γ and (ii) is verified. It only remains to prove (i). By the Euclidean property there exists λ ∈ Rd with |N(α − λγ)| < |N(γ)|. Now α ∈ I and λγ ∈ I so α − λγ ∈ I. But as γ was chosen to make |N(γ)| as small as possible for non-zero γ ∈ I then α − λγ = 0 and γ | α. A similar argument shows γ | β. 2 If γ has the properties described in the proof we call γ a greatest common divisor of α and β. Greatest common divisors are only unique up to multiplication by units. Given α and β we can find

9 γ by a generalization of the standard Euclidean algorithm, and also η and ξ by a generalization of the extended Euclidean algorithm. To prove unique factorization for the ordinary integers we need the result that if p is prime and p | ab then either p | a or p | b. The corresponding result is true for irreducibles in Euclidean Rd.

Proposition 8 Suppose that Rd is Euclidean and π is irreducible in Rd. If π | αβ with α, β ∈ Rd then either π | α or π | β.

Proof Suppose π - α, and consider a greatest common divisor γ of α and π. As π is irreducible and γ | π then either γ is a unit or an associate of π. But if γ were an associate of π we’d have π | γ | α and so π | α. Hence γ is a unit and we may assume that γ = 1. There exist η and ξ ∈ Rd with 1 = ηα + ξπ and so 1 ≡ ηα (mod π). But then β ≡ ηαβ ≡ 0 (mod π) and π | β as required. 2 Because of this result we define π ∈ Rd to be prime if π isn’t zero or a unit, and if π | αβ implies that π | α or π | β. This Proposition now states that if Rd is Euclidean, then every irreducible is prime. This is not necessarily√ true if R√d isn’t Euclidean.√ For instance 2 is irreducible but not prime in R−5. (Note that 2 | (1 + −5)(1 − −5) but 2 - (1 ± −5).) On the other hand it’s not hard to show that every prime is irreducible.

Lemma 6 If π is prime in Rd then it’s irreducible. Proof Suppose that π = αβ. Then either π | α or π | β. Let’s suppose it’s the former. Then π | α and α | π so α is an associate of β and β is a unit. This means that π is irreducible. 2 An easy induction argument shows that if π is prime and π | α1α2 ··· αn then π | αj for some j. If in Rd all irreducibles are primes then we have unique factorization. The proof follows the lines of the unique factorization proof for the integers.

Theorem 5 If every irreducible element of Rd is prime then Rd has unique factorization.

Proof Suppose that the non-zero element α of Rd has two factorizations

α = π1π2 ··· πr = ρ1ρ2 ··· ρs into irreducibles. If r = 1 then α is irreducible and we must have s = 1 and ρ1 = α. We may suppose that r > 1. As π1 is prime then π1 | ρ1ρ2 ··· ρs and so π1 | ρj for some j. We can re-arrange the ρjs so that π1 | ρ1. As ρ1 is irreducible then π1 must be an associate of ρ1. Put ρ1 = επ1 where ε is a unit. Dividing through by π1 we get

π2π3 ··· πr = ερ2ρ3 ··· ρs.

We can repeat the argument; π2 must divide one of the terms in the second product, and as it cannot divide the unit ε, then we may assume it divides ρ2 and so is an associate of ρ2. We divide through and keep going. Eventually we re-order the ρjs so that πj is an associate of ρj for all j, and r = s. 2

6 Ideal Theory

We can remedy the lack of unique factorization in the general Rd by considering factorizations of ideals rather than factorizations of elements. We’ve already seen an application of this idea in the proof of Proposition 7. The set I defined therein has the properties: (i) 0 ∈ I,

10 (ii) if α, β ∈ I then α + β ∈ I, (iii) if α ∈ I and λ ∈ Rd then λα ∈ I. We define an ideal of Rd to be a I of Rd satisfying the above properties. If I is an ideal and α ∈ I then −α = (−1)α ∈ I by (iii) and so (I, +) is a subgroup of (Rd, +). Obvious examples of ideals include both {0} and Rd. Sometimes we want to exclude these from consideration and so we say that an ideal I is proper if I 6= Rd and non-zero if I 6= {0}. Another basic class of ideals are the principal ideals; if α ∈ Rd the set

hαi = {αβ : β ∈ Rd} = {γ : α | γ} is an ideal, the generated by α. It’s easy to see that hαi = hβi if and only if α and β are associates; in particular hαi = Rd if and only if α is a unit. If Rd is Euclidean then all its ideals are principal.

Theorem 6 If Rd is Euclidean and I is an ideal of Rd then I is principal. Proof If I = {0} then I = h0i. If I 6= {0} then the proof follows the proof of Proposition 7 exactly. Choose a nonzero element γ of I making |N(γ)| as small as possible. I claim that I = hγi. As βγ ∈ I for all β ∈ Rd then hγi ⊆ I. For the reverse inclusion assume that α ∈ I. By the Euclidean property there exists λ ∈ Rd with |N(α − λγ)| < |N(γ)|. As I is an ideal λγ ∈ I and so α − λγ ∈ I. But as γ was chosen to make |N(γ)| as small as possible for non-zero γ ∈ I then α − λγ = 0 and γ | α. 2 We say Rd is principal if all its ideals are principal. The above theorem shows that every Euclidean Rd is principal. (The converse isn’t true; for instance R−19 is principal but not Euclidean.) If Rd is principal then it has unique factorization.

Theorem 7 If all ideals of Rd are principal, then Rd has unique factorization.

Proof By Theorem 5 it suffices to show that every irreducible in Rd is prime. Let π be irreducible, and suppose that π | αβ, but π - α. As in Proposition 7 the set

I = {ηπ + ξα : η, ξ ∈ Rd} is an ideal, and so I = hγi for some γ. As π ∈ I then γ | π and as π is irreducible then γ is either a unit or an associate of π. But if γ is an associate of π then π | γ | α as α ∈ I. This is contrary to hypothesis, so γ is a unit and I = hγi = Rd. Hence 1 ∈ I and so 1 = ηπ + ξα with η, ξ ∈ Rd. It follows that 1 ≡ ξα (mod π) and so β ≡ ξαβ ≡ 0 (mod π). This means that π | β and we’ve shown that π is prime. 2 (NB this proof is a rehash of that of Proposition 8.) The two basic operations on ideals are addition and multiplication. Addition is easy to define. If I and J are ideals of R then we define their sum to be

I + J = {α + β : α ∈ I, β ∈ J} which it’s easy to show is an ideal. The sum I + J is the smallest ideal containing both I and J. The sum has some familiar properties: I + J = J + I, I + (J + K) = (I + J) + K and I + h0i = I, and some unfamiliar ones: I + I = I and I + h1i = h1i. All of these make instructive exercises. The product is harder to define. We might try to define IJ to be

{αβ : α ∈ I, β ∈ J}

11 but this isn’t always an ideal. (There are examples where it’s not closed under addition.) Instead we consider the additive subgroup of Rd generated by the above set which is

IJ = {α1β1 + α2β2 + ··· + αnβn : α1, . . . , αn ∈ I, β1, . . . , βn ∈ J, n ∈ N} as the product of I and J. This is an ideal. The product IJ is a subset of both I and J. The product also has some familiar properties: IJ = JI, I(JK) = (IJ)K, I(J + K) = IJ + IK, Ih0i = h0i and Ih1i = I. Again these make good exercises! There’s also a conjugation operation on ideals. If I is an ideal then

I∗ = {α∗ : α ∈ I} is easily seen to be an ideal. Conjugation respects addition and multiplication: (I + J)∗ = I∗ + J ∗ and (IJ)∗ = I∗J ∗ (exercises!). Addition and multiplication preserve inclusions: if J ⊆ K then I + J ⊆ I + K and IJ ⊆ IK. We would like to be able to do arithmetic with ideals explicitly, as we do for elements of Rd. As it stands this is difficult as we don’t even have a way of specifying ideals (apart from principal ones). One case is easy though; multiplication by a principal ideal. If I is a principal ideal then multiplication by I is easy.

Lemma 7 Let α ∈ Rd and J be an ideal. Then

hαiJ = αJ = {αβ : β ∈ J}.

In particular if β ∈ Rd then hαihβi = hαβi.

Proof If β ∈ J then as α ∈ hαi we have αβ ∈ hαiJ and so αJ ⊆ hαiJ. Now the typical element of hαiJ has the form γ1β1 + γ2β2 + ··· + γnβn with each γj ∈ hαi. If we write γj = αλj we get

γ1β1 + γ2β2 + ··· + γnβn = α(λ1β1 + λ2β2 + ··· + λnβn) ∈ αJ as λ1β1 + ··· + λnβn ∈ J. Hence hαiJ ⊆ αJ and so these two ideals are equal. If J = hβi it’s easy to see that hαihβi = αhβi = hαβi. 2 We generalize our notation for principal ideals as follows. We shall denote the ideal

hα1i + hα2i + ··· + hαni by hα1, α2, . . . , αni. More explicitly

hα1, α2, . . . , αni = {β1α1 + β2α2 + ··· + βnαn : βj ∈ Rd}.

Using the associative and distributive laws we see that if I = hα1, . . . , αni and J = hβ1, . . . , βmi then

I + J = hα1, . . . , αn, β1, . . . , βmi and IJ = hα1β1, α1β2, . . . , α1βn, α2β1, . . . , αnβmi.

12 √ In general not all ideals are principal. For instance h2, 1 + −5i is a non-principal ideal of R−5. I’ll prove this and give other examples in the lectures. But although we can’t always write an ideal in the form hαi, can we always write ideals in the form hα1, α2, . . . , αni? The answer to this is yes. In fact every ideal of Rd can actually be written in the form hα, βi. To demonstrate this I’ll prove an even stronger result. Every ideal is a subgroup of (Rd, +) and it is convenient to first classify all possible subgroups of Rd and then to ask which of them are ideals. We introduce some notation. If α1, . . . , αn ∈ Rd then I denote by [ α1, . . . , αn ] the subgroup of Rd generated by the αjs, i.e.,

[ α1, . . . , αn ] = {b1α1 + ··· + bnαn : b1, . . . , bn ∈ Z}. In fact we only need two generators.

Proposition 9 Let G be a subgroup of Rd. There exist integers a, m and n such that m, n ≥ 0 and G = [ a + mτ, n ].

Proof A typical element of G has the form r + sτ where r and s are integers. We consider first the set H of all the s that occur, i.e., H = {s : r + sτ ∈ G}. It is easy to see that H is a subgroup of Z. Now every subgroup of Z has the form mZ for some m ≥ 0. As m ∈ H there’s an a with a + mτ ∈ G. Also G ∩ Z is a subgroup of Z and so G ∩ Z = nZ for some n ≥ 0. Certainly n ∈ G. I claim that G = [ a + mτ, n ]. It’s clear that [ a + mτ, n ] ⊆ G. For the reverse inclusion take r + sτ ∈ G. As s ∈ H there exists u ∈ Z with s = um. Now consider r − ua = r + sτ − u(a + mτ). As this is an element of G ∩ Z then r − ua = vn where v ∈ Z. Hence r + sτ = r − ua + u(a + mτ) = vn + u(a + mτ) ∈ [ a + mτ, n ].

It follows that G = [ a + mτ, n ] as required. 2

Corollary 1 If I is a non-zero ideal of Rd, then there exist integers a, m and n with m > 0 and n > 0 such that I = [ a + mτ, n ].

Proof Let G = I in the above proof. Choose α to be any non-zero element of I. Note that N(α) = αα∗ ∈ I and so N(α)τ ∈ I. It follows that the sets H and I ∩ Z of the above proof both contain N(α) and so are non-zero. This means we can take m > 0 and n > 0. 2 Given a subgroup G we can express it in the above form by the following algorithm. Suppose that G = [ b1 + c1τ, b2 + c2τ, . . . , bk + ckτ ].

Let m = gcd(c1, c2, . . . , ck). Then we can find integers r1, r2, . . . , rk with m = r1c1 + ··· + rkck. Now

r1(b1 + c1τ) + ··· + rk(bk + ckτ) = a + mτ say. Also each cj = sjm where sj is an integer. Put tj = (bj + cjτ) − sj(a + mτ) = bj − sja. Then n = gcd(t1, . . . , tk). Representing G as [ a + mτ, n ] is very convenient for computation for we can easily test whether a particular element α is an element of G. For if α = b + cτ, then α ∈ G if and only if there are integers r and s with b + cτ = r(a + mτ) + sn.

13 It follows that r must equal c/m and s must equal (b − ra)/n, i.e., α ∈ G if and only if these are both integers. Similarly we can test if two subgroups are equal for it is easy to show (exercise!) that provided m, m0, n, n0 > 0 then [ a + mτ, n ] = [ a0 + m0τ, n0 ] if and only if m0 = m, n0 = n and a0 ≡ a (mod n). Another consequence is that every ideal can be generated by two elements.

Corollary 2 If I is an ideal in Rd then I = hα, βi for some α, β.

Proof Certainly I = [ m, a + nτ ] where a, m and n are integers. Put α = a + mτ and β = n. Then α, β ∈ I and so hα, βi ⊆ I. Conversely if γ ∈ I then γ = bα + cβ with b and c integers, and so γ ∈ hα, βi. Hence I ⊆ hα, βi as required. 2 Given any ideal I = hα1, α2, . . . , αki then it’s easy to show (exercise!) that

I = [ α1, τα1, α2, τα2, . . . , αk, ταk ], and by applying the above algorithm we can write I√in the form [ a + mτ, n ]. √ As an example of this procedure let I = h2, 1 + −5i be an ideal of R−5. Then as τ = −5 we have √ √ √ √ I = [ 2, 1 + −5, 2 −5, −5(1 + −5) ] √ √ √ = [ 2, 1 + −5, 2 −5, −5 + −5 ]. √ √ As gcd(0√ , 1, 2, 1) =√ 1 then m =√ 1 and we can√ take a + mτ = 1 + −√5. Now 2 − 0(1√ + −5) = 2, (1 + −5) − 1(1 + −5) = 0, 2 −5 − 2(1 + −√5) = −2, and (−5 + −5) − 1(1 + −5) = −6, and so n = gcd(2, 0, −2, −6) =√ 2. Thus I = [ 2, 1 + −5 ]. We can show that this is a proper ideal, as if 1 ∈ I then 1 = r2 + s(1 + −5) (r, s ∈ Z) and so s = 0 and r = 1/2 which is impossible. Our next goal is to vindicate the study of ideals by proving that every ideal can be uniquely expressed as the product of prime ideals (we’ll find out what a prime ideal is later). We start by defining a notion of “size” of an ideal. When considering elements of Rd, |N(α)|, the absolute value of the norm of the element α gives an idea of the “size” of α—if |N(α)| = 0 then α = 0, if |N(α)| = 1 then α is a unit etc. As the norm is defined by N(α) = αα∗ we might consider the product II∗ for an ideal I. The following result is the cornerstone of our theory of ideals.

∗ Theorem 8 (Hurwitz’ Lemma) Suppose that α, β ∈ Rd and g ∈ N. If N(α), N(β) and T (αβ ) ∗ ∗ are all divisible by g then g | αβ and g | α β in Rd.

∗ ∗ ∗ ∗ Proof Let γ = αβ /g, so that γ = α β/g. We want to show that γ ∈ Rd for then also γ ∈ Rd. We use the criterion of Proposition 4, namely that γ ∈ Rd if and only if both T (γ) ∈ Z and N(γ) ∈ Z. Now αβ∗ + α∗β T (αβ∗) T (γ) = γ + γ∗ = = ∈ g g Z and αβ∗α∗β αα∗ ββ∗ N(α) N(β) N(γ) = γγ∗ = = = ∈ . g2 g g g g Z ∗ Hence γ ∈ Rd and also γ ∈ Rd. 2

∗ Corollary 3 If I is an ideal of Rd then II = hNi for some integer N ≥ 0.

14 Proof We know that I = hα, βi for some α and β and so

II∗ = hα, βihα∗, β∗i = hαα∗, αβ∗, βα∗, ββ∗i.

Hence the rational integers αα∗ = N(α), ββ∗ = N(β) and αβ∗ + α∗β = T (αβ∗) all lie in II∗. Let N be the greatest common divisor of these three integers. Clearly N ∈ II∗ and so hNi ⊆ II∗. As N | N(α), N | N(β) and N | T (αβ∗) then by Hurwitz’ Lemma N | αβ∗ and N | βα∗. Hence II∗ ⊆ hNi as required. 2 The integer N above is uniquely defined and is called the norm of the ideal I. We denote the norm of I by N(I). The above result gives us the formula

N(hα, βi) = gcd(N(α),N(β),T (αβ∗)).

If I = hαi is principal then II∗ = hαα∗i and so N(hαi) = |N(α)|. More generally if α ∈ I then α∗ ∈ I∗ and N(α) = αα∗ ∈ II∗ = hN(I)i, and so N(I) | N(α). Also the ideal norm is multiplicative as hN(IJ)i = (IJ)(IJ)∗ = IJI∗J ∗ = (II∗)(JJ ∗) = hN(I)ihN(J)i = hN(I)N(J)i and so N(IJ) = N(I)N(J). If N(I) = 0 then N(α) = αα∗ = 0 for all α ∈ I and so I = {0}; if ∗ ∗ N(I) = 1, then II = h1i = Rd and so I ⊇ II = Rd which means that I = Rd. Hence non-zero ideals have non-zero norm, and non-zero proper ideals have norm bigger than 1. This observation will be very useful! We can link the definition of norm with group theory by means of the following result.

Proposition 10 Let I = [ n, a + mτ ] be a non-zero ideal of Rd, with m, n ∈ N and a ∈ Z. Then N(I) = mn.

Proof For convenience let α = a + mτ. Now I = [ n, α ] and so I∗ = [ n, α∗ ]. Let N = N(I) so that II∗ = hNi. As nα = an + mnτ ∈ I∗I = hNi it follows that N | an and N | mn. To show that N = mn it suffices to show that mn | N. I claim that every element of II∗ has the form A + Bmnτ where A and B are integers. As every element of II∗ is a sum of elements of the form βγ where β ∈ I and γ ∈ I∗ it suffices to show that all such elements have this form. Now β = rn + sα and γ = un + vα∗ so

βγ = run2 + rvnα∗ + sunα + svαα∗.

Hence it suffices to show that nα and nα∗ have the form A + Bmnτ. But nα = na + mnτ and nα∗ = T (nα) − nα = T (nα) − na − mnτ, and the claim is established. Now Nτ ∈ II∗ and so mn | N as required, concluding the proof. 2 Recall from group theory the definition of the index [ G : H ] of a subgroup H of a group G as the number of left (or right) cosets of H in G. It is not hard to show that the index of I = [ a + mτ, n ] in Rd is mn (the cosets are (r + sτ) + I where 0 ≤ r < n and 0 ≤ s < m). Hence we get the formula, valid for non-zero ideals I, N(I) = [ Rd : I ] which is taken by many books as the definition of norm. The fact that II∗ is always principal allows us to “cancel” by a non-zero ideal.

Proposition 11 Let I, J and K be ideals of Rd with I non-zero and IJ = IK. Then J = K.

15 Proof Suppose first that I = hαi is principal. Then IJ = αJ and so J = α−1(IJ) = {α−1β : β ∈ IJ}. Similarly K = α−1(IK) = α−1(IJ) = J. Now suppose that I is any non-zero ideal. From IJ = IK it follows that

(II∗)J = (IJ)I∗ = (IK)I∗ = (II∗)K and as II∗ is principal and nonzero, then by the previous paragraph it follows that J = K. 2 We next investigate divisibility of ideals. We say that I divides J (and write I | J) if there is an ideal K with J = IK. If I | J then I ⊇ J as we can write J = IK ⊆ I. A crucial result is that the converse holds.

Theorem 9 If I and J are non-zero ideals of Rd then I | J if and only if I ⊇ J. Proof We have already noted that if I | J then I ⊇ J. Conversely suppose that I ⊇ J. Now JI∗ ⊆ II∗ = hN(I)i. Then 1 K = JI∗ = {N(I)−1α : α ∈ JI∗} ⊆ R N(I) d and it is easy to see that K is an ideal. Now 1 1 1 IK = I(JI∗) = J(II∗) = JhN(I)i = J N(I) N(I) N(I) and so I | J, as required. 2 This theorem is summarized by the maxim to contain is to divide. Learn it by heart! Naturally we say that an ideal I is irreducible if I is non-zero, I 6= h1i and if J | I then J = I or J = h1i. If I is a non-zero proper ideal that isn’t irreducible, then we can write I as a product of two ideals of smaller norm. By repeating the process if necessary we can write I as a product of irreducible ideals. (If this argument doesn’t convince you, write out a formal proof.) Each irreducible ideal I is also maximal. This means that the only ideals containing I are I itself and Rd. This follows as to contain is to divide. More importantly irreducible ideals are also prime i.e., if I is irreducible and I | JK then either I | J or I | K.

Theorem 10 Let I, J and K be ideals of Rd with I irreducible. If I | JK then either I | J or I | K.

Proof Suppose that I - J. As to contain is to divide then I + J. The ideal I + J contains I, and so divides I but cannot equal I for then I = I + J ⊇ J. Hence as I is irreducible we have I + J = Rd, and so 1 = α + β for some α ∈ I and β ∈ J. If γ ∈ K then γ = (α + β)γ = αγ + βγ. Now αγ ∈ I as α ∈ I; also βγ ∈ JK but as I | JK then I ⊇ JK and βγ ∈ I. It follows that γ ∈ I and I ⊇ K, or equivalently I | K. 2 An obvious extension of this result is that if I is irreducible and I | J1J2 ··· Jk then I | Jj for some j. From now on we use the term prime ideal as a synonym for irreducible ideal. We can now prove the most important result of the theory.

Theorem 11 (Unique Factorization) If I is a non-zero proper ideal of Rd and

I = P1P2 ··· Pr = Q1Q2 ··· Qs are two factorizations of A into prime ideals, then r = s and we can re-order the Qjs so that Pj = Qj for all j.

16 Proof Use induction on r. If r = 1, I is irreducible and so s = 1 and Q1 = I = P1. Assume that r > 1. As P1 | Q1Q2 ··· Qs then P1 | Qj for some j. Relabel the Qjs so that j = 1, so now P1 | Q1. As Q1 is prime then P1 = Q1. We cancel P1 to get

P2P3 ··· Pr = Q2Q3 ··· Qs and using the inductive hypothesis now gives the result. 2 Note that this proof is similar to, but easier than, the unique factorization theorem for elements when Rd is principal, as we don’t have to worry about units. To understand factorization into prime ideals we need to find out what the prime ideals are. The next result is the first step in this direction. It shows a connection between prime ideals and (ordinary) prime numbers.

Proposition 12 If P is a prime ideal of Rd then P | hpi for some p. Proof Certainly P | PP ∗ = hNi where N = N(P ) > 1 is P as non-zero and proper. Write N = p1p2 ··· pk with each pj an (ordinary) prime number. Then P | hp1ihp2i · · · hpki, but as P is prime then P | hpji for some j and this establishes the proposition. 2 This proposition means that to find all prime ideals all we need is to factorize all ideals of the form hpi, where p is a prime number, into prime ideals. Now the ideal hpi has norm p2, and so any proper ideal dividing it has norm p or p2. This means that either hpi is itself prime, or hpi is the product of two prime ideals each of norm p. If hpi is already prime we say that p is inert in Rd, if hpi is the product of two equal prime ideals of norm p we say that p is ramified in Rd, and if hpi is the product of two distinct prime ideals we say that p is split in Rd. All three possibilities do occur: 2 in R−1, 3 is inert, 2 is ramified (h2i = h1 + ii ) and 5 is split (h5i = h2 + iih2 − ii). To find out whether a given prime p is inert, ramified or split it suffices to count the number of ideals of norm p. For if N(P ) = p then PP ∗ = hpi and P | hpi. If p is inert there are no ideals of norm p. If p ramifies there is exactly one ideal of norm p, and if p splits there are two ideals of norm p. To classify ideals of norm p we employ Proposition 10.

Proposition 13 Let p be a prime number, and let P = [ a + mτ, n ] where a, m and n are integers with m > 0 and n > 0. Then P is an ideal of norm p if and only if m = 1, n = p and p | N(a + τ).

Proof Suppose that P is an ideal of norm p. By Proposition 10 mn = p and so either m = 1 and n = p, or m = p and n = 1. But in the latter case 1 ∈ P so P = Rd which does not have norm p. Hence m = 1 and n = p. Let α = a + τ ∈ P . Now N(α) = αα∗ ∈ PP ∗ = hpi so p | N(α). Now suppose that P = [ a + τ, p ] with p | N(a + τ). If P is an ideal it will have norm p, and so it suffices to show that P is an ideal. As P is a subgroup of Rd all we need to show is that βγ ∈ P whenever β ∈ P and γ ∈ Rd. Now β = rp + s(a + τ) and γ = u + vτ with r, s, u and v integers, and so βγ = rup + rvpτ + su(a + τ) + svτ(a + τ). As p ∈ P and a + τ ∈ P it suffices to show that pτ ∈ P and τ(a + τ) ∈ P . But

pτ = p(a + τ) − ap ∈ P and

τ(a + τ) = (T (τ) − τ ∗)(a + τ) = (a + T (τ) − a − τ ∗)(a + τ) = (a + T (τ))(a + τ) − N(a + τ) ∈ P as p | N(a + τ). This completes the proof. 2

17 We can now find all ideals of prime norm p. It’s easy to see that [ a + τ, p ] = [ b + τ, p ] if and only if a ≡ b (mod p). Hence the number of ideals of norm p is the number of distinct solutions modulo p of the√ congruence N(a+τ) ≡ 0 (mod p). What does this congruence√ look like? If d 6≡ 1 (mod 4) then 2 1 2 1 τ = d and N(a+τ) = a −d; if d ≡ 1 (mod 4) then τ = 2 (1+ d) and N(a+τ) = a +a− 4 (d−1). Hence this congruence is quadratic, and we can use the theory of quadratic residues to count the number of solutions.

Proposition 14 (i) If p is an odd prime then p is inert, ramified or split in Rd according to whether  d  the Legendre symbol p = −1, 0 or 1. (ii) If p = 2 then p is ramified in Rd if d 6≡ 1 (mod 4), is inert in Rd if d ≡ 5 (mod 8), and split in Rd if d ≡ 1 (mod 8). √ Proof (i) Suppose first that d 6≡ 1 (mod 4) so that τ = d. Then the congruence is

N(a + τ) = a2 − d ≡ 0 (mod p)   which has zero, one or two solutions according to whether d = −1, 0 or 1. On the other hand if √ p 1 d ≡ 1 (mod 4) then τ = 2 (1 + d) and the congruence is and N(τ) = (1 − d)/4. d − 1 N(a + τ) = a2 + a − ≡ 0 (mod p), 4 or equivalently 4a2 + 4a + 1 − d ≡ 0 (mod p).

2  d  This reduces to b ≡ d (mod p) where b = 2a + 1. We thus have 1 + p solutions for b modulo p  d  giving rise to 1 + p solutions for a modulo p. (ii) Let p = 2. If d 6≡ 1 (mod 4) then the congruence is

a2 ≡ d (mod 2) which has the unique solution a ≡ 0 (mod 2) if d is even and the unique solution a ≡ 1 (mod 2) if d is odd. If d ≡ 1 (mod 4) then the congruence is 1 − d a2 + a + ≡ 0 (mod 2). 4 If d ≡ 5 (mod 8) then (1 − d)/4 is odd and this congruence is insoluble. If d ≡ 1 (mod 8) then (1 − d)/4 is even and this congruence has the solutions a ≡ 0 and a ≡ 1 (mod 2). 2 We have the easy corollary.

Corollary 4 In each Rd only a finite number of primes are ramified. Proof The only ramified primes are the odd prime factors of d, and possibly also 2. 2 In more advanced ramification theory becomes very important.

7 Ideal Classes and the Class Group

If every ideal of Rd is principal, then Rd has unique factorization. But more often than not Rd has non-principal ideals. It is easy to classify principal ideals, but can we do the same for non-principal ideals? The notion of equivalence of ideals gives us a handle on this problem.

18 √ ∗ We say that two non-zero ideals I and J of Rd are equivalent if I = λJ for some λ ∈ Q( d) . We write I ∼ J if I and J are equivalent. It is an easy exercise to show that ∼ is an equivalence relation: i.e., I ∼ I for all I, I ∼ J implies that J ∼ I, and I ∼ J ∼ K implies that I ∼ K. We write the equivalence class of I as [ I ], i.e., [ I ] = {J : I ∼ J} and we call [ I ] an ideal class of Rd. The ideal class [ Rd ] is the collection of all principal ideals. Hence Rd is principal if and only if Rd has exactly one ideal class. 0 0 0 0 Equivalence of ideals is respected√ by multiplication: if I ∼ I and J ∼ J then IJ ∼ I J , as I0 = αI and J 0 = βJ,(α, β ∈ Q( d)∗) and so I0J 0 = (αβ)IJ. This means that the ideal class [ IJ ] depends only on the ideal classes [ I ] and [ J ]. We can therefore define multiplication of ideal classes by [ I ][ J ] = [ IJ ]. This multiplication is obviously commutative and associative. As [ I ][ Rd ] = [ I ] ∗ and [ I ][ I ] = [ hN(I)i ] = [ Rd ] the set Cl(Rd) of ideal classes of Rd forms an —the identity being the class of principal ideals, and inversion being the operation of conjugation. We call Cl(Rd) the class group of Rd. The most important fact about Cl(Rd) is that it is a finite group. The size of this group shows how far away Rd is from having unique factorization. To prove that the class group is finite we need the following lemma, which shows that in a particular Rd the minimum size of an element of an ideal I isn’t much bigger than N(I).

Lemma 8 If d is a squarefree integer and d 6= 1 then there is a constant Cd, depending only on d, such that given any non-zero ideal I of Rd there is a non-zero element α of I with |N(α)| ≤ CdN(I). √ Proof Let N = N(I) and choose K to be the biggest integer with K ≤ N. Then K2 ≤ N < (K + 1)2. Write I = [ a + mτ, n ] with a, m, n ∈ Z and m > 0, n > 0. Then by Proposition 10 N = mn. Consider the set S = {r + sτ : r, s ∈ Z, 0 ≤ r ≤ K, 0 ≤ s ≤ K}. 2 2 This set has (K + 1) elements, and (K + 1) > mn. Define S0, S1,...,Sm−1 of S by

Sj = {r + sτ ∈ S : s ≡ j (mod m)}.

As S is the union of the Sjs at least one of these sets must have more than n elements. Suppose Sj is such a subset so Sj = {r1 + s1τ, r2 + s2τ, . . . , rk + skτ} where k > n. Put ti = ri − a(si − j)/m. Note that tj is an integer as si ≡ j (mod m). As k > n there must exist i1 < i2 with ti1 ≡ ti2 (mod n). Let α = ri1 + si1 τ and β = ri2 + si2 τ. Then α 6= β, α, β ∈ S and

α − β = ri1 − ri2 + (si1 − si2 )τ

= ri1 − ri2 + (si1 − j − si2 + j)τ a + mτ a = r − r + (s − j − s + j) − ((s − j) − (s − j)) i1 i2 i1 i2 m m i1 i2 t − t s − s = i1 i2 n + i1 i2 (a + mτ) ∈ I. n m We must estimate the size of γ = α − β. Now γ = u + vτ where |u|, |v| ≤ K, and so |N(u + vτ)| = |(u + vτ)(u + vτ ∗)| = |u2 + uv(τ + τ ∗) + v2ττ ∗| ≤ u2 + |uvT (τ)| + v2|N(τ)| 2 ≤ K (1 + |T (τ)| + |N(τ)|) ≤ CdN(I)

19 if we put Cd = 1 + |T (τ)| + |N(τ)|. 2 (One can accelerate the proof of the above Lemma by using the idea of the index of a subgroup—I may expand on this in lectures.) We can now show the finiteness of Cl(Rd).

Proposition 15 If Cd is as in the above Lemma, then every ideal class in Cl(Rd) contains an ideal of norm ≤ Cd. Proof Let I be an ideal, and consider its class [ I ]. By the Lemma the ideal I∗ contains a non-zero ∗ ∗ ∗ ∗ element α with |N(α)| ≤ CdN(I ). As hαi ⊆ I then I | hαi and we can write hαi = I J for an ∗ ∗ ideal J. As [ I ][ J ] = [ hαi ] = [ Rd ] it follows that the class [ J ] is the inverse of [ I ] which is the ∗ ∗ ∗ inverse of [ I ] and so [ I ] = [ J ]. Now |N(α)| = Nhαi = N(I )N(J), and so N(I )N(J) ≤ CdN(I ) and N(J) ≤ Cd as required. 2

Corollary 5 The group Cl(Rd) is finite.

Proof By the Proposition every ideal class in Rd has the form [ I ] where N(I) ≤ Cd. But for each particular norm N there are only a finite number of ideals of norm N, as such an ideal must be a product of prime ideals of norm p or p2 where p runs through the prime factors of N. As there are only finitely many such prime ideals, there are only finitely many ideals of norm N. Hence there are only finitely many ideals of norm ≤ Cd and so there are only finitely many ideal classes. 2 We call the order of Cl(Rd) the class number of Rd and denote it as h(d). To compute Cl(Rd) and so h(d) one finds all prime ideals of norm ≤ Cd, then uses these to find all ideals of norm ≤ Cd, and then decides which of these are equivalent. Once we have done this we have found all ideal classes in Rd. In practice one finds that there are usually a lot of ideals of norm ≤ Cd and this procedure takes a long time. However the following theorem, due to Minkowski, gives a lower value for the constant Cd and so enables us to calculate the class group faster.

Theorem 12 (Minkowski) Let d be a squarefree integer different from 1. Let D = d if d ≡ 1 (mod 4) and D = 4d if d 6≡ 1 (mod 4). Define √  D  if d > 0,  2 Md = q  2 |D|  if d < 0. π

Then every non-zero ideal I of Rd contains a non-zero element α with |N(α)| ≤ MdN(I).

Proof Omitted. See [1] or [3]. The proof isn’t hard but it involves a technique——which I haven’t developed. 2

References

[1] H. Cohn, Advanced Number Theory (a.k.a. A Second Course in Number Theory), Dover, 1962, 1980

[2] I.N. Stewart, Galois Theory, Chapman and Hall, 1973, 1989

[3] I.N. Stewart & D.O. Tall, Algebraic Number Theory, Chapman and Hall, 1979

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