Algebraic Number Theory summary of notes Robin Chapman 3 May 2000, revised 28 March 2004, corrected 4 January 2005 This is a summary of the 1999–2000 course on algebraic number theory. Proofs will generally be sketched rather than presented in detail. Also, examples will be very thin on the ground. I first learnt algebraic number theory from Stewart and Tall’s textbook Algebraic Number Theory (Chapman & Hall, 1979) (latest edition retitled Algebraic Number Theory and Fermat’s Last Theorem (A. K. Peters, 2002)) and these notes owe much to this book. I am indebted to Artur Costa Steiner for pointing out an error in an earlier version. 1 Algebraic numbers and integers We say that α ∈ C is an algebraic number if f(α) = 0 for some monic polynomial f ∈ Q[X]. We say that β ∈ C is an algebraic integer if g(α) = 0 for some monic polynomial g ∈ Z[X]. We let A and B denote the sets of algebraic numbers and algebraic integers respectively. Clearly B ⊆ A, Z ⊆ B and Q ⊆ A. Lemma 1.1 Let α ∈ A. Then there is β ∈ B and a nonzero m ∈ Z with α = β/m. Proof There is a monic polynomial f ∈ Q[X] with f(α) = 0. Let m be the product of the denominators of the coefficients of f. Then g = mf ∈ Z[X]. Write g = Pn a Xj. Then a = m. Now j=0 j n n X h(X) = mn−1g(X/m) = mn−1+ja Xj j j=0 1 is monic with integer coefficients (the only slightly problematical coefficient is that of Xn which equals m−1A = 1). Also h(mα) = mn−1g(α) = 0. m Hence β = mα ∈ B and α = β/m. (cid:3) Let α ∈ A. Then there is a monic polynomial f ∈ Q[X] of least degree such that f(α) = 0. This polynomial is uniquely determined. Proposition 1.1 Let α ∈ A. Then there is precisely one monic polynomial f ∈ Q[X] of minimum degree with f(α) = 0. This polynomial f has the property that if g ∈ Q[X] and g(α) = 0 then f | g. Proof Note first that if h ∈ Q[X] is a nonzero polynomial with deg(h) < deg(f), then h(α) 6= 0 since otherwise h = a−1h is a monic polynomial, 1 where a is the leading coefficient of h, with the property that deg(h ) < 1 deg(f) and h (α) = 0. That would contradict the definition of f. Now 1 f is unique, since if f had the same degree as f and also satisfied the 1 same conditions then h = f − f , if nonzero, has h ∈ Q[X], h(α) = 0 and 1 deg(h) < deg(g) which is impossible. Now let g ∈ Q[X] and suppose that g(α) = 0. By the division algorithm (Proposition A.1), g = qf +h where q, h ∈ Q[X], and either h = 0 (which means that f | g as we want) or h 6= 0 and deg(h) < deg(f). But as h(α) = g(α)−f(α)q(α) = 0 this is impossible. (cid:3) We call this f the minimum polynomial of α and call its degree the degree of α. Minimum polynomials are always irreducible. Lemma 1.2 Let f be the minimum polynomial of α ∈ A. Then f is irreducible over Q. Proof If f is not irreducible then f = gh where g, h ∈ Q[X] are monic polynomials of degree less than that of f. Then 0 = f(α) = g(α)h(α) and so either g(α) = 0 or h(α) = 0. We may assume g(α) = 0. Then f | g which is impossible since deg(g) < deg(f). (cid:3) Suppose the minimum polynomial f of α lies in Z[X]. Then, since f is monic and f(α) = 0, α is an algebraic integer. In fact the converse holds: if α ∈ B then its minimum polynomial lies in Z[X]. We need to study integer polynomials in more detail to prove this. A nonzero polynomial f ∈ Z[X] is primitive if the greatest common divisor of its coefficients is 1. Equivalently f is primitive if there is no prime number dividing all its coefficients. Lemma 1.3 (Gauss’s Lemma) Let f, g ∈ Z[X] be primitive polynomials. Then fg is also a primitive polynomial. 2 Proof Write f(X) = a +a X +a X2 +···+a Xm 0 1 2 m and g(X) = b +b X +b X2 +···+b Xn. 0 1 2 n We show that there is no prime p dividing all the coefficients of fg. Take a prime p. As f is primitive there is a coefficient of f not divisible by p; let a be the first such. Similarly let b be the first coefficient of g not divisible r s by p. Then p | a for i < r and p | b for j < s. The coefficient of Xr+s in fg i j is X c = a b . r+s i j i+j=r+s This sum contains the term a b , which is not divisible by p. Its other terms r s a b are all divisible by p, since they either have i < r or j < s. Hence c i j r+s is not divisible by p. As there is no prime dividing all the coefficients of fg, the polynomial fg is primitive. (cid:3) If f ∈ Z[X] is nonzero, let a be the greatest common divisor of the coefficients of f. Then f = af where f is primitive. We call a the content 1 1 of f and denote it by c(f). More generally, let g be a nonzero element of Q[X]. Then bg ∈ Z[X] where the positive integer b is the product of the denominators of the coefficients of f. Then bg = cg where c is the content 1 of bg and g is primitive. Hence g = (c/b)g where q is primitive polynomial 1 1 1 in Z[X] and c/b is a positive rational. We can write any nonzero g ∈ Z[X] as g = rg with r ∈ Q, r > 0 and g ∈ Z[X] being primitive. It’s an instructive 1 1 exercise to show that this r is uniquely determined; hence it makes sense to call r the content of g. Putting s = 1/r we see that there is a positive rational s with sg a primitive element of Z[X]. WenowshowthatifamonicpolynomialinZ[X]factorsovertherationals then it factors over the integers. Proposition 1.2 Let f and g be monic polynomials with f ∈ Z[X] and g ∈ Q[X]. If g | f then g ∈ Z[X]. Proof Suppose that g | f. Then f = gh where h ∈ Q[X]. Then h is monic, as both f and g are. There are positive rationals r and s with rg and sh primitive elements of Z[X]. The leading coefficients of rg and sh are r and s respectively, so that r, s ∈ Z. By Gauss’s lemma, (rg)(sh) = (rs)f is primitive. But since f ∈ Z[X] all coefficients of (rs)f are divisible by rs. 3 Hence rs = 1 (as rs is a positive integer) and so r = s = 1 (as r and s are positive integers). Thus g = rg ∈ Z[X] as required. (cid:3) An immediate corollary is this important characterization of algebraic integers. Theorem 1.1 Let α ∈ A have minimum polynomial f. Then α ∈ B if and only if f ∈ Z[X]. Proof Suppose f ∈ Z[X]. Since f is monic and f(α) = 0 then α ∈ B. Converselysupposethatα ∈ B. Thereisamonicg ∈ Z[X]withg(α) = 0. Then f | g, since f is the minimum polynomial of α. By Proposition 1.2, f ∈ Z[X]. (cid:3) Another corollary is this useful criterion for irreducibility. Proposition 1.3 (Eisenstein’s criterion) Let p be a prime number and let n−1 X f(X) = Xn + a Xj ∈ Z[X]. j j=0 If • p | a when 0 ≤ j < n, and j • p2 - a 0 then f is irreducible over Q. Proof Suppose that f is reducible over Q. Then f = gh where g, h ∈ Q[X], g and h are monic, and also 0 < r = deg(g) < n and deg(g) = s = n−r. By Proposition 1.2, g, h ∈ Z[X]. Write r s X X g(X) = b Xi and h(X) = c Xj. i j i=0 j=0 Note that b = 1 = c . Certainly p - b and p - c . Let u and v be the least r s r s nonnegative integers with p - b and p - c . Then u ≤ r and v ≤ s. I claim u v that u = r and v = s. Otherwise u+v < r+s = n and X a = b c . u+v i j i+j=u+v This sum contains the term b c which which is not divisible by p. The u v remaining terms have the form b c with either i < u or j < v. In each case i j 4 one of b and c is divisible by p. Hence a is the sum of a nonmultiple of i j u+v p with a collection of multiples of p and so p - a contrary to hypothesis. u+v Hence u = r and v = s. As r, s > 0 both b and c are divisible by p so that 0 0 a = b c is divisible by p2 again contrary to hypothesis. This contradiction 0 0 0 shows that f is irreducible over Q. (cid:3) Example Let p be a prime number and let p−1 X Xp −1 f(X) = 1+X +X2 +···+Xp−1 = Xj = . X −1 j=0 We cannot apply Eisenstein to f directly, but if we set f (X) = f(X+1) we 1 get (X +1)p −1 (X +1)p −1 Xp−1 (cid:18)p(cid:19) f (X) = = = Xp−j−1. 1 (X +1)−1 X j j=0 This is a monic polynomial, but its remaining coefficients have the form (cid:0)p(cid:1) j for0 < j < pandsoaredivisiblebyp. Thefinalcoefficientis(cid:0) p (cid:1) = pwhich p−1 is not divisible by p2. By Eisenstein’s criterion, f is irreducible over Q. It 1 followsthatf isirreducibleoverQ, foriff(X) = g(X)h(X)wereanontrivial factorization of f, then f (X) = g(X + 1)h(X + 1) would be a nontrivial 1 factorization of f . 1 We now show that A is a subfield of C and B is a subring of C. Theorem 1.2 (i) Let α, β ∈ A. Then α+β, α−β, αβ ∈ A, and if α 6= 0 then α−1 ∈ A. (ii) Let α, β ∈ B. Then α+β, α−β, αβ ∈ B. Proof We first prove (ii) in detail, since the bulk of the proof of (i) follows mutatis mutandis. Let α and β have minimum polynomials f and g of degrees m and n respectively. Write m−1 n−1 X X f(X) = Xm + a Xi and g(X) = Xn + b Xj. i j i=0 j=0 Then the a and b are integers and i j m−1 n−1 X X αm = − a αi and βn = − b βj. (∗) i j i=0 j=0 5 Let v be the column vector of height mn given by v> = (1 α α2 ··· αm−1 β αβ α2β ··· αm−1β β2 ··· αm−1βn−1). In other words the entries of v are the numbers αiβj for 0 ≤ i < m and 0 ≤ j < n. I claim that there are (mn-by-mn) matrices A and B with entries in Z such that Av = αv and Bv = βv. The typical entry in αv has the form αiβj where 1 ≤ i ≤ m and 0 ≤ j < n. If i < m this already is an entry of v while if i = m, (∗) gives m−1 X αmβj = − a αkβj. k k=0 In any case this entry αiβj of αv is a linear combination, with integer coefficients, of the entries of v. Putting these coefficients into a matrix A we get αv = Av. Similarly there is a matrix B with integer entries with βv = Bv. Now (A+B)v = (α+β)v, (A−B)v = (α−β)v and (AB)v = (αβ)v. As v 6= 0 the numbers α+β, α−β and αβ are eigenvalues of the matrices A+B, A − B and AB each of which has integer entries. But if the matrix C has integer entries, its eigenvalues are algebraic integers, since the characteristic polynomial of C is a monic polynomial with integer coefficients. It follows that α+β, α−β and αβ are all algebraic integers. If we assume instead that α, β ∈ A, the above argument shows (when we replace ‘integer’ by ‘rational’ etc.) that α+β, α−β, αβ are all algebraic numbers. Finally suppose that α is a nonzero algebraic number with minimum polynomial n−1 X f(X) = Xn + a Xi. i i=0 Then a 6= 0 (why?) and dividing the equation f(α) = 0 by a αn gives 0 0 n−1 X a 1 α−n + n−iα−i + = 0 a a 0 0 i=1 so that α−1 ∈ A. (cid:3) √ Example Let us see what the matrices A and B are for say α = 2 √ and β = 1(1+ 5). The minimum polynomials of α and β are X2 −2 and 2 X2−X−1 respectively so that α2 = 2 and β2 = β+1. Let v> = (1 α β αβ). 6 Then        α α 0 1 0 0 1  α2   2   2 0 0 0  α  αv =   =   =     αβ   αβ   0 0 0 1  β  α2β 2β 0 0 2 0 αβ and        β β 0 0 1 0 1  αβ   αβ   0 0 0 1  α  βv =   =   =   .  β2   1+β   1 0 1 0  β  αββ2 α+αβ 0 1 0 1 αβ We can take     0 1 0 0 0 0 1 0  2 0 0 0   0 0 0 1  A =   and B =  .  0 0 0 1   1 0 1 0  0 0 2 0 0 1 0 1 Then, for instance, αβ is an eigenvalue of   0 0 0 1  0 0 2 0  AB =    0 1 0 1  2 0 2 0 so that h(αβ) = 0 where h is the characteristic polynomial of AB. 2 Number fields The set A of algebraic numbers is too large to handle all at once. We restrict our consideration to looking at smaller subfields of A which contain all the algebraic numbers “generated” from a given one. For instance consider K = Q(i) = {a+bi : a,b ∈ Q} 1 and √ √ √ K = Q( 3 2) = {a+b 3 2+c 3 4 : a,b,c ∈ Q}. 2 It is apparent that both K and K are rings, being closed under addition, 1 2 subtraction and multiplication. It’s not hard to see that K is a field since if 1 a+bi is a nonzero element of K then 1 1 a b = − i ∈ Q(i). a+bi a2 +b2 a2 +b2 7 √ √ But it is not so obvious that 1/(a + b 3 2 + c 3 4) is an element of K . But 2 this is in fact so, and is an example of a general phenomenon. Let α be an algebraic number of degree n. Define Q(α) = {a +a α+a α2 +···+a αn−1 : a ,a ,...,a ∈ Q}. 0 1 2 n−1 0 1 n−1 Proposition 2.1 For each α ∈ A, Q(α) is a subfield of A. Proof Since A is closed under addition and multiplication, and α ∈ A and Q ⊆ A then it is apparent that Q(α) ⊆ A. Let α have degree n and minimum polynomial f. Then by definition Q(α) = {g(α) : g ∈ Q[X], and either g = 0 or deg(g) < n}. I claim that in fact Q(α) = {g(α) : g ∈ Q[X]}. Certainly Q(α) ⊆ {g(α) : g ∈ Q[X]} so that to prove equality we need to show that g(α) ∈ Q(α) whenever g ∈ Q[X]. By the division algorithm (Proposition A.1) there is q ∈ Q[X] such that h = g −qf either vanishes or has deg(h) < n. Then h(α) ∈ Q(α) but h(α) = g(α)−f(α)q(α) = g(α) as f(α) = 0. This proves that Q(α) = {g(α) : g ∈ Q[X]}. It is now clear that, since Q[X] is closed under addition, subtraction and multiplication then so is Q(α). Hence Q(α) is a subring of A. (Alternatively, one sees that the map g 7→ g(α) from Q[X] to A is a ring homomorphism with image Q(α) which must therefore be a subring of A.) TocompletetheproofthatQ(α)isafield, wemustshowthat1/β ∈ Q(α) whenever β is a nonzero element of Q(α). Write β = g(α) with g ∈ Q[X] and note that f - g since otherwise g(α) = 0. Let h be the greatest common divisor of f and g. By Proposition A.2, there exist u, v ∈ Q[X] with h = uf+vg. But f is irreducible, and so either h = 1 or h = f. But this latter is impossible since h - f. Hence 1 = uf+vg and so 1 = u(α)f(α)+v(α)g(α) = v(α)β. It follows that 1/β = v(α) ∈ Q(α) and so K is a field. (cid:3) The numbers 1,α,α2,...,αn−1, where n is the degree of α, form a basis of Q(α) as a vector space over Q. Thus the degree n is also the dimension of Q(α) as a vector space over Q, and so we call n the degree of Q(α). In general when we speak of a basis for a number field K = Q(α) me mean a basis for K as a vector space over Q. Given α ∈ A of degree n, its minimum polynomial f factors over C as n Y f(X) = (X −α ) j j=1 8 where α = α say. The numbers α ,...,α are the conjugates of α. They 1 1 n are all algebraic numbers with minimal polynomial f. It is important to note that the conjugates of α are all distinct. This follows from the following lemma. Lemma 2.1 Let f ∈ Q[X] be a monic polynomial and suppose that f is irreducible over Q. Then f(X) = 0 has n distinct roots in C. Proof Suppose that α is a repeated root of f(X) = 0. Then f(X) = (X −α)2g(X) where g(X) ∈ C[X]. Consequently f0(X) = (X −α)2g0(X)+ 2(X − α)g(X) and so f(α) = f0(α) = 0. Let h be the greatest common divisor of f and f0. Then h = uf + vf0 for some u, v ∈ Q[X]. Thus h(α) = u(α)f(α) + v(α)f0(α) = 0. But as h | f and f is irreducible, then h = 1 of h = f. Since h(α) = 0, h = f. But then f | f0 and as f0 has leading term nXn−1 this is impossible. (cid:3) The field Q(α) forms the set of numbers which can be expressed in terms of rational numbers and α using the standard arithmetic operations. We might instead consider what happens when we take two algebraic numbers α and β and consider which numbers can be expressed in terms of both. Suppose α and β have degrees m and n respectively, and define ( ) m−1n−1 XX Q(α,β) = c αjβk : c ∈ Q . jk jk j=0 k=0 It is readily apparent that Q(α,β) is a ring; less apparent but nonetheless true that it is a field. However this field can be expressed in terms of one generator. Theorem 2.1 (Primitive element) Let α, β ∈ A. Then there is γ ∈ A with Q(α,β) = Q(γ). Proof We show that for a suitable rational number c, γ = α+cβ suffices. Let α and β have degrees m and n respectively, and let their minimum polynomials be m n Y Y f(X) = (X −α ) and g(X) = (X −β ) j k j=1 k=1 respectively, with α = α and β = β . Suppose that 1 ≤ j ≤ m and 1 1 2 ≤ k ≤ n. The equation α+xβ = α +xβ j k 9 can be rewritten as (β −β )x = α −α 1 k 1 j and so has exactly one solution x = x as β 6= β . Choose c to be a nonzero jk 1 k rational which is not equal to any of the x . This is possible as Q is an jk infinite set. Then α+cβ 6= α +cβ j k whenever k 6= 1, by the choice of c. Let γ = α + cβ. For convenience put K = Q(γ). I claim that Q(α,β) = K. Certainly γ ∈ Q(α,β) and as Q(α,β) is a ring, then K ⊆ Q(α,β). To prove that K ⊇ Q(α,β) it suffices to show that α ∈ K and β ∈ K. Let h(X) = f(γ − cX). Then h has degree m, as c 6= 0, and h ∈ K[X]. Also h(β) = f(γ−cβ) = f(α) = 0. But of course g(β) = 0 so that g and h have β as a common zero. Suppose that it had another one, so that g(δ) = h(δ) = 0. Then δ = β for some k ≥ 2 as g(δ) = 0. But then 0 = h(β ) = f(γ −cβ ) k k k so that γ − cβ = α for some j. Thus γ = α + cβ which is false by the k j j k choice of c. The greatest common divisor of g(X) and h(X) must be X − β. As g ∈ Q[X] ⊆ K[X] and h ∈ K[X] there exists u, v ∈ K[X] with u(X)g(X)+ h(X)v(X) = X −β. Thus β = −(u(0)g(0)+h(0)v(0)) ∈ K, and it follows that α = γ −cβ ∈ K. This completes the proof. (cid:3) More generally we can consider fields Q(β ,...,β ) generated by any 1 n finite number of algebraic numbers. But by using the primitive element theorem and induction we see that each such field still has the form Q(γ). We call a field of the form Q(α) for α ∈ A an algebraic number field or simply a number field. Let α ,...,α be the conjugates of α. The fields Q(α ) are very similar 1 n j to Q(α) each being generated by an element with minimum polynomial f. In fact they are all isomorphic. We define an isomorphism σ : Q(α) → Q(α ) j j by setting σ (g(α)) = g(α ) when g ∈ Q[X]. It is perhaps not immediately j j evident that σ is well-defined. But this follows since if g , g ∈ Q[X] and j 1 2 g (α) = g (α)theng (α ) = g (α ). Thisisaconsequenceofα andα having 1 2 1 j 2 j j the same minimum polynomial. Once σ is seen to be well-defined, then it j is straightforward to prove it is an isomorphism. As α = α then σ is the 1 1 identity map on Q(α). Let β ∈ Q(α). We define the norm N(β) and trace T(β) of β as follows. Let n Y N(β) = σ (β) j j=1 10

Algebraic Number Theory: summary of notes [Lecture notes]

Similar Algebraic Number Theory: summary of notes [Lecture notes]