DISCRIMINANTS IN TOWERS

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1 DISCRIMINANTS IN TOWERS JOSEPH RABINOFF Let A be a Dedekind domain with fraction field F, let K/F be a finite searable extension field, and let B be the integral closure of A in K. In this note, we will define the discriminant ideal B/A and the relative ideal norm N B/A (b). The goal is to rove the formula C/A = N B/A C/B [L:K] B/A, where C is the integral closure of B in a finite searable extension field L/K. See Theorem 6.1. The main tool we will use is localizations, and in some sense the main urose of this note is to demonstrate the utility of localizations in algebraic number theory via the discriminants in towers formula. Our treatment is self-contained in that it only uses results from Samuel s Algebraic Theory of Numbers, cited as [Samuel]. Remark. All finite extensions of a erfect field are searable, so one can relace Let K/F be a searable extension by suose F is erfect here and throughout. Note that Samuel generally assumes the base has characteristic zero when it suffices to assume that an extension is searable. We will use the more general fact, while quoting [Samuel] for the roof. 1. Notation and review. Here we fix some notations and recall some facts roved in [Samuel]. Let K/F be a finite field extension of degree n, and let x 1,..., x n K. We define D(x 1,..., x n ) = det Tr K/F (x i x j ) n i,j=1. We write D K/F instead of D if the field extension is not clear from context. If K/F is searable and x 1,..., x n are an F-basis for K, then D(x 1,..., x n ) 0 by [Samuel, Proosition 2.7.3]. If M is an n n matrix with coefficients in F then (1.1) D(M x 1,..., M x n ) = det(m) 2 D(x 1,..., x n ) by [Samuel, Proosition 2.7.1]. If M is a ermutation matrix then det(m) = ±1, so D(x 1,..., x n ) is indeendent of the ordering of its arguments. Let A be an integrally closed domain with fraction field F, let K/F be a finite extension of degree n, and let B K be a subring which is integral over A. Then for x 1,..., x n B, we have Tr K/F (x i x j ) A, so D(x 1,..., x n ) A. If B is a free A-module of rank n, with basis x 1,..., x n, the discriminant ideal is the rincial ideal (1.2) B/A := D(x 1,..., x n ) A. This is indeendent of the choice of basis by (1.1), as if M is a change of basis matrix for B, then det(m) A. If K/F is searable then B/A is nonzero. 1

2 2 JOSEPH RABINOFF Now we secialize to the case of Dedekind domains. See [Samuel, Theorems and 3.4.1] for roofs of the following facts. Proosition 1.3. Let A be a Dedekind domain with fraction field F, let K/F be a finite searable extension of degree n, and let B be the integral closure of A in K. Then: (1) B is a Dedekind domain with fraction field K. (2) B is an A-submodule of a free A-module of rank n. (3) B is a finitely generated A-module. (4) If q B is a maximal ideal then q A is a maximal ideal. (5) If A is a rincial ideal domain, then B is a free A-module of rank n, and any A-basis for B is an F-basis for K. Let A be a Dedekind domain with field of fractions F. For a nonzero fractional ideal a F, we have a unique factorization a = v(a), where the roduct is taken over all nonzero rime ideals of A. Of course, all but finitely many of the integers v (a) are zero. For fixed we have v (ab) = v (a) + v (b), where a, b F are nonzero fractional ideals. For x F we write v (x) = v ((x)). Remark 1.4. We will follow Samuel s convention that a field is a Dedekind domain. This has the advantage that any localization of a Dedekind domain is Dedekind (see Proosition 2.1 below), but the disadvantage that one has to distinguish between maximal ideal and nonzero rime ideal throughout. 2. Localizations. For our uroses, the main advantage of localizations is that they roduce rincial ideal domains from Dedekind domains. This allows us to use the classification of modules over a PID, and facilitates our constructions and roofs. The localization is well-behaved, in that it resects most ring-theoretic constructions, and one can recover information about a ring from its localizations. Let A be an integral domain with fraction field F, and let S A\{0} be a multilicatively closed subset. Recall that the localization of A at S is S 1 A := a s F a A, s S This is a subring of F containing A, and the fraction field of S 1 A is F. For a rime ideal A, the comlement A \ is multilicatively closed, and we write a A := (A \ ) 1 A = F a, s A, s /. s We call A the localization of A at. If A is a subring of an integral domain B and A is a rime ideal, then A \ is also a multilicatively closed subset of B, and we write B = (A \ ) 1 B. Note that B in general is not the localization of B at a rime ideal of B. Let A be an integral domain, let S A\ {0} be a multilicatively closed subset, and let A = S 1 A. For an ideal a A, we define its extension a A to be the ideal of A generated by a. The contraction of an ideal a A is the ideal a A of A. An ideal of A of the form a A is called contracted..

3 DISCRIMINANTS IN TOWERS 3 We have the following ideal corresondence for localizations from [Samuel, Proosition 5.1.1]. Note the similarity with the ideal corresondence for quotients. Proosition 2.1. Let A be an integral domain, let S A \ {0} be a multilicatively closed subset, and let A = S 1 A. (1) For every ideal a A we have (a A) A = a. (2) Extension and contraction give rise to an inclusion-reserving bijection contracted ideals of A ideals of A. (3) The bijection of (2) restricts to a bijection rime ideals A such that S = rime ideals of A. In other words, a rime ideal A is contracted if and only if S =. Exercise 2.2. Let A be an integral domain, let a A be an ideal, let S A \ {0} be a multilicatively closed subset, and let A = S 1 A. Prove that: a (1) a A = s a a, s S. (2) If A is rime with S =, a A and s S, then a/s A if and only if a. [Careful: remember that there are multile ways of exressing an element of S 1 A as a fraction.] (3) The contraction of a nonzero ideal is nonzero. (4) Extension is comatible with roducts, in that for any two ideals a, b A, (5) a A = A if and only if S a. (6) A = S 1 A if and only if S A. (a A )(b A ) = (ab) A. Exercise 2.3. Let A be an integral domain, let a A be a roer ideal, and let S = 1 + a. Verify that S is a multilicatively closed subset, and show that for a rime ideal A, we have S if and only if + a = A. 3. Localizations of Dedekind domains. The localization is an extremely well-behaved construction, in that it is comatible with most of the ring-theoretic constructions we have already encountered. For instance, we have [Samuel, Proosition and 5.1.3]. Proosition 3.1. Let A be a Dedekind domain with fraction field F, let S A \ {0} be a multilicatively closed subset, let K/F be a finite searable extension, and let B be the integral closure of A in K. Then (1) S 1 B is the integral closure of S 1 A in K, and (2) S 1 B is a Dedekind domain. By Exercise 2.2, nonzero rime ideals in S 1 A corresond to nonzero rime ideals in A which are disjoint from S. The following lemma relates rime factorizations in A to rime factorizations in S 1 A. Lemma 3.2. Let A be a Dedekind domain, let S A\{0} be a multilicatively closed subset, and let A := S 1 A.

4 4 JOSEPH RABINOFF (1) If a A is a nonzero ideal then a = v(a) = a A = (A ) v(a). S= S= (2) If a A is a nonzero ideal then a = (A ) v A (a ) = a A = v A (a ). S= (3) The nonzero contracted ideals in A are those ideals whose rime factors are disjoint from S. In other words, to find the rime factorization of a A, one simly deletes the rime factors which do not give rise to rime ideals of A. By Proosition 2.1(3), every nonzero rime ideal of A has the form A for a nonzero rime ideal A, and Lemma 3.2 imlies the comatibility (3.3) v A (a A ) = v (a). Proof. By Exercise 2.2(4), extension is comatible with ideal roducts, so a A = (A ) v(a) = (A ) v (a) S= because A = A when S. This gives (1). Now we rove (2). By Proosition 2.1(1), the ideal a := a A extends to a, and by (1), the ideal b := S= v A (a ) extends to a as well. Clearly b (b A ) A = a A = a, so a b. This imlies v (a) v (b) for all rimes A. By construction and (3.3), if S = then v (b) = v A (a ) = v (a), and since v (b) = 0 for S, for such we have 0 v (a) v (b) = 0. This means v (a) = v (b) for all, so a = b. This roves (2); art (3) follows from (1) and (2). We will roduce rincial ideal domains from Dedekind domains in the following way. Definition 3.4. A nonzero ring is called semi-local if it has finitely many maximal ideals. A Dedekind domain with exactly one nonzero rime ideal is a called a discrete valuation ring, or DVR. Lemma 3.5. A semi-local Dedekind domain is a PID. In articular, a DVR is a PID. Proof. Let A be a semi-local Dedekind domain with nonzero rime ideals 1,..., r. These are distinct maximal ideals, so they are airwise corime: i + j = (1) for i j. Moreover, since 2 1 does not share any common factors with j for j 2, we have likewise j = gcd( 2 1, j) = (1). Alying the Chinese remainder theorem gives a surjection c : A A A A r

5 DISCRIMINANTS IN TOWERS 5 with kernel r. Let x 1 / 2 be a nonzero element, and choose x A such that 1 c(x) = (x, 1,..., 1). Then x 1 but not 2, and for j 2 we have x 1 mod 1 j for j 2, so x / j. In terms of ideals, this says 1 (x) but 2 (x), and 1 j (x) for j 2. Hence (x) = 1, as this is the only ossible rime factorization. It follows that 1 is rincial, and by the same argument, that i is rincial for all i. As any nonzero ideal in A is a roduct of rime ideals, A is a rincial ideal domain. Remark 3.6. Let A be a DVR with maximal ideal. Then = () is rincial, and every nonzero ideal of A has the form n for some n 0, since there are no other ossible ideal factorizations. Hence the ideals of A are (1), (), ( 2 ), ( 3 ),, (0). Examle 3.7. The ring A = Ct is a DVR with maximal ideal (t). Lemma 3.8. If A is a Dedekind domain, then A is a DVR, and A is its maximal ideal. Proof. We know that A is Dedekind by Proosition 3.1(2), and by Proosition 2.1(3) that every nonzero rime ideal of A has the form q A for a nonzero rime ideal q A such that q (A \ ) =. But q (A \ ) = if and only if q, so since and q are maximal, we must have = q. Lemma 3.9. Let A be a discrete valuation ring with fraction field F, let K/F be a finite searable extension, and let B be the integral closure of A in K. Then B is semi-local, hence a PID. Proof. Let q be a maximal ideal of B. Then q A is a maximal ideal by Proosition 1.3(4), so q A =, the unique maximal ideal of A. Thus q, so B q, and hence q B. But B has only finitely many rime factors. It follows from Lemmas 3.8 and 3.9 that if A is a Dedekind domain with fraction field F, A is a nonzero rime, K/F is a finite searable extension, and B is the integral closure of A in K, then both A and B are PIDs. We will also use localizations to test if two ideals are equal. Lemma Let A be a Dedekind domain, and let a, a A be nonzero ideals. Then a = a if and only if a A = a A for all nonzero rime ideals A. Proof. If a A = a A then v (a) = v (a ) by (3.3). Hence a and a have the same rime factorizations. Exercise Let A be a Dedekind domain, let S A \ {0} be a multilicatively closed subset, and let A = S 1 A. Prove that for any two ideals a, b A, we have (a A) (b A) = (a b ) A. In other words, contraction is comatible with roducts in the case of localizations of Dedekind domains. 1 Exercise Let A be a Dedekind domain, let S A \ {0} be a multilicatively closed subset, and let A = S 1 A. 1 I strongly believe this to be false in general, but I cannot find a counterexamle when A is an integral domain.

6 6 JOSEPH RABINOFF (1) Let A be a nonzero rime ideal such that S =. Generalize [Samuel, Proosition 5.1.5] to rove that for any n 0, the natural homomorhism A/ n A /( A ) n is an isomorhism. (2) Let a A be a nonzero contracted ideal. Prove that the natural homomorhism is an isomorhism. A/a A /a A Exercise Let A be a Dedekind domain and let a A be a nonzero ideal. (1) Prove that every ideal of A/a is rincial. [Localize at S = 1+a as in Exercise 2.3. Show that a is contracted and S 1 A is semi-local, then use Exercise 3.12(2).] (2) Prove that a can be generated by two elements. [Aly (1) to the ideal a/aa of A/aA for any nonzero element a a.] Exercise Let K be a number field and let A = K. Prove that there exists f A\{0} such that A f := {1, f, f 2,...} 1 A is a rincial ideal domain. [Choose f to kill all elements of the class grou of A.] Exercise Generalize Lemma 3.9 as follows. Let A be a semi-local Dedekind domain with fraction field F, let K/F be a finite searable extension, and let B be the integral closure of A in K. Show B is semi-local. Exercise Suose that there were finitely many rime numbers. Use Exercise 3.15 to rove that Z[ 5] is a PID, and derive a contradiction. Thus there are infinitely many rime numbers The relative ideal norm. In this section we fix a Dedekind domain A with fraction field F. Let K/F be a finite searable extension of degree n, and let B be the integral closure of A in K. For a nonzero ideal b B we will define its relative norm N B/A (b), which is an ideal of A. This will generalize the (absolute) ideal norm N(b) defined when A = Z, and it will also generalize the norm of an element N K/F (x) for x B. See Remark 4.2 and Proosition 4.4. By Proosition 1.3, B is a Dedekind domain and a finitely generated A-module, and for every maximal ideal q B, the contraction := q A is a maximal ideal. Hence A/ is a subfield of B/q. Any set of generators for B as an A-module also generates B/q as a vector sace over A/, so the degree [B/q : A/] is finite. Definition 4.1. Let b = q qvq(b) be a nonzero ideal of B. The relative ideal norm of b is defined to be N B/A (b) := (q A) [B/q : A/] vq(b), where = q A. q For ideals b, b B, we have N B/A (bb ) = N B/A (b) N B/A (b ) since v q (bb ) = v q (b) + v q (b ). 2 Brian Conrad attributes this ridiculous roof to Larry Washington.

7 DISCRIMINANTS IN TOWERS 7 Remark 4.2. Suose that A = Z, so F = Q and K is a number field with ring of integers B = K. The absolute ideal norm N(b) of a nonzero ideal b B is defined to be the (finite) number of elements of the quotient ring, i.e. N(b) := #(B/b). We claim that (4.3) N B/Z (b) = N(b)Z. The absolute ideal norm is multilicative in b by [Samuel, Proosition 3.5.2], so it suffices to rove N(q)Z = N B/Z (q) = (q Z) [B/q : Z/q Z] for q B rime. But it is easily verified that where q Z = Z. N(q) = #(B/q) = [B/q : F ], The following roosition says that the relative ideal norm is comatible with the norm of an element. It is surrising because the two kinds of norm are defined in comletely different ways: indeed, the norm of an element b is defined as the determinant of multilication by b. Proosition 4.4. Let b B \ {0}. Then N B/A (bb) = N K/F (b) A. In order to rove Proosition 4.4, we first need to show that the relative ideal norm is comatible with localizations. Lemma 4.5. Let S be a multilicatively closed subset of A, let A := S 1 A, and let B := S 1 B. Then for every nonzero ideal b B, we have N B/A (b) A = N B /A (bb ). Proof. As norms and extensions are both multilicative in b, we may assume b = q is rime. Let = q A, and note that S = q S because S A. If q S then qb = B, so N B /A (qb ) = A. In this case, S as well, so A = A, and N B/A (q) A = [B/q:A/] A = (A ) [B/q:A/] = A. Now suose q S = S =. Then B/q = B /(qb ) and A/ = A /A by [Samuel, Proosition 5.2.5]. Moreover, ((qb ) A ) A = (qb ) A = ((qb ) B) A = q A =, so (qb ) A = A because they contract to the same ideal of A. Hence we have a commutative square A/ B/q = A /A so [B/q : A/] = [B /qb : A /A ]. It follows that = B /qb N B/A (q) A = [B/q : A/] A = (A ) [B /qb : A /A ] = N B /A (qb ).

8 8 JOSEPH RABINOFF Proof of Proosition 4.4. By Lemma 3.10, it suffices to check that for all nonzero rime ideals A, we have N B/A (bb) A = N K/F (b) A. Lemma 4.5 gives that N B/A (bb) A = N B /A (bb ), so we may relace A by A and B by B to assume A is a discrete valuation ring. Thus A and B are rincial ideal domains by Lemmas 3.8 and 3.9. As ideal and element norms are multilicative, and since B is a unique factorization domain, it suffices to show that N B/A (qb) = N K/F (q) A for q B a rime element. The contraction (qb) A is the unique maximal ideal of A; let be a generator. By the classification of finitely generated modules over a PID, we have (4.6) B/qB = A/ e 1 A A/ e r A for some 1 e 1 e 2 e r, since A is the only maximal ideal of A. But qb, so annihilates B/qB, so e 1 =... = e r = 1, and B/qB = (A/A) r. This means that B/qB is an r-dimensional A/A-vector sace, so by definition, N B/A (B) = (A) r = r A. On the other hand, equation (4.6) shows that the Smith normal form for the multilication-by-q homomorhism m q : B B is the diagonal matrix with r diagonal entries being and the rest being 1. Hence N K/F () = det(m q ) = r, u to units. Exercise 4.7. Let b, b be nonzero ideals of B, with b b. Show that N B/A (b) N B/A (b ), and that if N B/A (b) = N B/A (b ) then b = b. Exercise 4.8. Let b be a nonzero ideal of B. (1) Prove that N B/A (b) = N K/F (x) x b. [Localize at a nonzero rime ideal of A.] (2) Prove by examle that N B/A (b) is not necessarily generated by the norms of a given set of generators for b. [Take b = B, and suose that there are two distinct nonzero rime ideals of B which contract to the same ideal of A.] Exercise 4.9. Let L/K be a finite searable extension and let C be the integral closure of B (or A) in L. Prove that for any nonzero ideal c C, we have N B/A NC/B (c) = N C/A (c). [It is temting to use Exercise 4.8(1), but this does not work because of Exercise 4.8(2).] Exercise Let a be a nonzero ideal of A. Prove that N B/A (ab) = a [K:F]. [Localize to reduce to the case where a is rincial.]

9 DISCRIMINANTS IN TOWERS 9 Exercise Suose that K/F is Galois with Galois grou G = Gal(K/F). Note that σ(b) = B for all σ G. Prove that for a nonzero ideal b B, N B/A (b) B = σ(b). [The left side is included in the right by Exercise 4.8. Take norms of both sides.] Remark Suose that F = Q and A = Z, and that K/F is Galois with Galois grou G = Gal(K/F). By Exercise 4.11 and Remark 4.2, for any nonzero ideal b B = K, we have N(b) K = σ(b), where N(b) = #( K /b) Z 1 is the absolute ideal norm. This useful observation gives a concrete formula for the inverse fractional ideal: b 1 = 1 σ(b). N(b) The situation becomes articularly simle when K = Q( d) is a quadratic extension of Q. In this case K/Q is automatically Galois, with Gal(K/Q) = {1, τ}, where τ( d) = d. Writing b = τ(b) for a nonzero ideal b K, we obtain σ G σ G σ 1 N(b)Z = bb. 5. The discriminant. In this section we fix a Dedekind domain A with fraction field F. Let K/F be a finite searable extension of degree n, and let B be the integral closure of A in K. By Proosition 1.3, B is a Dedekind domain with fraction field K, and B is an A-submodule of a free A-module of rank n. However, if A is not a rincial ideal domain then B may not itself be a free A-module, so that we cannot use (1.2) to define the discriminant B/A. We will use localizations in order to define the discriminant B/A in this case. By Proosition 3.1(1), for any rime ideal A, the integral closure of A in K is B := (A\ ) 1 B. Since A is a rincial ideal domain, B is a free A -module of rank n, so the discriminant ideal B /A of (1.2) is a well-defined nonzero ideal of A. Since A has only one maximal ideal A, the rime factorization of B /A has the form (5.1) B /A = (A ) n (B) for a unique nonnegative integer n (B). Lemma 5.2. We have n (B) = 0 for all but finitely many. Proof. Since Frac(B) = K, by clearing denominators we can find an F-basis y 1,..., y n for K contained in B. Let d = D(y 1,..., y n ) A. It is enough to show that n (B) v (d) for all, since the latter is zero for all but finitely many. For any rime ideal, we have y 1,..., y n B B. Let x 1,..., x n be an A -basis for B, and let M be the unique matrix with entries in A such that M x i = y i. Then det(m) A and d = D(y 1,..., y n ) = D(M x 1,..., M x n ) = det(m) 2 D(x 1,..., x n )

10 10 JOSEPH RABINOFF by (1.1). Thus d B /A, and B /A da. It follows from Lemma 3.2 that da = (A ) v(d), and by definition B /A = (A ) n(b), so n (B) v (d), as desired. Definition 5.3. Let A be a Dedekind domain with fraction field F, let K/F be a finite searable extension of degree n, and let B be the integral closure of A in K. The discriminant ideal of B/A is defined to be (5.4) B/A := n(b) A, where the roduct is taken over all nonzero rime ideals of A, and where n (B) is defined in (5.1). This roduct has only finitely many factors by Lemma 5.2. If B haens to be a free A-module, we have now made two a riori different definitions of B/A in (1.2) and (5.4). The following lemma imlies that they are comatible. Lemma 5.5. Suose that B is a free A-module, with A-basis x 1,..., x n. Then for all nonzero rime ideals A, we have n (B) = v D(x1,..., x n ). Proof. We claim that x 1,..., x n are an A -basis of B for any nonzero rime ideal A. Indeed, the x i are A -linearly indeendent (as they are K-linearly indeendent by Proosition 1.3(5)), and clearly x 1,..., x n B, so we only need to show B A x A x n. Let b/s B, where b B and s /. Then there exist a 1,..., a n A such that b = n a i=1 i x i, so as claimed. It follows that b s = 1 s n a i x i = i=1 n i=1 so n (B) = v (D(x 1,..., x n )) by Lemma 3.2. a i s x i A x A x n, (A ) n (B) = B /A = D(x 1,..., x n ) A, Remark 5.6. Suose that B/A is a monogenic extension, i.e. that there exists x B such that B = A[x]. Then 1, x, x 2,..., x n 1 is an A-basis for B, so B/A = D(1, x, x 2,..., x n 1 ) A. Hence if B/A is not a rincial ideal, then A is not a PID and B/A is not monogenic. These conditions are commonly satisfied for number fields, but it takes work to write down an exlicit examle where B/A is not rincial. In order to rove the discriminants in towers formula (6.2), we will (not surrisingly) reduce to the case where A is a PID by localizing. Hence we will need to know that discriminants are comatible with localizations, as the following lemma shows. Comare with Lemma 4.5. Lemma 5.7. Let S A \ {0} be a multilicatively closed subset, let A = S 1 A, and let B := S 1 B. Then B /A = B/A A.

11 DISCRIMINANTS IN TOWERS 11 Proof. This amounts to showing that for every nonzero rime ideal of A such that S =, we have n (B) = n (B ), where = A. Observe that a/s A = b/t a, b, s, t A, s, t /, b / t at = a, b, s, t A, b, s, t / = A, bs where b/t / if and only if b / by Exercise 2.2(2). Similarly, B = B, so Exercise 5.8. Prove that [Comare to Exercise 4.8.] (A ) n (B) = B /A = B /A = ( A ) n (B ). B/A = D(x 1,..., x n ) x 1,..., x n B. Exercise 5.9. Let B B be a subring which is a free A-module of rank n = [K : F]. Prove that B/A B /A. 6. Discriminants in towers. We are now in a osition to state and rove the discriminants in towers formula. Theorem 6.1 (Discriminants in towers). Let A be a Dedekind domain with fraction field F. Let K/F and L/K be finite searable extensions, and let B (res. C) be the integral closure of A in K (res. L). Then we have an equality of ideals of A: (6.2) C/A = N B/A C/B [L:K] B/A. We will localize to reduce the roof of Theorem 6.1 to the case when A and B are both PIDs. In this case, the discriminant and norm ideals in (6.2) are rincial, so we can work on the level of elements. This then becomes a matrix determinant calculation, carried out in Proosition 6.3 below. Proosition 6.3. Let F be a field, and let K/F and L/K be finite searable extensions of degrees n and m, resectively. Let x 1,..., x n be an F-basis for K and let y 1,..., y m be a K-basis for L, so {x i y j i = 1,..., n, j = 1,..., m} is an F-basis for L. Then (6.4) D L/F (x 1 y 1,..., x n y m ) = N K/F DL/K (y 1,..., y m ) D K/F (x 1,..., x n ) m. Proof. Fix an algebraic closure C of F. Let σ 1,..., σ n : K C be the distinct F-homomorhisms. For i = 1,..., n let τ i1,..., τ im : L C be the distinct embeddings restricting to σ i on K. Then {τ i j i = 1,..., n, j = 1,..., m} is the comlete set of F-embeddings of L into C. Define the following matrices with entries in C: S = (σ i x j ) n i,j=1 T k = (τ ki y j ) m i, j=1 (k = 1,..., n) M = (τ i j (x k y l )) (n,m) (i,j), (k,l)=(1,1).

12 12 JOSEPH RABINOFF Here and below we order the airs (i, j) lexicograhically. By [Samuel, Proosition 2.7.3], we have D K/F (x 1,..., x n ) = det(s) 2 σ k D L/K (y 1,..., y m ) = det(t k ) 2 D L/F (x 1 y 1,..., x n y m ) = det(m) 2. Let I m be the m m identity matrix. We have τ i j (x k y l ) = σ i (x k )τ i j (y l ) because x k K, so we can write M in block form as (6.5) (σ 1 x 1 )T 1 (σ 1 x 2 )T 1 (σ 1 x n )T 1 (σ M = 2 x 1 )T 2 (σ 2 x 2 )T 2 (σ 2 x n )T (σ n x 1 )T n (σ n x 2 )T n (σ n x n )T n T (σ 1 x 1 )I m (σ 1 x 2 )I m (σ 1 x n )I m 0 T = (σ 2 x 1 )I m (σ 2 x 2 )I m (σ 2 x n )I m T n (σ n x 1 )I m (σ n x 2 )I m (σ n x n )I m The square of the determinant of the first matrix in the roduct (6.5) is det(t 1 ) 2 det(t n ) 2 = σ 1 DL/K (y 1,..., y m ) σ n DL/K (y 1,..., y m ) = N K/F DL/K (y 1,..., y m ). By erforming a series of row and column exchanges, we transform the second matrix in the roduct (6.5) into the block matrix S S S Hence the square of the determinant of the second matrix is (det(s) 2 ) m = D K/F (x 1,..., x n ) m. Taking roducts yields (6.4).

13 DISCRIMINANTS IN TOWERS 13 Examle 6.6. Let us illustrate the roof of Proosition 6.3 by taking n = m = 2. In this case, σ 1 x 1 τ 11 y 1 σ 1 x 1 τ 11 y 2 σ 1 x 2 τ 11 y 1 σ 1 x 2 τ 11 y 2 σ M = 1 x 1 τ 12 y 1 σ 1 x 1 τ 12 y 2 σ 1 x 2 τ 12 y 1 σ 1 x 2 τ 12 y 2 σ 2 x 1 τ 21 y 1 σ 2 x 1 τ 21 y 2 σ 2 x 2 τ 21 y 1 σ 2 x 2 τ 21 y 2 σ 2 x 1 τ 22 y 1 σ 2 x 1 τ 22 y 2 σ 2 x 2 τ 22 y 1 σ 2 x 2 τ 22 y 2 τ 11 y 1 τ 11 y σ 1 x 1 0 σ 1 x 2 0 τ = 12 y 1 τ 12 y σ 0 0 τ 21 y 1 τ 21 y 1 x 1 0 σ 1 x 2 2 σ 2 x 1 0 σ 2 x τ 22 y 1 τ 22 y 2 0 σ 2 x 1 0 σ 2 x 2 = T1 0 0 T 2 (σ1 x 1 )I 2 (σ 1 x 2 )I 2. (σ 2 x 1 )I 2 (σ 2 x 2 )I 2 By making one row exchange and one column exchange, we transform: σ 1 x 1 0 σ 1 x 2 0 σ 1 x 1 σ 1 x σ 1 x 1 0 σ 1 x 2 σ 2 x 1 σ 2 x σ 2 x 1 0 σ 2 x σ 1 x 1 σ 1 x. 2 0 σ 2 x 1 0 σ 2 x σ 2 x 1 σ 2 x 2 Proof of Theorem 6.1. By Lemma 3.10 it suffices to check that both sides of (6.2) extend to the same ideal in A for all nonzero rime ideals A. By Lemmas 5.7 and 4.5, both sides of (6.2) are comatible with localizations, so we may relace A by A, B by B, and C by C to assume A is a discrete valuation ring. Then A and B are PIDs by Lemmas 3.8 and 3.9, so B is a free A-module and C is a free B-module. Let x 1,..., x n be an A-basis for B and y 1,..., y m a B-basis for C. Then {x i y j i = 1,..., n, j = 1,..., m} is an A-basis for C, and By Proosition 4.4, N B/A C/B [L:K] B/A C/A = D L/F (x 1 y 1,..., x n y m ) A C/B = D L/K (y 1,..., y m ) B B/A = D K/F (x 1,..., x n ) A. = N B/A DL/K (y 1,..., y m ) B D K/F (x 1,..., x n ) m A = N K/F DL/K (y 1,..., y m ) D K/F (x 1,..., x n ) m A, so the theorem follows from Proosition 6.3. Corollary 7. With the notation in Theorem 6.1, we have B/A C/A. 8. Examles and alications. Here we resent some situations in which Theorem 6.1 becomes useful. For a number field K we let D K Z denote the absolute discriminant, so K /Z = D K Z.

14 14 JOSEPH RABINOFF 8.1. Multiquadratic extensions of Q. Let d 1,..., d r be squarefree integers, let K = Q d 1,..., d r, and let F i = Q( d i ) for i = 1,..., r. Then F i K for each i, so by Corollary 7, we have D Fi D K for all i. The discriminant of a quadratic field is di if d 1 mod 4, (8.2) D Fi = 4d i otherwise. Therefore d i D K for each i. Conversely, we claim that every rime factor of D K divides 2d 1 d r. We roceed by induction on r, the case r = 1 being handled above. Suose that the claim is true for r. Let d = d r+1 be a squarefree integer and let L = K( d), so we need to show that every rime factor of D L divides 2d 1 d r d. If L = K then we are done by the inductive hyothesis, so suose L K. Then [L : K] = 2, so the conjugates of a + b d over K are a ± b d. Thus Tr L/K (1) = 2 Tr L/K (d) = 2d Tr L/K ( d) = d d = 0, where the first two identities hold since 1, d K and [L : K] = 2. We comute TrL/K (1) Tr D L/K (1, d) = det L/K ( d) Tr L/K ( 2 0 = det = 4d. d) Tr L/K (d) 0 2d By Exercise 5.8, we have 4d = D L/K (1, d) L / K, so L / K 4d K. Alying Theorem 6.1, D L/Q Z = L /Z = N K /Z L / K 2 NK K /Z /Z 4dK 2 K /Z = N K/Q (4d) 2 K /Z = (4d)[K:Q] 2 K /Z = (4d)[K:Q] D 2 K/Q Z, where we used Exercise 4.7 for N K /Z L / K NK /Z 4dK, and where NK/Q (4d) = (4d) [K:Q] because 4d K. Hence D L/Q (4d) [K:Q] D 2, so by induction, every rime K/Q factor of D L/Q divides 2d 1 d r d, as claimed. To summarize, we have roved: Proosition 8.3. Let d 1,..., d r be squarefree integers, and let K = Q d 1,..., d r. Then the odd rime factors of D K are exactly the odd rime factors of d 1 d r. Corollary 8.4. Let d 1,..., d r be squarefree integers, let K = Q d 1,..., d r, and let d be a squarefree integer which has a rime factor which is corime to 2d 1 d r. Then d is not a square in K. Proof. If d were a square in K then F = Q( d) K, so D F D K by Corollary 7. But d D F and d D K by Proosition 8.3, since is an odd rime dividing d but not d 1 d r. This is a contradiction. For examle, 13 is not a square in Q( 2, 3, 5, 7, 11).

15 DISCRIMINANTS IN TOWERS The quadratic subfield of Q(ζ ). Let be an odd rime and let ζ be a rimitive th root of unity. The minimal olynomial for ζ over Q is the -cyclotomic olynomial Φ (X ) = X 1 X 1 = X 1 + X X + 1 by [Samuel, 2.9]. In articular, K = Q(ζ ) has degree 1 over Q. Moreover, K = Z[ζ ] by [Samuel, Theorem 2.9.2], so D K = D(1, ζ, ζ 2,..., ζ 2 ) = ± N K/Q (Φ (ζ )) by [Samuel, 2.8], where Φ is the derivative of Φ. We have and therefore Φ (X ) = X 1 X 1 (X 1) = Φ 1 (X ) X (X 1) 2 X 1 ζ 1 D K = ± N K/Q ζ 1 = ± N K/Q ( ) N K/Q (ζ ) 1 N K/Q (ζ 1) 1 = ± 1 1 = ±, where we used N K/Q (ζ ) = 1 and N K/Q (ζ 1) = ±, as shown in [Samuel, 2.9]. The extension K/Q is Galois, being the slitting field of X 1, and it has Galois grou isomorhic to (Z/Z). This grou is cyclic of order 1, which is an even number. Therefore it admits a unique subgrou of index 2, so by the Galois corresondence, there is a unique subfield F K of degree two over Q. As F/Q is quadratic, there is a unique squarefree integer d such that F = Q( d). By Corollary 7 we have D F D K = ±, so D F = ±. But 2 D K, so D F 1 mod 4 and d = ± = D F by (8.2). Observe that 1 mod 4 3 mod 4 and 3 mod 4 1 mod 4, so we have shown: Proosition 8.6. Let be an odd rime. Then 1 mod 4 = Q(ζ ) and 3 mod 4 = Q(ζ ).