First and raw version september 2013 klokken 13:45

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1 The discriminant First and raw version september 2013 klokken 13:45 One of the most significant invariant of an algebraic number field is the discriminant. One is tempted to say, apart from the degree, the most fundamental invariant. In simplest case of a quadratic equation, the discriminant tells us the behavior of solution, and of course, even its square roots gives us the solutions. To some extent the same is true for cubic equations, and the higher the degree of an equation less the influence of the discriminant is, but it always plays an important role. For a field extension we shall define a discriminant for any basis. Of course this depends on the basis, but in a very perspicuous way. In the case of a number field, the ring of algebraic integers A is a free module over Z, andifthebasisesareconfined to be Z-basis for A, thediscriminantisanintegerindependentofthethebasis,isan invariant of the number field. The discriminant serves several purposes at. Its main main feature is that it tells us in which primes a number fields ramify, or more generally, in which prime ideals an extension ramifies. Additionally the discriminant is a valuable tool to find Z-basis for the ring A of algebraic integers in a number field. To describe A is in general a difficult task, and the discriminant is some times helpful. In relative situation, the situation is somehow more complicated, and the discriminant is well-defined just as an ideal. The discriminant of a basis The scenario is the usual one: A denotes a Dedekind ring with quotient field K, andl afinite,separableextensionofk. The integral closure of A in L is B. InSeparability we defined the trace form tr L/K (xy). ItisasymmetricandquadraticformonL taking values in K, andsincel is separable over K, itisnon-degenerateaftertheorem1 on page 10 of Separability. Let 1,..., n be a K-basis for L and denote it by B.Onemayform asonemaydo for any quadratic form the matrix M of the trace form by putting M =(tr L/K ( i j )). If B and B are the coordinate vectors relative to the basis B of two elements and of L, thevalueofthetraceformtr L/K ( ) is given as is tr KQ ( )= t BM B We define the discriminant of the basis B as the determinant of the matrix M. That B = ( 1,..., n ) = det(tr K/Q ( i j )). The discriminant depends of course on the basis, but in a very clean way. Assume that B 0 is another basis and denote by M 0 the matrix of the trace form in that basis. 1

2 If V denotes the transition matrix between the two basis that is B = V B 0 for all then BM t B =(V B 0) t M(V B 0)= B t 0(V t MV) B 0, and as this holds for all and determinants, one obtains:, one draws the conclusion that M 0 = V t MV.Taking Lemma 1 Assume that B and B 0 are two K-basises for the field L with transitions matrix V (from B 0 to B). Then the relation between the corresponding discriminants is given as: B0 =(detv )2 B An important observation is, that if B is an integral basis, i.e., the basis elements are all lying in B, thenthediscriminantbelongstoa. Indeed,alreadytheallthetraces tr L/K ( i j ) lie in A. Example. The case of a primitive basis. Assume that the field L has a primitive element x, i.e., L = K(x), andletn be the degree of x. Then L has the special basis 1,x,...,x n 1 formed by the n first powers of x. There is a nice expression for the discriminant of L over K in that particular basis, it equals the discriminant of the minimal polynomial of x as defined in Separability. Torealizethis,let! 1,...,! n be the roots of the minimal polynomial of x in some big field. Inotherwords,the values the different embeddings of L in take at x. The! k s are the eigenvalues in of the multiplication map x,andthereforetr L/K (x i )= P k!i k. Let V =(! j 1 i ) be the van der Monde matrix formed by the! is. 0 Then, forming the product of V and its transposed, we get V t V =(! i 1 j )(! j 1 i )=( X k! i 1 k! j 1 k )=( X k! i+j 2 k )=(tr L/K (x i 1 x j 1 )), and therefore 1,x,...,x n 1 =(detv )2. Combining this with xxxx, we obtain Proposition 1 Given a field extension L = K(x) of degree n with primitive element x. Assumethatx has the minimal polynomial f(t) over K. Thenthediscriminantof the power basis 1,x...,x n 1 is equal to the discriminant of the polynomial f. Thatis f = 1,x,...,x n 1. e The case of extensions of a pid Assume now that in the standard setup A is a principal ideal domain, and n =[L : K]. The two main cases that concerns us are the case of a number field whose with ring of integers extends the principal ideal domain Z and the case that A is a dvr.ifa is 2

3 a pid,anyfinitelygeneratedandtorsionfreemoduleovera is a free module and thus has an A-basis. The transition matrix between two such Z-basises belongs to group Gl(n)A of invertible n n-matrices with coefficients in A, anditsdeterminantbelongs therefore to groups A of units. The case of a number field In the all-important case of a number field the ring of integers A is a free Z-module, and the determinant of a transition matrix between two basises equals ±1. The formula in lemma 1 above shows that the discriminant B is independent of B as long as we restrict B to be a Z-basis of A. This number i called the discriminant of K and it is denoted by K. Proposition 2 Assume that K is a number field and that A is the ring of integers in K. Thenthediscriminant B is independent of the basis B as long as B is a Z-basis for A. Itiscalledthe discriminant of K and denoted by K. Example. The discriminant of the quadratic field Q( p d) depends on the residue class of d mod 4. The discriminant is 4d in case d 6 1 mod 4 and equals d if d 1 mod 4. e The discriminant and the ramified primes The most important application of the discriminant is that it detects ramification. In the case that A has a primitive element, say A = Z[ ], thisisreasonable,sincethenaprimep ramifies exactly when the minimal polynomial f(x) of has a multiple root modulo p, andthediscriminant of f(x) is made to detect this. However, in general A does not have a primitive element and this complicates the matter substantially, but still the discriminant behaves well: Theorem 1 Assume that K is a number field with discriminant is ramified in K if and only if p divides the discriminant K. K. Thenaprimep Proof: There are three points, the first one being that the trace is functorial in the following sense. Let p 2 Z be a prime and let a 1,...,a n be a Z basis for A. Then of course ā 1,...,ā n is a k(p) basis for A/pA. Clearly if one reduces the multiplication map a in A modulo p one obtains the multiplication map ā in A/pA. NowusinganZ-basis for A to compute tr K/Q (a), oneseesthattheintegertr K/Q (a) reduces to tr A/pA/k(p) (a) mod p. Applyingthistoallthecoefficientsofthematrix(tr K/Q (a i a j )), oneobtainsthat det(tr K/Q (a i a j )) reduces to det(tr (A/pA)/k(p) (ā i ā j )) mod p, andas K =det(tr K/Q (a i a j )), it follows that p divides K if and only if the trace form on the k(p)-algebra A/pA is degenerate. The second point is that trace form on A/pA is non-degenerate if and only if A/pA is a product of separable field extensions of k(p). Butask(p) is a finite field, all field extensions are separable, so this happens if and only if A/pA is a product of fields. That is, if and only if it has no nilpotent elements. The third and final point is that A/pA = Q 1appleiapples A/qe i i is without nilpotents if and only if all the e i s are equal to one, that is if and only if p is non-ramified. o 3

4 The discriminant and the ring of integers. The discriminant can sometimes be of great help to find the ring of integers in a number field. Assume we have at our disposal an integral basis B of K, thatisaq-basis for K contained in A. Assumethat B 0 is a Z-basis for A and let V be the transition matrix between these two basises. The known basis B will be a Z-basis for A if and only if V is invertible in Gl(Z)n, thatis, if and only if det V is plus or minus one. But by lemma 1 above the two discriminant B and B 0 differ by the factor (det V ) 2,henceif B is a square free integer, itmustbe so that det V = ±1, andourbasisb is a Z-basis for A. Proposition 3 Assume that K is a number field with ring of integers is A. Ifan integral basis for K has a square free determinant, then it is a Z-basis for A. Example. Let f(x) =x 3 x 1, thenf(x) is irreducible (only possible roots are ±1). Let K = Q( ) where is one of the roots of f. The polynomial f(x) has discriminant 4( 1) = 23 which coincides with the diskriminant of the basis 1,, 2. This is obviously an integral basis, and 23 is square free. It follows that 1,, 2 is a basis for the ring of integers. e Problem 1. Let f(x) =x 3 x +3.Showthatf(x) is irreducible over Q and if denotes a root of f(x), findabasisfortheringofintegersk = Q( ). Hint: Compute the discriminant by the formula 4a 3 27b 2 (see Separability for this formula) X Problem 2. Let g(x) =x 3 3x +9.Showthatg is irreducible and compute the discriminant of f. Let be a root of g(x). ShowthatQ( ) =Q( ) =K (where is as in the previous problem). Show that Z[ ] is not the ring of integers in K. Hint: 3/ is a root of g(x). X Problem 3. Assume that a 1,...,a n is an integral basis for A. Showthatifd = (a 1,...,a n ),thenda a 1 Z + + a n Z. X Problem 4. Let 1,..., n be the different embeddings of the number field K into some big field. Let 1,..., n be a basis for K over Q. Showthat ( 1,..., n )=(det( i ( j ))) 2 X Problem 5. Let ( i ( j )) be the n n-matrix from problem 4.LetP = P even (1)( 1 ) (n)( n ) and N = P odd (1)( 1 ) (n)( n ) where runs through the permutations of {1,...,n}. Showthatdet(a ij )=P N, andthatbothp + N and PN are invariant under the action of that permutes the different embeddings i. Then show that P and N both lie in Z. X Problem 6. (Stickelberger). Let K be the discriminant of the number field K. Show that K is a square modulo 4, hence K 0 or K 1 modulo 4. Hint: Write =(P N) 2 =(P + N) 2 4PN with a smart choice of P an N. X 4

5 The general case and the diskriminant ideal In the general case, there is no way of getting one element in A that plays the role of the discriminant as for a number field. There are two reasons for this. First of all, the ring B is not necessarily is a free A-module and we do have A-basises for B at our disposal. Secondly, even if B were a free A-module, the group A of units in A is in general much more complicated than the one of Z and (A ) 2 can be highly non-trivial. However one may define a discriminant ideal d B/A that does the job. It is defined as the ideal in A generated by (b 1,...,b n ) as the b 1,...,b n run through all integral basises for L over K. Recallthatthismeansthatb 1,...,b n is a basis for L contained in B. In case K is number field with ring of integers A, thediscriminant K is a slightly more subtle invariant than the discriminant ideal d A/Z.Ofcourse, K will be one of the two generators of d A/Z,thesubtilityliesinthefactthatoneofthemispreferredover the other. Just like for the discriminant, the interest of the discriminant deal lies in the fact that it detects ramification: Theorem 2 (Dedekind) Let A be a Dedekind ring with quotient field K and let B be the integral closure of A in a finite and separable extension L of K. Thenaprime p of A ramifies in B if and only if d B/A p. Proof: The proof has three ingredient. The first one is that the discriminant ideal localizes well. That is If S is a multiplicative system in A, wehaved BS /A S =(d B/A ) S Clearly if 1,..., n is a K-basis for L contained in B,itisaK-basis for L contained in B S.Hencetheinclusion(d B/A ) S d BS /A S.Ontheotherhand,if 1,..., n is a K basis for L contained in the localization B S,thens 1,...,s n is contained in B for a suitable s 2 S, anditisstillak basis for L. Obviously (s 1,...,s n )=s 2n ( 1,..., n ). This shows that d BS /A S (d B/A ) S. The second ingredient is that the ramification behavior of a prime ideal localizes. This somehow vague statement, means the following: If p is a prime ideal i A, thenp ramifies in B if and only if pa p ramifies in B p. Indeed, this is an immediate consequence of the isomorphism B/p ' B p /pb p and the fact that p ramifies (respectively pa p )ifandonlyifb/pb (resp. B p /pb p )has non-trivial nilpotent elements. The third and final ingredient is that the theorem holds for the local rings A p,in fact it holds whenever A is a pid : If A is a pid,thenp ramifies if and only if d B/A p. Indeed, it follows from lemma 1 on page 2 that if 1,..., n is an integral basis for L, then ( 1,..., n ) is a generator for the discriminant ideal d B/A.AndasB is a free A- module (any torsion free and finitely generated A-module is since A is, a pid )theproof of theorem 1 shows mutatis mutandi that p ramifies if and only if ( 1,..., n ) /2 p. o 5

6 An observation which is not aprioriclear, is that in the standard situation, only finitely many prime ideals ramify in B. Versjon: Monday, September 23, :45:34 PM 6