The discriminant MAT4250 — Høst 2013 The discriminant First and raw version 0.1 — 23. 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 discrimi- nant 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 trL/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 .Onemayform—asonemaydo B for any quadratic form—the matrix M of the trace form by putting M =(trL/K (↵i↵j)). If ↵ and β are the coordinate vectors relative to the basis of two elements ↵ and B B B β of L,thevalueofthetraceformtrL/K (↵) is given as t trKQ(↵)=↵ Mβ B B We define the discriminant of the basis as the determinant of the matrix M. That B is ∆ =∆(↵1,...,↵n) = det(trK/Q(↵i↵j)). B The discriminant depends of course on the basis, but in a very clean way. Assume that 0 is another basis and denote by M 0 the matrix of the trace form in that basis. B —1— The discriminant MAT4250 — Høst 2013 If V denotes the transition matrix between the two basis—that is ↵ = V↵ for all ↵ B B0 —then t t t t ↵ Mβ =(V↵ 0 ) M(Vβ 0 )=↵ (V MV)β 0 , B B B B B0 B t and as this holds for all ↵ and β, one draws the conclusion that M 0 = V MV.Taking determinants, one obtains: Lemma 1 Assume that and 0 are two K-basises for the field L with transitions B B matrix V (from 0 to ). Then the relation between the corresponding discriminants is B B given as: ∆ =(detV )2∆ B0 B An important observation is, that if is an integral basis, i.e., the basis elements B are all lying in B,thenthediscriminantbelongstoA.Indeed,alreadytheallthetraces trL/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 n 1 special basis 1,x,...,x − 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 ⌦ i i of the multiplication map ⇢x,andthereforetrL/K (x )= k !k. j 1 Let V =(! − ) be the van der Monde matrix formed by the !0s. Then, forming i P i the product of V and its transposed, we get t i 1 j 1 i 1 j 1 i+j 2 i 1 j 1 V V =(!j− )(!i − )=( !k− !k− )=( !k − )=(trL/K (x − x − )), Xk Xk 2 n 1 and therefore ∆1,x,...,x − =(detV ) . 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 n 1 the power basis 1,x...,x − is equal to the discriminant of the polynomial f.Thatis n 1 ∆f =∆1,x,...,x − . 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— The discriminant MAT4250 — Høst 2013 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 ∆ is independent of as long as we B B restrict to be a Z-basis of A. This number i called the discriminant of K and it is B denoted by δK . Proposition 2 Assume that K is a number field and that A is the ring of integers in K.Thenthediscriminant∆ is independent of the basis as long as is a Z-basis B B B for A.Itiscalledthe discriminant of K and denoted by δK . Example . The discriminant of the quadratic field Q(pd) depends on the residue class of d mod 4. The discriminant is 4d in case d 1mod4and equals d if d 1 6⌘ ⌘ 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 δK .Thenaprimep is ramified in K if and only if p divides the discriminant δK . Proof: There are three points, the first one being that the trace is functorial in the following sense. Let p Z be a prime and let a1,...,an be a Z basis for A. Then of 2 course a¯1,...,a¯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 ⇢a¯ in A/pA.NowusinganZ-basis for A to compute trK/Q(a),oneseesthattheintegertrK/Q(a) reduces to trA/pA/k(p) (a) mod p.Applyingthistoallthecoefficientsofthematrix(trK/Q(aiaj)),oneobtainsthat det(trK/Q(aiaj)) reduces to det(tr(A/pA)/k(p)(¯aia¯j)) mod p,andasδK =det(trK/Q(aiaj)), it follows that p divides δK 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. ei The third and final point is that A/pA = 1 i s A/qi is without nilpotents if and only if all the ei’s are equal to one, that is if and only if p is non-ramified. o Q —3— The discriminant MAT4250 — Høst 2013 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 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 B known basis will be a Z-basis for A if and only if V is invertible in Gl(Z)n,thatis, B if and only if det V is plus or minus one. But by lemma 1 above the two discriminant 2 ∆ and 0 differ by the factor (det V ) ,henceif∆ is a square free integer,itmustbe B B B so that det V = 1,andourbasis is a Z-basis for A. ± B 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.