Norms and Traces MAT4250 — Høst 2013 Norms and Traces
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Norms and traces MAT4250 — Høst 2013 Norms and traces Version 0.2 — 10. september 2013 klokken 09:55 We start by recalling a few facts from linear algebra that will be very useful for us, and we shall do it over a general base field K (of any characteristic). Assume that V is afinitedimensionalvectorspaceoverK.Toanylinearoperatorφ: V V there are ! associated two polynomials, both in a canonical way. Firstly, there is the characteristic polynomial Pφ(t), which is, as we know, defined by Pφ(t)=det(t id φ). It is a monic polynomial whose degree n equals the dimension · − dimK V .TwoofthecoefficientsofPφ(t) play a special prominent role, the trace and the determinant. They are up to sign, respectively the coefficient of the subdominant term and the constant term of the characteristic polynomial Pφ(t): n n 1 n P (t)=t tr φt − + +( 1) det φ φ − ··· − Secondly, there is the minimal polynomial of φ.Itisdefinedasthemonicgenerator of the annihilator of φ,thatisthemonicgeneratoroftheidealIφ = f(t) K[t] { 2 | f(φ)=0 . The two polynomials arise in very different ways, but they are intimately } related. The Cayley-Hamilton theorem states that Pφ(φ)=0,sothecharacteristic polynomial of φ belongs to the ideal Iφ,andtheminimalpolynomialdividesthecha- racteristic polynomial. In general they are different. For example, the degree of the characteristic poly is always equal to dimK V ,butthedegreeoftheminimalpolycan be anything between one and dimK V (degree one occurring for scalar multiples of the identity map). We often will work over extensions ⌦ of the base field K.AvectorspaceV and alinearoperatorφ may accordingly be extended as the tensor products V K ⌦ and ⌦ φ idK .Ifv1,...,vn is a basis for V ,thenv1 1,...,vn 1 will be a basis for V K ⌦— ⌦ ⌦ ⌦ ⌦ the only thing that changes, is that we are allowed to use elements from ⌦ as coefficients. The matrix of the map φ id in the basis v1 1,...,vn 1 is identical to the matrix of ⌦ ⌦ ⌦ φ in the basis v1,...,vn. Hence the characteristic polynomial does not change when we extend the base field, neither does the minimal polynomial. However the factorizations of these polynomials may change, e.g., φ might have more eigenvaluews in ⌦ than in K.If⌦ is a splitting field for Pφ(t),boththecharacteristicandtheminimalpolynomial factor as a products of linear terms, the roots being the eigenvalues of φ. Suppose now K L is a finite field extension; that is, L is a field containing K which ✓ is of finite dimension [K : L] as a vector space over K.Atypicalsituationbeingan algebraic number field which by definition is a finite extension of the fied Q of rational numbers. Multiplication by an element ↵ K is of course a K-linear map which we denote 2 by ⇢↵ : L L, i.e., ⇢↵ acts as x ↵x. The determinant of this linear map we call ! 7! the norm of ↵ over K and write NL/K (↵).Itstracewecallthetrace of ↵ over K and it is denoted trL/K (↵) . The norm is multiplicative and the trace is additive,justasfor linear maps—in fact, the trace is even K-linear. We have —1— Norms and traces MAT4250 — Høst 2013 ⇤ NK/L(↵)=NK/L(↵)NL/K (β) ⇤ trL/K (a↵ + bβ)=a trL/K (↵)+b trL/K (β) where a and b are scalars, in this context this means that a, b K. 2 If a lies in K,thenthemap⇢a is just scalar multiplication by a so it holds that [L: K] ⇤ NL/K (↵)=↵ , and hence [L: K] ⇤ NL/K (a↵)=a NL/K (↵) whenever a K.Sothenormisahomogenous function of ↵ of degree [L: K]. 2 Both the norm and the trace behaves very well in towers. However we postpone the proof until some more machinery has been developed. Proposition 1 Assume that K E L is a tower of finite field extensions and let ✓ ✓ x L Then 2 ⇤ NL/K (x)=NE/K(NK/L(x)) ⇤ trL/K (x)=trE/K(trK/L(x)) The case of extensions with a primitive element An illustrative and important special case is when ↵ is a primitive element in L;that is, ↵ generates L over K,soL = K(↵).Letthedegreeof↵ over K be n and let the minimal polynomial of ↵ be n n 1 f(t)=t + an 1t − + + a1t + a0 − ··· whose coefficients are in K.Inthiscasetheminimalandthecharacteristicpolynomial of ⇢↵ coincide, being of the same degree n,andwecanthenreadoffthenormand trace of ↵ immediately: ⇤ trK(↵)/K (↵)= ↵n 1 − − N =( 1)na . ⇤ K(↵)/K − 0 If the minimal polynomial f(t) factors as f(t)= 1 i n(t xi) in some extension ⌦ − of K,thecoefficientsoff(f) are elementary symmetric functions of the xi’s, and in Q particular we have the formulas ⇤ trK(↵)/K (↵)= i xi P ⇤ NK(↵)/K (↵)= i xi. Q —2— Norms and traces MAT4250 — Høst 2013 Be aware that these formulas only holds for the element ↵ in the extension L = K(↵),soifK(↵) ( L,expressionsforNL/K (↵) and trL/K (↵) are slightly more compli- cated, see peoposition 3 below. 2 n 1 One may also compute the matrix of ⇢↵.IntheK-basis 1,↵,↵ ,...,↵ − of L = K(↵) our multiplication map ⇢↵ has as matrix the companion matrix of f (which is the origin of the name): 0 ... ... ... a − 0 10... ... a1 0 . − 1 01 .. ... a2 B. −. C B. ... C B. C B0 ... ... 1 an 1C B − − C i @i+1 A n 1 n 1 This is clear since ⇢↵(↵ )=↵ if 0 i<n 1 and ⇢↵(↵ − )= a − ↵n 1 − − − −···− a1↵ a0. − Example . Another special and practicable case, is the case of the quadratic fields Q(pd), where as usual d is a square free rational integer. This field has the natural Q-basis 1, pd.Inthisbasisoneeasilycomputesthatmultiplicationby↵ = a + bpd (where a and b are in Q)hasthematrix ab . bpda ✓ ◆ Hence tr (a + bpd)=2a and N (a + bpd)=a2 db2. e Q(pd)/Q Q(pd)/Q − Problem 1. Assume that d 1mod4and let q =(d 1)/4.Let↵ = a+b(1+pd)/2 ⌘ − be an integral element of the quadratic field Q(pd), i.e., a and b are both rational integers. Show that N = a2 + ab qb2 ⇤ Q(↵)/Q − ⇤ trQ(pd)/Q =2a + b. X Norm and trace of integral elements Let K be an algebraic number field, that is, a finite extension of Q. Then the rational numbers NK/Q(↵) and trK/Q(↵) are called respectively the absolute norm and the absolute trace of the algebraic number ↵ K. 2 It is a very important property that both the absolute norm and the absolute trace of an algebraic integer are (rational) integers. More generally, there is the following result, whose proof we shall give later on: Proposition 2 Let the domain A be integrally closed in its field of fractions K and let L be a finite, separable extension of K.If↵ L is integral over A,thenboththe 2 trace trL/K ↵ and the norm NL/K (↵) belong to A. —3— Norms and traces MAT4250 — Høst 2013 Problem 2. Assume that K is an algebraic number field whose ring of integers denoted by A.Showthatalgebraicintegera A is a unit if and only if N (a)= 1. X 2 K/Q ± Problem 3. Generalise the preceding exersise as follows. Let A K be a domain in ✓ its fields of fractions. Suppose that L is a finte extension of K and let B be the integral closure of A in L.Showthatthenorminducesagrouphomorphismbetweenthegroups of units, that is, the norm restricts to a multilicative map N : B⇤ A⇤. X L/K ! Problem 4. Assume that d>0 is a square free integer, and let A Q(p d) be ✓ − the ring of algebraic integers in Q(p d).FindtheunitsinA. Hint: the answer is − different for d =1and d =3than in the other cases. X Problem 5. Show that if d>0 is a square free integer, then the algebraic integer ↵ = a + bpd is a unit if and only if a and b satisfies Pell’s equation: a2 db2 = 1 − ± X The characteristic polynomial is a power of the minimal polynomials Coming back to the general case, we are going give two different descriptions of the characteristic polynomial of ⇢↵,atleastinthecasethatL is a separable extension of K.Inthisparagraphweexpressthecharacteristicpolynomialof↵ as a power of the minimal polynomial, and in the next we shall express it in terms of the different embeddings of L over K in any algebraically closed field ⌦ (or any field being big enough, e.g., asplittingfieldforthecharacteristicpoly).Inourmaincaseofalgebraic number fields this typically will be embeddings into the complex field C. The setting is as follows: We are given a finite field extension K L and an element ✓ ↵ L. The subfield of L generated by ↵ fits into the tower K K(↵) L.Forsimplicity 2 ✓ ✓ of notation we give names to the different degrees involved, that is, we put [L: K]=n, [K(↵): K]=m and [L: K(↵)] = r. Then of course it holds that n = mr. The setting may be summarized with the diagram L r n K(↵) m K i Chose a basis vj 1 j r for L over K(↵).Asthepowers↵ ,for1 i m,isbasis { } i for K(↵) over k,theproducts↵ vj form a basis for L over K.Wehave i+1 i ↵ vj for all 1 j r and 0 i<n 1 ↵ ↵ vj = n 1 − · ( a − ↵n 1vj a1↵vj a0vj for all 1 j r and i = n 1. − − −···− − − —4— Norms and traces MAT4250 — Høst 2013 Hence the matrix of ⇢↵ is an n n-matrix (remember m = nr) with r blocks ⇥ along the diagonal, each block being equal to the companion matrix of ↵ (which is an m m-matrix). Computing the characteristic poly gives us ⇥ Proposition 3 Assume that K L is a finite field extension of degree n and let ↵ K ✓ m m 1 2 be an element whose minimal polynomial over K is Q(t)=t + an 1t − + + a0.