# 4 Étale algebras, norm and trace, Dedekind extensions - PDF

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Number theory I Lecture #4 Fall /22/ Étale algebras, norm and trace, Dedekind extensions 4.1 Separability We recall some standard facts about separable and inseparable field extensions

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Number theory I Lecture #4 Fall /22/ Étale algebras, norm and trace, Dedekind extensions 4.1 Separability We recall some standard facts about separable and inseparable field extensions and define the more general notion of an étale algebra (or separable algebra). This is optional background material that may be skipped by those already familiar with it. Definition 4.1. A nonzero polynomial f over a field K is separable if the zeros of f are distinct in every extension of K; equivalently, gcd(f, f ) = 1. 1 Warning 4.2. Older texts (such as Bourbaki) define a polynomial in K[x] to be separable if all of its irreducible factors are separable (under our definition); so (x 1) 2 is separable under this definition but not under ours. The older definition has the disadvantage that it is not preserved under field extension (for example, a polynomial that is inseparable as an element of K[x] becomes separable when viewed as an element of K[x], since it splits into linear factors in K[x] and every linear polynomial is separable). This discrepancy does not impact the definition of separable element or field extensions. Definition 4.3. Let L/K be an algebraic field extension. An element α L is separable over K if it is the root of a separable polynomial in K[x] (in which case its minimal polynomial is separable). The extension L/K is separable if every α L is separable over K; otherwise it is inseparable. Lemma 4.4. Let K be a field. If f K[x] is irreducible but not separable then f = 0. Proof. Let f K[x] be irreducible, and let L be its splitting field over K. If f is not separable then f and f have a common root α L. The minimal polynomial g K[x] of α over K divides f, but f is irreducible and nonzero, so deg g = deg f. But g also must divide f, and deg f deg f, so f = 0. Corollary 4.5. Let p be the characteristic of the field K and let f K[x] be irreducible. If p = 0 then f is separable, and if p 0 then f is not separable if and only if it is of the form f = g(x p ), where the polynomial g K[x] is uniquely determined by f. Proof. Easy exercise. Corollary 4.6. If char K = 0 then every algebraic extension of K is separable. Corollary 4.7. Let K be a field of characteristic p 0. Every irreducible f K[x] is of the form g(x pn ) for some irreducible separable polynomial g K[x] and integer n 0 that are uniquely determined by f. Proof. We proceed by induction on deg f. If f is separable the corollary holds with g = f and n = 0. Otherwise p 0 and f = g(x p ) for some unique irreducible g K[x] with deg g deg f. By induction on n = deg f, we have g = h(x pn ) for some unique irreducible separable polynomial h K[x] and integer n 0 and then f = h(x pn+1 ). 1 Here f denotes the formal derivative of f in K[x] and the gcd is defined only up to a unit. Lecture by Andrew Sutherland Lemma 4.8. Let L = K(α) be an algebraic extension contained in an algebraic closure K of K and let f K[x] be the minimal polynomial of α over K. Then # Hom K (L, K) = #{β K : f(β) = 0} [L : K], with equality if and only if α is separable over K. Proof. Each element of Hom K (L, K) is uniquely determined by the image of α, which must be a root β of f(x) in K. The number of these roots is equal to [L : K] = deg f precisely when f and therefore α is separable over K. Definition 4.9. Let L/K be a finite extension. The separable degree of L/K is The inseparable degree of f is [L : K] s := # Hom K (L, K). [L : K] i := [L : K]/[L : K] s We will shortly show that [L : K] i is always a positive integer (in fact a power of the characteristic of K), but it follows immediately from the definition that [L : K] = [L : K] s [L : K] i. Remark As with the degree of a field extension (defined as the dimension of a vector space), one can define the separable degree of an arbitrary algebraic extension as the cardinality of the set Hom K (L, K). Multiplication of degrees is then multiplication of cardinals (if S and T are sets the product of their cardinalities is the cardinality of the set Hom(S, T ) = {f : S T }). Lemma 4.8 implies that for simple algebraic extensions L = K(α) we always have [L : K] s [L : K], with equality if and only if α is separable (which we will shortly show is equivalent to L/K being separable). Theorem Let φ K : K Ω be a homomorphism from a field K to an algebraically closed field Ω, and let L/K be algebraic. Then φ K extends to a homomorphism φ L : L Ω. Proof. We use Zorn s lemma. Define a partial ordering on the set F of pairs (F, φ F ) for which F/K is a subextension of L/K and φ F : F Ω extends φ K by defining (F 1, φ F1 ) (F 2, φ F2 ) whenever F 2 contains F 1 and φ F2 extends φ F1. Given any totally ordered subset C of F, let E = {F : (F, φ F ) C} and define φ E : E Ω by φ E (x) = φ F (x) for x F E (this does not depend on the choice of F because C is totally ordered). Then (E, φ E ) is a maximal element of C, and by Zorn s lemma, F contains a maximal element (M, φ M ). We claim that M = L. If not, then pick α L M and consider the field F = M[α] L properly containing M, and extend φ M to ϕ F : F Ω be letting φ F (α) by any root of α M (f) in Ω, where f M[x] is the minimal polynomial of α over M and α M (f) is the image of f in Ω[x] obtained by applying ϕ M to each coefficient. Then (M, φ M ) is strictly dominated by (F, φ F ), contradicting its maximality Fall 2015, Lecture #4, Page 2 Lemma Let L/F/K be a tower of finite extensions. Then # Hom K (L, K) = # Hom K (F, K)# Hom F (L, K). Proof. We decompose L/F/K into a tower of simple extensions and proceed by induction. The result is trival if L = K and otherwise it suffices to consider K F F (α) = L, where K = F in the base case. Theorem 4.11 allows us to define a bijection Hom K (F, K) Hom F (F (α), K) Hom K (F (α), K) that sends (φ 1, φ 2 ) to φ: L K defined by φ F = φ 1 and φ(α) = ( ˆφ 1 ˆφ2 ˆφ 1 1 )(α), where ˆφ 1, ˆφ 2 Aut(K) denote arbitrary extensions of φ 1, φ 2 to K; note that φ(α) does not depend on these choices and is a root of φ(f), where f F [x] is the minimal polynomial of α and φ(f) is its image in φ(f )[x]. The inverse bijection is φ 1 = φ F and φ 2 (α) = ( ˆφ ˆφ 1 )(α). Corollary Let L/F/K be a tower of finite extensions. Then [L : K] s = [L : F ] s [F : K] s [L : K] i = [L : F ] i [F : K] i Proof. The first equality follows from the lemma and the second follows from the identities [L : K] = [L : F ][F : K] and [L : K] = [L : K] s [L : K] i. Corollary Let L = K(α) be an algebraic extension. Then L/K is separable if and only if α is separable over K, equivalently, if and only if [L : K] s = [L : K]. Proof. If L/K is separable then α is separable over K and [L : K] s = [L : K], by Lemma 4.8. Now suppose α is separable; then [L : K] s = [L : K], by Lemma 4.8. For any β L we can write L = K(β)(α), and note that α is separable over K(β), since its minimal polynomial over K(β) divides it minimal polynomial over K, which is separable. So we also have [L : K(β)] s = [L : K(β)]. The equalities [L : K] = [L : K(β)][K(β) : K] [L : K] s = [L : K(β)] s [K(β) : K] s then imply [K(β) : K] s = [K(β) : K], so β is separable over K, by Lemma 4.8. Corollary Let L/K be a finite extension. Then [L : K] s [L : K] with equality if and only if L/K is separable. Proof. We have alredy established this for simple extensions, and otherwise we my decompose L/K into a finite tower of simple extensions and proceed by induction on the number of extensions, using the previous two corollaries at each step. Corollary If L/F/K is a tower of finite extensions with L/F and F/K separable then L/K is separable. Proof. This follows from Corollaries 4.13 and Corollary If L/F/K is a tower of algebraic extensions with L/F and F/K separable, then L/K is separable. ˆφ Fall 2015, Lecture #4, Page 3 Proof. Let β L. If β F the β is separable over K, since F/K is separable. Otherwise, β is separable over F and we may consider the subextension M/K of F/K generated by the coefficients of the minimal polynomial f F [x] of β over F. This is a finite separable extension of K, and M(β) is also a finite separable extension of M, since the minimal polynomial of β over M(β) is f, which is separable. By the previous corollary, M(β), and therefore β, is separable over K. Corollary Let L/K be an algebraic extension, and let Then F is a separable field extension of K. F = {α L : α is separable over K}. Proof. This is clearly a field, since if α and β are both separable over K then K(α) and K(α, β) are separable extensions of K (by the previous corollary), thus every element of K(α, β), including αβ and α + β, is separable over K and lies in F. The field F is then separable by construction. The field F in the corollary is the maximal separable extension of K in L. When L = K it is called a separable closure of K and denoted K sep. If K has characteristic zero then K sep = K is algebraically closed. This also holds when K is a perfect field. Definition A field K is perfect if every algebraic extension of K is separable. All fields of characteristic zero are perfect, as are all finite fields. Theorem Every finite field is a perfect field. Proof. It suffices to consider a finite field of prime order F p, since every finite field is an algebraic extension of its prime field, and any algebraic extension of a perfect field is perfect. Let f F p [x] be irreducible, and use Corollary 4.7 to write f(x) = g(x pn ) with g F p [x] irreducible and separable, and n 0. If n 0 then f(x) = g(x pn ) = g(x pn 1 ) p, since h(x p ) = h(x) p for any h F p [x], but this contradicts the irreducibility of f. So n = 0 and f = g is separable. Definition A field K is separably closed if K has no nontrivial finite separable extensions. Equivalently, K is equal to its separable closure in any algebraic closure. Definition An algebraic extension L/K is purely inseparable if [L : K] s = 1. Remark The trivial extension K/K is both separable and purely inseparable. Example If K = F p (t) and L = F p (t 1/p ) then L/K is a purely inseparable extension of degree p. Proposition Let K be a field of characteristic p 0. If L/K is purely inseparable of degree p then L = K(a 1/p ) for some a K\K p ; equivalently, L K[x]/(x p a). Proof. Every α L\K is inseparable over K, and by Corollary 4.5 its minimal polynomial over K is of the form f(x) = g(x p ) with f monic. We have 1 deg f [L : K] = p, so g(x) must be a monic polynomial of degree 1, which we can write as g(x) = x a. Then f(x) = x p a, and we must have a K p since f is irreducible. We have [L : K(α)] = 1, so L = K(α) K[x]/(x p a) as claimed Fall 2015, Lecture #4, Page 4 Theorem Let L/K be an algebraic extension and let F be the maximal separable extension of K in L. Then L/F is purely inseparable. Proof. If L = F the theorem holds, so we assume otherwise and let p 0 be the characteristic of K. Fix an algebraic closure K of K that contains L. Let α be an element of L that is not in F and let f be its minimal polynomial over F. Use Corollary 4.7 to write f(x) = g(x pn ) with g F [x] irreducible and separable, and n 0. We must have deg g = 1, since otherwise the roots of g would be separable over F and therefore over K but not in F. Thus f(x) = x pn c for some c F (since f is monic and deg g = 1). Since we are in characteristic p 0, we can factor f in F (α)[x] as f(x) = x pn α pn = (x α) pn. There is thus only one F -homomorphism from F (α) to K. The same statement applies to any extension of F obtained by adjoining any set of elements of L (even an infinite set). Therefore # Hom F (L, K) = 1, so [L : F ] s = 1 and L/F is purely inseparable. Corollary Every algebraic extension L/K can be decomposed into a tower of extensions L/F/K with F/K separable and L/F purely inseparable. Corollary The inseparable degree of any finite extension is a power of the characteristic. Proof. This follows from the proof of the theorem above. 4.2 Étale algebras Definition An étale K-algebra is a (necessarily commutative) K-algebra that is isomorphic to a finite product of separable extensions of K. A finite étale K-algebra is a K-algebra that is isomorphic to a finite product of finite separable extensions of K. By the dimension of an étale K-algebra we mean its dimension as a K-vector space. Every separable field extension L/K is an étale K-algebra, and if an étale K-algebra L is a field, then it is necessarily isomorphic to a separable extension of K. An étale K- algebra L need not be a field, but every α L is separable (note that when L is not a field the minimal polynomial of α need not be irreducible, but it will be separable). Example If K is a separably closed field then every étale K-algebra of dimension n is isomorphic to K n = K K. Our main interest in étale algebras is that they naturally arise via base change, a notion we now recall (this is not the most general definition but suffices for our purposes). Definition Let ϕ: A B be a ring homomorphism (so B is an A-module), and let M be an A-module. The tensor product of A-modules M A B is a B-module (with multiplication defined by b(m b ) := m bb ) called the base change (or extension of scalars) of M from A to B. If M is an A-algebra then its base change to B is a B-algebra. Remark Each ϕ: A B determines a functor from the category of A-modules to the category of B-modules via base change. It has an adjoint functor called restriction of scalars which, given a B-module M turns it into an A-module by the rule am = ϕ(a)m Fall 2015, Lecture #4, Page 5 The ring homomorphism ϕ: A B will often be an inclusion, in which case we have a ring extension B/A (we may also take this view in whenever ϕ is injective, which is necessarily the case if A is a field). We are specifically interested in the case where B/A is a field extension M is a finite étale A-algebra. We first recall the primitive element theorem. Theorem Let L/K be a finite separable extension. Then L = K(α) for some α L; equivalently L K(x)/(f) for some monic, irreducible, separable f K[x]. Proof. See [1, 15.8] or [4, V.7.4]. Proposition Suppose L is a finite étale K-algebra and K /K is any field extension. Then L K K is a finite étale K -algebra of the same dimension as L. Proof. Without loss of generality we assume that L is actually a field; if not apply the following argument to each factor of L. By the primitive element theorem, L K[x]/(f) for some irreducible separable polynomial f K[x]. Suppose f = f 1 f 2 f m is the irreducible factorization of f in K [x]. Each f i is necessarily separable (no repeated roots over any extension), since f is. We have an isomorphism of K -algebras L K K K [x]/(f), and by the Chinese Remainder Theorem, K [x]/(f) i K [x]/(f i ). Each field K [x]/(f i ) is a finite separable extension of K, thus L K K is a finite étale algebra over K. We have dim L = deg f = dim K [x]/(f), so the dimension is clearly preserved. Example Any finite dimensional real vector space V is a finite étale R-algebra (with coordinate-wise multiplication with respect to some basis); the complex vector space V R C is then a finite étale C-algebra of the same dimension. Note that even when an étale K-algebra L is a field, the base change L K K will often not be a field. For example, if K = Q and L Q is a number field, then the base L K C will never be a field, it will be isomorphic to a C-vector space of dimension [L : K] 1. We record the following corollary of Proposition 4.34, which is implied by its proof. Corollary Let L K[x]/(f) be a finite separable extension of a field K defined by an irreducible separable polynomial f K[x]. Let K /K be any field extension, and let f = f 1 f m be the factorization of f into distinct irreducible polynomials f i K [x]. We have an isomorphism of étale K -algebras L K K i K [x]/(f i ) where each K [x]/(f i ) is a finite separable field extension of K. Proposition Suppose L is a finite étale K-algebra and Ω is a separably closed field extension of K.Then L K Ω Ω. σ Hom K (L,Ω) In particular, the map L K Ω Ω sends y 1 to (σ(y)) σ for each y L. Proof. Without loss of generality we may assume L = K[x]/(f) is a simple field extension. Then f factors as f(x) = (x α 1 ) (x α n ) over Ω, with the α i are distinct. We have a bijection between Hom K (K[x]/(f), Ω) and the set {α i }: each σ Hom K (K[x]/(f), Ω) Fall 2015, Lecture #4, Page 6 is determined by σ(x) {α}, and for each α i, the map x α i determines a K-algebra homomorphism σ i Hom K (K[x]/(f), Ω). As in the proof of Proposition 4.34 we have Ω-algebra homomorphisms given by the maps K[x] (f) K Ω Ω[x] (f) α i Ω[x] (x α i ) σ i Ω i. x 1 x (α i ) i (σ i (x)) i. The element x 1 generates L K Ω as an Ω-algebra, and has image (σ i (x)) i in σ i. It follows that y 1 (σ i (y)) i for every y L. Remark The proof of Proposition 4.37 does not actually require Ω to be separably closed, we only needed f(x) to split into linear factors in Ω[x]. Thus the proposition holds whenever all the irreducible polynomials f K[x] for which the field K[x]/(f) is a isomorphic to one of the finite separable field extensions of K whose product is L split completely in Ω[x] (for example, when L is a field, one could take Ω to be its normal closure). Example Let L/K = Q(i)/Q and Ω = C. We have Q(i) Q[x]/(x 2 + 1) and Q(i) Q C Q[x] x Q C C[x] x C[x] x i C[x] x + i C C. As C-algebra isomorphisms, the corresponding maps are determined by i 1 x 1 x (x, x) (i, i) (i, i). Taking the base change of Q(i) to C lets us see the two distinct embeddings of Q(i) in C, which are determined by the image of i. Note that (i 1) 2 = i = 1 1 = (1 1) and (i, i) 2 = ( 1, 1) = (1, 1), so we also have 1 1 (1, 1). Thus as an isomorphism of C- vector spaces, the C-basis (1 1, 1 i) for Q(i) Q C is mapped to the C-basis ( (1, 1), (i, i) ) for C C. It follows that for any (α, β) C C, the inverse image of (α, β) = α + β 2 in Q(i) C under this isomorphism is α + β 2 (1, 1) + α β (i, i) 2i (1 1) + α β (i 1) = 1 α + β 2i 2 + i α β. 2i Now R/Q is an extension of rings, so we can also consider the base change of the Q-algebra Q(i) to R. But note that R is not separably closed and in particular, it does not contain a subfield isomorphic to Q(i), thus Proposition 4.37 does not apply. Indeed, as an R-module, we have Q(i) Q R R 2, but as an R-algebra, Q(i) Q R C R Norms and traces We now introduce the norm and trace of a finite extension B/A. These are often defined only for field extensions, but in fact the same definition works without modification whenever B is a free A-module of finite rank. One can generalize further to projective modules (with some restrictions), but this is somewhat more involved and not needed here Fall 2015, Lecture #4, Page 7 Definition Let B/A be ring extensions in which B is a free A-module of finite rank. The (relative) norm N B/A (b) and trace T B/A (b) of b (down to A) are the determinant and trace of the A-linear multiplication-by-b map B B defined by x bx. As a special case, note that if B/A is a finite extension of fields, then B is an A-vector space of finite dimension, hence a free A-module of finite rank. In practice one computes the norm and trace by picking a basis for B as an A-module and computing the matrix of the multiplication-by-b map with respect to this basis; this is an n n matrix with entries in A whose determinant and trace do not depend on the choice of basis. It follows immediately from the definition that N B/A is multiplicative and T B/A is additive. Moreover, the norm map N B/A defines a group homomorphism from B to A and the trace map T B/A defines a group homomorphism from B to A (as additive groups). Example Consider A = R and B = C, which has the A-module basis {1, i}. For b = 2 + 3i the matrix of B b B with respect to this basis can be written as ( ) , thus ( ) 2 3 N C/R (2 + 3i) = det = 13, 3 2 ( ) 2 3 T C/R (2 + 3i) = tr =
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