Selected Texts. Contents. Pierre-Yves Gaillard. Monday 26 th June, 2017, 10:36. I put together three short texts I have written.

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1 Selected Texts Pierre-Yves Gaillard Monday 26 th June, 2017, 10:36 I put together three short texts I have written. Contents 1 The Fundamental Theorem of Galois Theory 2 2 Function of a Matrix 4 3 Zorn s Lemma 6 1

2 1 The Fundamental Theorem of Galois Theory Abstract. We give a short and self-contained proof of the Fundamental Theorem of Galois Theory (FTGT) for finite degree extensions. We derive the FTGT (for finite degree extensions) from two statements, denoted (a) and (b). These two statements, and the way they are proved here, go back at least to Emil Artin (precise references are given below). The argument is essentially taken from Chapter II of Emil Artin s Notre Dame Lectures [A]. More precisely, statement (a) below is implicitly contained in the proof Theorem 10 page 31 of [A], in which the uniqueness up to isomorphism of the splitting field of a polynomial is verified. Artin s proof shows in fact that, when the roots of the polynomial are distinct, the number of automorphisms of the splitting extension coincides with the degree of the extension. Statement (b) below is proved as Theorem 14 page 42 of [A]. The proof given here (using Artin s argument) was written with Keith Conrad s help. Theorem Let E/F be an extension of fields, let a 1,..., a n be distinct generators of E/F such that p := (X a 1 ) (X a n ) is in F [X]. Then the group G of automorphisms of E/F is finite, there is a bijective correspondence between the sub-extensions S/F of E/F and the subgroups H of G, and we have S H H = Aut(E/S) S = E H = [E : S] = H, where E H is the fixed subfield of H, where [E : S] is the degree (that is the dimension) of E over S, and where H is the order of H. We claim: Proof (a) If S/F is a sub-extension of E/F, then [E : S] = Aut(E/S). (b) If H is a subgroup of G, then H = [E : E H ]. Proof that (a) and (b) imply the theorem. Let S/F be a sub-extension of E/F and put H := Aut(E/S). Then we have trivially S E H, and (a) and (b) imply [E : S] = [E : E H ]. 2

3 Conversely let H be a subgroup of G and set H := Aut(E/E H ). trivially H H, and (a) and (b) imply H = H. Then we have Proof of (a). Let 1 i n. Put K := S[a 1,..., a i 1 ] and d := [K[a i ] : K]. It suffices to check that any F -embedding φ of K in E has exactly d extensions to an F -embedding Φ of K[a i ] in E. Let q K[X] be the minimal polynomial of a i over K. It is enough to verify that φ(q) (the image under φ of q) has d distinct roots in E. But this is clear since q divides p, and thus φ(q) divides φ(p) = p. Proof of (b). In view of (a) it is enough to check H [E : E H ]. Let k be an integer larger than H, and pick a b = (b 1,..., b k ) E k. We must show that the b i are linearly dependent over E H, or equivalently that b (E H ) k is nonzero, where denotes the vectors orthogonal to in E k with respect to the dot product on E k. Any element of b (E H ) k is necessarily orthogonal to hb for any h H, so b (E H ) k = (Hb) (E H ) k, where Hb is the H-orbit of b. We will show that (Hb) (E H ) k is nonzero. Since the span of Hb in E k has E-dimension at most H < k, (Hb) is nonzero. Choose a nonzero vector x in (Hb) such that x i = 0 for the largest number of i as possible among all nonzero vectors in (Hb). Some coordinate x j is nonzero in E, so by scaling we can assume x j = 1 for some j. Since the subspace (Hb) in E k is stable under the action of H, for any h in H we have hx (Hb), so hx x (Hb). Since x j = 1, the j-th coordinate of hx x is 0, so hx x = 0 by the choice of x. Since this holds for all h in H, x is in (E H ) k. [A] Emil Artin, Galois Theory, Lectures Delivered at the University of Notre Dame, Chapter II: 3

4 2 Function of a Matrix Abstract. Let a be a square matrix with complex entries and f a function holomorphic on an open subset U of the complex plane. It is well known that f can be evaluated on a if the spectrum of a is contained in U. We show that, for a fixed f, the resulting matrix depends holomorphically on a. The following was explained to me by Jean-Pierre Ferrier, and Ahmed Jeddi made useful comments. For any matrix a in A := M n (C), write Λ(a) for the set of eigenvalues of a. For each λ in Λ(a), write 1 λ C[a] for the projector onto the λ-generalized eigenspace parallel to the other generalized eigenspaces, and put a λ := a1 λ, and z λ := z1 λ for z in C. Let U be an open subset of C, let U be the subset of A, which is open by Rouché s Theorem, defined by the condition Λ(a) U, let a be in U, let X be an indeterminate, let O(U) be the C-algebra of holomorphic functions on U, and equip O(U) and C[a] with the C[X]-algebra structures associated respectively with the element z z of O(U) and the element a of C[a]. Theorem 1. (i) There is a unique C[X]-algebra morphism from O(U) to C[a]. We denote this morphism by f f(a). (ii) The map U a f(a) A is holomorphic. (iii) For any a in U we have f(a) = λ Λ(a),0 k<n f (k) (λ) λ (a λ) k. Proof of (i) and (iii). By the Chinese Remainder Theorem, C[a] is isomorphic to the product of C[X]-algebras of the form C[X]/(X λ) m, with λ C. So we can assume that C[a] is of this form, and the lemma follows from the fact that, for any f in O(U), there is unique g in O(U) such that for all z in U. q.e.d. f(z) = m 1 k=0 f (k) (λ) (z λ) k + (z λ) m g(z) 4

5 Say that a cycle is a formal finite sum of smooth closed curves. Let γ be a cycle in U \ Λ(a) such that I(γ, λ) = 1 for all λ Λ(a) (where I(γ, λ) is the winding number of γ around λ), and let N be the set of those b in A such that Λ(b) U, and that γ is a cycle in U \ Λ(b) satisfying I(γ, λ) = 1 for all λ Λ(b). As already observed, Rouché s Theorem implies that N is an open neighborhood of a in A. Theorem 2 below will imply Part (ii) of Theorem 1. Theorem 2. We have f(b) = 1 2πi γ f(z) z b dz for all f in O(U) and all b in N. In particular the map b f(b) from U to A is holomorphic. Proof. We have 1 f(z) 2πi γ z b dz = 1 f(z) 1 λ dz 2πi γ z b = 1 λ f(z) 2πi γ z b dz = 1 λ f(z) dz 2πi (z λ) (b λ) = = ( ) = 1 λ 2πi,0 k<n γ,0 k<n = 1 λ 2πi γ,0 k<n f(z) γ n 1 k=0 (b λ) k dz (z λ) k+1 f(z) dz (b λ)k (z λ) k+1 I(γ, λ) f (k) (λ) λ f (k) (λ) λ ( ) = f(b), (b λ) k (b λ) k where Equality ( ) follows from the Residue Theorem, and Equality ( ) from Part (iii) of Theorem 1. q.e.d. 5

6 3 Zorn s Lemma Zorn s Lemma Let P be a poset all of whose well-ordered subsets have an upper bound. Then P has a maximal element. Proof. We assume that P has no maximal element and we try to get a contradiction. For any well-ordered subset W of P there is an element p(w ) in P >W, that is, an element p(w ) in P such that p(w ) > w for all w in W. Indeed, W has an upper bound b, and, b being not maximal, there is a p(w ) such that p(w ) > b. Let W be the set of those well-ordered subsets W of P such that p(w <w ) = w (self-explanatory notation) for all w in W. If W is in W, then, clearly, so is W {p(w )}. Thus W (ordered by inclusion) has no maximal element. We will reach the contradiction we are after by showing that W does have a maximal element, namely the union of W, that is, the set of all elements of P which belong to some member of W. We denote this set by U. Obviously, it suffices to show (a) U is in W. For any pair I S of subsets of P, say that I is an initial segment of S if S s < i I implies s I. To prove (a) we will need: (b) If W and X are in W, then W is an initial segment of X, or X is an initial segment of W. This implies immediately that W and X are initial segments of W X, and that U is totally ordered. (c) Any W in W is an initial segment of U. To prove (b) let I be the set of those elements of P which belong to some initial segment common to W and X. We claim (d) I is an initial segment of W. Let W w < i I and let us show w I. There is an initial segment J common to W and X containing i, and thus w. As J I, this implies w I, proving (d). In particular I is well-ordered and p(i) is defined. Setting I := I {p(i)}, we claim (e) if I W, then I is an initial segment of W. 6

7 Let W w < i I and let us show w I. We can assume i = p(i). We have W <w0 = I, where w 0 is the least element of W \ I. This implies w 0 = p(w <w0 ) = p(i) = i, and thus w I, proving (e). Now (e) and (e) with X instead of W imply (b). To prove (c) let U u < w W and let us verify that u is in W. We have u X for some X in W. By (b) W is an initial segment of W X. This entails u W. We prove (a). To check that U is well-ordered, let A be a nonempty subset of U, choose a W in W which meets A, let m be the minimum of W A, and let a be in A. It suffices to show m a. If such was not the case, (b) would imply a < m, then (c) would imply a W, in contradiction with the minimality of m. To verify that p(u <u ) = u for u in U, let W be a member of W containing u. It suffices to show U <u = W <u. The inclusion W <u U <u being obvious, it is enough to check U <u W. If v is in U <u, then v is in W by (c). This completes the proof of (a), and thus, as explained above, the proof of Zorn s Lemma. QED Murray Eisenberg pointed out to me the article of Hellmuth Kneser, in which the above argument seems to appear for the first time, and the text dan/shortproofs/zorn.pdf of Daniel Grayson. 7

ZEROS OF SPARSE POLYNOMIALS OVER LOCAL FIELDS OF CHARACTERISTIC p

ZEROS OF SPARSE POLYNOMIALS OVER LOCAL FIELDS OF CHARACTERISTIC p BJORN POONEN 1. Statement of results Let K be a field of characteristic p > 0 equipped with a valuation v : K G taking values in an ordered