Coefficient fields and scalar extension in positive characteristic

arXiv:math/0411323v2 [math.AC] 19 May 2005 Coefficient fields and scalar extension in positive characteristic M. Fern´andez-Lebr´on, L. Narv´aez-Macarro∗ Department of Algebra, Faculty of Mathematics University of Seville c/ Tarfia s/n, 41012 Sevilla, Spain E-mail: [email protected]s, narv[email protected] Abstract Let kbe a perfect field of positive characteristic, k(t)per the perfect closure of k(t) and A=k[[X1,...,Xn]]. We show that for any maximal ideal nof A′=k(t)per ⊗kA, the elements in c A′ nwhich are annihilated by the “Taylor” Hasse-Schmidt derivations with respect to the Xiform a coefficient field of c A′ n. Keywords: Complete local ring; Coefficient field; Hasse-Schmidt derivation. 2000 Mathematics Subject Classification: 13F25, 13N15, 13B35, 13A35. Introduction Let kbe a perfect field, k(∞)=k(t)per the perfect closure of k(t) and A=k[[X1,...,Xn]]. If kis of characteristic 0, then k(∞)=k(t) and A(t) = A⊗kk(t) is obviously noetherian. Actually, A(t) is an n-dimensional regular non-local ring (see Example (2.3)) whose maximal ideals have the same height (= n). In [8] the second author proved that there is a uniform way to obtain a coefficient field in the completions \ (A(t))n, for all maximal ideals nin A(t). Namely, the elements in \ (A(t))nwhich are annihilated by the partial derivatives ∂ ∂Xiform a coefficient field of \ (A(t))n. In this paper, we generalize the above result to the positive characteristic case. At first sight, in positive characteristic it seems natural to consider Hasse-Schmidt derivations instead of usual derivations (see [4, Theorem 3.17]), but Example (2.3) shows that the question is not so clear. Consequently, in the characteristic p > 0 case we take the scalar extension k→k(∞) instead of k→k(t), but a new problem appears: it is not obvious that the ring A(∞)= A⊗kk(∞)is noetherian. We have proved that result in [3]. ∗Both authors are partially supported by MTM2004-07203-C02-01 and FEDER. The main result in this paper says that, for every maximal ideal nin A(∞), the elements in \ (A(∞))nwhich are annihilated by the “Taylor” Hasse-Schmidt derivations with respect to the Xiform a coefficient field of \ (A(∞))n. Let us now comment on the content of this paper. In Section 1 we introduce our basic notations and recall some results, mainly from [3]. In Section 2 we prove our main result and give the (counter)Example (2.3). In the Appendix we give a complete proof of the Normalization Lemma for power series rings over perfect fields, which is an important ingredient in the proof of Theorem (2.1) and that we have not found in the literature. Our proof closely follows the proof in [1], but the latter works only for infinite perfect fields. 1 Preliminaries and notations All rings and algebras considered in this paper are assumed to be commutative with unit element. If Bis a ring, we shall denote by dim(B) its Krull dimension and by Ω(B) the set of its maximal ideals. We shall use the letters K, L, k to denote fields and Fpto denote the finite field of pelements, for a prime number p. If p∈Spec(B), we shall denote by ht(p) the height of p. Remember that a ring Bis said to be biequidimensional if all its saturated chains of prime ideals have the same length. If Bis an integral domain, we denote by Qt(B) its quotient field. If kis a ring and Bis a k-algebra, the set of all derivations (resp. of all Hasse-Schmidt derivations) of Bover k(cf. [5] and [6], §27) will be denoted by Derk(B) (resp. HSk(B)). Now, we recall the notations and some results of [3] which are used in this paper. For any Fp–algebra B, we denote B♯:= \ e≥0 Bpe. Let kbe a field of characteristic p > 0 and consider the field extension k(∞):= [ m≥0 kt1 pm⊃k(t). If kis perfect, k(∞)coincides with the perfect closure of k(t). For each k-algebra A, we denote A(t) := k(t)⊗kA. For the sake of brevity, we will write tm=t1 pmand denote A(m):= A(tm) := A⊗kk(tm) = A(t)⊗k(t)k(tm), A[m]:= A[tm], A(∞):= A⊗kk(∞)=[ m≥0 A(m), A[∞]:= [ m≥0 A[tm]. Each A(m)(resp. A[m]) is a free module over A(t) (resp. over A[t]) of rank pm. 2 For each prime ideal Nof A(∞)we denote N[∞]:= N∩A[∞],N[m]:= N∩A[m]and N(m):= N∩A(m). Similarly, if Pis a prime ideal of A[∞]we denote P[m]:= P∩A[m]. (1.1) We have the following properties [3, 4, 8]: (i) N=[ m≥0 N(m),N[∞]=[ m≥0 N[m], (resp. P=[ m≥0 P[m]). (ii) N(n)∩A(m)=N(m)and N[n]∩A[m]=N[m]for all n≥m(resp. P[n]∩A[m]=P[m]for all n≥m). (iii) The following conditions are equivalent: (a) Nis maximal (resp. Pis maximal). (b) N(m)(resp. P[m]) is maximal for some m≥0. (c) N(m)(resp. P[m]) is maximal for all m≥0. (iv) ht(N) = ht(N[∞]) = ht(N(m)) = ht(N[m]) for all m≥0. Moreover, dim(A(∞)) = dim(A(m)). (v) [8, Proposition (1.4) and Theorem (1.6)] Let us assume that Ais noetherian and that for every maximal ideal mof A, the residue field A/mis algebraic over k. Then for every m≥0 we have dim(A(∞)) = dim(A(m)) = dim(A(t)). Moreover, if Ais biequidimensional, universally catenarian of Krull dimension n, then every maximal ideal of A(∞)(or of A(m)) has height n. (vi) [3, Proposition 2.2] If kis perfect and B=k[[X1,...,Xn]], then Qt(B)♯=k. (vii) [3, Proposition 3.4] If kis perfect, Ais an integral k-algebra, K=Qt(A) and K♯is algebraic over k, then any prime ideal P∈Spec(A[∞]) with P∩k[t] = 0 and P∩A= 0 is the extended ideal of some P[m0],m0≥0. (viii) [3, Corollary 3.10] If kis perfect, Ais noetherian and for every maximal ideal mof A, the residue field A/mis algebraic over k, then A(∞)is also noetherian. In particular k[[X1,...,Xn]](∞)is noetherian. (ix) [6, Theorem 30.6] Let (R, m) be an equicharacteristic n-dimensional regular local ring containing a quasi-coefficient field k0, and D1,...,Dn∈Derk0(R), a1,...,an∈R such that Di(aj) = δij. Then, Derk0(R) is a free R-module with basis {D1,...,Dn}. (x) [4, Theorem 3.17] Let (R, m) be an equicharacteristic n-dimensional regular local ring containing a quasi-coefficient field k0, and let D1,...,Dn∈HSk0(R) such that their degree 1 components {D1 1,...,Dn 1}form a basis of Derk0(R). Let b D1,..., b Dnbe the extensions of D1,...,Dnto b R. Then, the set {a∈b R|b Dj i(a) = 0 ∀j= 1, . . . , n, i ≥1} is a coefficient field of b R(the only one containing k0). 3 (1.2) Taylor expansions (cf. [7]). Let n≥1 be an integer. We write X= (X1,...,Xn), T= (T1,...Tn), X+T= (X1+T1,...,Xn+Tn) and, for α∈Nn,Xα=Xα1 1···Xαn n. Let Abe the formal power series ring k[[X]] (or the polynomial ring k[X]). For any f(X) = X α∈Nn λαXα∈Awe define ∆(α)(f(X)) by: f(X+T) = X α∈Nn ∆(α)(f(X))Tα. One has ∆(α)(f·g) = X β+σ=α ∆(β)(f)∆(σ)(g) (1) and α!∆(α)= ( ∂ ∂X1)α1···(∂ ∂Xn)αn. For i∈N, 1 ≤j≤nand α= (0,..., j ⌣ i,...,0) we denote ∆j i= ∆((0,..., j ⌣ i,...,0)). From (1) we obtain ∆j i(f·g) = X r+s=i ∆j r(f)∆j s(g), i.e. the sequences ∆j:= (1A,∆j 1,∆j 2,...), 1 ≤j≤n, are Hasse-Schmidt derivations of A (over k) (cf. [6], §27). Now, let us recall the following basic well known result (cf. [2, Propositions 5.5.3 and 5.5.6]). (1.3) Proposition. Let Bbe a noetherian ring, Pbe a prime ideal of B[t]and p=P∩B. Then, one of the following conditions holds: (a) P=p[t],ht(P) = ht(p)and B[t]/P ≃(A/p)[t]. (b) P⊃p[t],ht(P) = ht(p) + 1 and B[t]/P is an algebraic extension of B/p(generated by tmod P). 2 Coefficients fields and the extension k→k(∞). Let kbe a perfect field of characteristic p > 0 and Aak-algebra. For every Hasse-Schmidt derivation D∈HSk(A), we also denote by D∈HSk(∞)(A(∞)) the extended Hasse-Schmidt derivation. If n⊂A(∞)is a maximal ideal, we denote by Dnand c Dnthe extended Hasse- Schmidt derivations to (A(∞))nand \ (A(∞))n, respectively. The following theorem generalizes Theorem 2.3 of [8] to the positive characteristic case. (2.1) Theorem. Let kbe a perfect field of positive characteristic p > 0,A=k[[X1,...,Xn]] the power series ring and let us consider the Hasse-Schmidt derivations ∆j∈HSk(A), j= 1,...,n, defined in (1.2). Then, for each maximal ideal n⊂A(∞)the set K0=a∈\ (A(∞))n|\ (∆j i)n(a) = 0 ∀j= 1,...,n;∀i≥1 4 is a coefficient field of the complete local ring \ (A(∞))n. Proof. We proceed in two steps, as in the proof of Theorem 2.3 of [8]: reduction to the case n= 1 and treatment of this case. Step 1: the reduction. Let us write P=n∩A[∞],p=n∩A=P∩A=P(m)∩A. From (1.1) (iii), (iv) we know that the ideals n(m)are maximal and ht(n(m)) = ht(n) for all m≥0. By Remark (1.8) of [8], there are only two possibilities for the prime ideal p: (i) ht(p) = n, and then p= (X1,...,Xn) and n=pe. (ii) ht(p) = n−1. In case (i), k(∞)is a coefficient field of (A(∞))nas well as of its completion, and \ (∆j i)n(k(∞)) = 0 for every j= 1,...,n,i≥1. The theorem is then a consequence of (1.1) (ix), (x). Let us suppose we are in case (ii). By Theorem (A.6) (Normalization Lemma) there exists a new set of variables X′ 1,...,X′ nin Asuch that •p∩k[[X′ 1]] = (0), •k[[X′ 1]] ֒→A/pis a finite extension, and since A/pis finitely generated over k[[X′ 1]], A/pis a finite k[[X′ 1]]–module, •k((X′ 1)) ֒→Qt(A/p) is a separable finite extension. Since the Hasse-Schmidt derivations of Aover kwith respect to the variables X′ ican be expressed in terms of the ∆j([4, Theorem 2.8]), we can suppose X′ i=Xi. Let us write K=A(∞)/n= Qt A[∞]/P,R=A/p,A′=k[[X1]], n′=n∩A′ (∞), P′=P∩A′ [∞]=n′∩A′ [∞]and K′=A′ (∞)/n′= Qt A′ [∞]/P′. We have R[m]=A[m] pA[m] ,R[∞]=A[∞] pA[∞] ,K=[ m≥0 A(m) n(m) and K′=[ m≥0 A′ (m) n′ (m) . Let us consider the following commutative diagram of inclusion A′[t]/P′ [0] ր ց A′A[t]/P[0]. ց ր R=A/p The bottom inclusions are algebraic (Ris a finite A′-module and P[0] ∩A=p), hence the top ones must be so. In particular A′[t]/P′ [0] is algebraic over A′, which implies (Proposition (1.3)) that P′ [0] 6= 0, then n′ (0) 6= 0 and n′6= 0. Therefore n′is maximal since dim(A′) = 1. 5 Let us show that the inclusion K′⊂Kis separable algebraic. For that, it is enough to prove that the extensions A′ (m) n′ (m) ⊂A(m) n(m) are finite and separable. Let us write L′= Qt(A′) = k((X1)), L= Qt(A/p) and consider the following diagram of field extensions L′= Qt(A′)⊂Qt A′ [m] P′ [m]!=A′ (m) n′ (m) ∩ ∩ L= Qt(R)⊂Qt A[m] P[m]=A(m) n(m) . These extensions satisfy the following properties: i) L′⊂Lis finite and separable. Hence, there is a primitive element e,L=L′[e], whose minimal polynomial f(X)∈L′[X] satisfies f′(X)6= 0. ii) By Proposition (1.3), the extensions L⊂Qt A[m] P[m],L′⊂Qt A′ [m] P′ [m]!are finite and generated by the class tof t. Therefore, A(m) n(m) = Qt A[m] P[m]=L[t] = L′[e][t] = Qt A′ [m] P′ [m]!![e] = A′ (m) n′ (m)![e] and the extension A′ (m) n′ (m) ⊂A(m) n(m) is finite and separable for all m≥0. Hence, K′⊂Kis separable algebraic. Let us assume that the theorem is proved for n= 1. Then K′ 0=na∈\ (A′ (∞))n ′|\ (∆1 i)n ′(a) = 0 ∀i≥1o is a coefficient field of \ (A′ (∞))n ′. We can consider K′ 0as a subfield of \ (A(∞))nvia the inclusion \ (A′ (∞))n ′֒→\ (A(∞))n. Since K′ 0 ∼ −→ K′and K′⊂Kis separable algebraic, we deduce that K′ 0is a quasi-coefficient field of \ (A(∞))n. It is clear that for all a∈K′ 0 \ (∆j i)n(a) = 0 ∀j= 1,...,n, ∀i≥1. 6 In particular, the \ (∆j)nare Hasse-Schmidt derivations over K′ 0, and by (1.1) (ix), the {∆1 1,...,∆n 1}form a basis of DerK′ 0(\ (A(∞))n). Now, by applying (1.1) (x), we obtain that a∈\ (A(∞))n|\ (∆j i)n(a) = 0 ∀j= 1,...,n, ∀i≥1 is a coefficient field of \ (A(∞))nand the theorem is proved. Step 2: the case n=1. Let us write A=k[[X]], L= Qt(A) = k((X)) and let nbe a maximal ideal of A(∞)=A⊗kk(∞). Let us denote P=n∩A[∞]. By (1.1) (iv), we know that ht(n) = ht(n(m)) = ht(P[m]) = ht(P) = 1. As in the first step, we focus on the case n∩A= (0) (and then P∩A= (0)). Since each A[m]=A[tm] is a unique factorization domain and each P[m]is a prime ideal of A[m]of height 1, P[m]is generated by an irreducible polynomial Fm(tm)∈A[tm] of degree d≥1 and with some non-constant coefficient, since P[m]∩k[tm] = (0). By irreducibility, at least one of the coefficients of Fm(tm) must be a unit, so we may assume that it is 1. Let us write K=A(∞) nand Km=A(m) n (m). Since A(∞)and A(m)are localizations of A[∞] and A[m]respectively, it follows that K=A(∞) n= Qt A[∞] P, Km=A(m) n(m) = Qt A[m] P[m]. The minimal polynomial of θm:= tmmod P[m]over Lis Fm(tm). We have Km= L[θm], K=[ m≥0 Km=[ m≥0 L[θm] = L[θ0, θ1, θ2,...], where θm=θp m+1, and the inclusion k(∞)֒→Kis a k-morphism which sends each tmonto θm. By (1.1) (vi), it follows that L♯=k((X))♯=k, and we can apply (1.1) (vii) to conclude that there exists m0≥0 such that Pis the extended ideal of P[m0]= (Fm0(tm0)). Then, P(resp. n) is the ideal of A[∞](resp. of A(∞)) generated by µ=Fm0(tm0). Moreover, for every j≥1, P[m0+j]is the extended ideal of P[m0]and some of the coefficients of µis not a p–th power. Hence, we can take Fm0+j(tm0+j) = Fm0(tm0) = Fm0(tpj m0+j), j ≥1. Since k(∞)is perfect, the field extension k(∞)⊂Kis separable and, by Cohen structure theorem, there exists a k(∞)–isomorphism ϕ:\ (A(∞))n ∼ →K[[s]] (2) 7 which induces the identity on residue fields and sends the regular parameter µof \ (A(∞))n onto s. One has: ϕ(µ) = s ϕ(tm) = θm ϕ(X) = X+ξwith ξ∈(s). Let us denote by ∆X= (1,∆X 1,∆X 2,...)∈HSk(k[[X]]) the Hasse-Schmidt derivation defined in (1.2) and let us assume, for the moment, that ϕ satisfies the relation ϕ(a(X)) = a(X+ξ)⊆k[[X, ξ]] ⊆K[[ξ]] ⊆K[[s]] (3) for all a(X)∈A=k[[X]]. Then, writing µ=ad(X)td m0+···+a0(X), s=ϕ(µ) = ϕ d X r=0 ar(X)tr m0!= d X r=0 ϕ(ar(X)) θr m0= d X r=0 ar(X+ξ)θr m0 (1.2) = = d X r=0 ∞ X i=0 ∆X i(ar(X))ξi!θr m0= ∞ X i=0 d X r=0 ∆X i(ar(X))θr m0!ξi∈K[[ξ]], and ξmust be of order one in s. Hence, ξis a new variable in K[[s]] and K[[s]] = K[[ξ]]. Let us denote by ∆′the unique extension of ∆Xto K[[s]] through Ascalar ext. −−−−−−→ A⊗kk(∞) local. −−−→ (A(∞))n compl. −−−−→ \ (A(∞))n ϕ≃ −−→ K[[s]], which belongs to HSk(∞)(K[[s]]), and let us denote by ∆ξ= (1,∆ξ 1,∆ξ 2,...)∈HSK(K[[ξ]]) = HSK(K[[s]]) the Hasse-Schmidt derivation defined in (1.2), this time with respect to the variable ξ. We will show that relation (3) implies that ∆ξ= ∆′, i.e. (ϕ◦∆X i)(a) = (∆ξ i◦ϕ)(a)∀i≥0,∀a∈k[[X]],(4) and then ϕ−1(K) = ϕ−1nc∈K[[s]] |∆ξ i(c) = 0,∀i > 0o=na∈\ (A(∞))n|\ (∆X i)n(a) = 0,∀i > 0o is a coefficient field of \ (A(∞))nand the step 2 would be finished. Let ϕ0:A=k[[X]] →k[[X, ξ]] be the local k-homomorphism defined by ϕ0(X) = X+ξ. Relation (3) says that ϕ(a(X)) = ϕ0(a(X)) for all a(X)∈A. 8 Let Ybe a new variable and consider the local k-homomorphisms δ:k[[X]] →k[[X, Y ]], ε:k[[X, ξ]] →k[[X, ξ, Y ]] and fϕ0:k[[X, Y ]] →k[[X, ξ, Y ]] defined by: δ(X) = X+Y, ε(X) = X, ε(ξ) = ξ+Y, fϕ0(Y) = Y, fϕ0(X) = X+ξ. Let us also consider the local K-homomorphism Θ : K[[ξ]] →K[[ξ, Y ]] defined by Θ(ξ) = ξ+Y. Then, the following diagram k[[X]ϕ0 −−−−→ k[[X, ξ]] ⊂ −−−−→ K[[ξ]] δ  yε  y  yΘ k[[X, Y ]] fϕ0 −−−−→ k[[X, ξ, Y ]] ⊂ −−−−→ K[[ξ, Y ]] is commutative and we have ∞ X i=0 ∆ξ i(ϕ(a))Yi= Θ(ϕ(a)) = ε(ϕ0(a)) = fϕ0(δ(a)) = fϕ0 ∞ X i=0 ∆X i(a)Yi!= ∞ X i=0 ϕ0(∆X i(a))Yi= ∞ X i=0 ϕ(∆X i(a))Yi for all a∈k[[X]]. Therefore relation (4) is proved and ∆ξ= ∆′. The point now is to construct a ϕin (2) satisfying (3). We first find ϕ(X) = X+ξ∈ K[[s]], and for this we state and prove the following lemma which is a generalization of Lemma (2.3.3) of [8]. (2.2) Lemma. There exists a unique ξ∈K[[s]] such that ξ(0) = 0 of order 1satisfying ad(X+ξ)θd m0+···+a0(X+ξ) = s. Proof. The lemma is a consequence of the implicit function theorem. Let G(s, σ) = ad(X+σ)θd m0+···+a0(X+σ)−s∈K[[s, σ]], with G(0,0) = ad(X)θd m0+···+a0(X) = Fm0(θm0) = 0. We have to check that ∂G ∂σ |s=σ=0=a′ d(X)θd m0+···+a′ 0(X)6= 0 in K. Assume the contrary: then a′ d(X)td m0+···+a′ 0(X) should be a multiple of Fm0(tm0) in k((X))[tm0] and there would be an α∈k((X)) such that a′ r(X) = α(X)ar(X) for every r= 0,1,...,d. 9