Conservation of the noetherianity by perfect transcendental field extensions

PREPUBLICACIONES DEL DEPARTAMENTO DE ´ ALGEBRA DE LA UNIVERSIDAD DE SEVILLA Conservation of the noetherianity by perfect transcendental field extensions M. Fern´andez-Lebr´on & Luis Narv´aez-Macarro Prepublicaci´on no8 (Marzo-2001) Departamento de ´ Algebra. Universidad de Sevilla Conservation of the noetherianity by perfect transcendental field extensions M. Fern´andez-Lebr´on and L. Narv´aez-Macarro∗ Departamento de Algebra, Facultad de Matem´aticas, Universidad de Sevilla, P.O. Box 1160, 41080 Sevilla, Spain. [email protected], narv[email protected] March, 2001 Abstract Let kbe a perfect field of characteristic p > 0, k(t)per the perfect closure of k(t) and Aak-algebra. We characterize whether the ring A⊗kk(t)per =[ m≥0 (A⊗kk(t 1 pm)) is noetherian or not. As a consequence, we prove that the ring A⊗k k(t)per is noetherian when Ais the ring of formal power series in n indeterminates over k. Keywords: perfect–power series ring–noetherian ring– perfect extension– complete local ring. Introduction Motivated by the generalization of the results in [7] (for the case of a perfect base field kof characteristic p > 0) in this paper we study the conservation of noetherianity by the base field extension k→k(t)per, where k(t)per is the ∗Both authors are partially supported by DGESIC, PB97-0723. 1 perfect closure of k(t). Since this extension is not finitely generated, the conservation of noetherianity is not clear a priori for k-algebras which are not finitely generated. Our main result states that k(t)per ⊗kAis noetherian if and only if Ais noetherian and for every prime ideal p ⊂Athe field \ m≥0 Qt(A/ p )pmis algebraic over k(see theorem 3.6). In particular, we are able to apply this result to the case where Ais the ring of formal power series in nindeterminates over k We are indebted to J. M. Giral for giving us the proof of proposition 2.5 and for other helpful comments. 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 F pto denote the finite field of pelements, for p a prime number. If p ∈Spec(B), we shall denote by ht( p ) the height of p . Remember that a ring Bis said to be equicodimensional if all its maximal ideals have the same height. Also, Bis said to be biequicodimensional if all its saturated chains of prime ideals have the same length. If Bis an integral domain, we shall denote by Qt(B) its quotient field. For any F p-algebra B, we denote B]=\ m≥0 Bpm. We shall first study the contraction-extension process for prime ideals relative to the ring extension K[t]⊂K[t1 p], Kbeing a field of characteristic p > 0. Let us recall the following well known result (cf. for example [4], th. 10.8): Proposition. 1.1 Let Kbe a field of characteristic p > 0. Let g(X)be a monic polynomial of K[X]. Then, the polynomial f(X) = g(Xp)is irreducible in K[X]if and only if g(X)is irreducible in K[X]and not all its coefficients are in Kp. From the above result, we deduce the following corollary. 2 Corollary. 1.2 Let Kbe a field of characteristic p > 0. Let Pbe a non zero prime ideal in K[t1 p]and let F(t)∈K[t]be the monic irreducible generator of the contraction Pc=P∩K[t]. Then the following conditions hold: 1. If F(t) = ap 0+ap 1t+· · · +td∈Kp[t], then P= (a0+a1t1 p+· · · +td p). 2. The equality P=Pce holds if and only if F(t)6∈ Kp[t]. Proof: 1. Consider the polynomial G(τ) = a0+a1τ+···+τd∈K[τ](τ=t1 p) and the ring homomorphism µ:K[τ]→K[t] defined by µ(Xaiτi) = Xap iti. From the identity µ(G) = Fwe deduce that G(τ) is irreducible. Since G(t1 p)p=F(t)∈P, we deduce that G(t1 p)∈Pand then P= (G(t1 p)). 2. The equality P=Pce means that F(t) = F(τp)∈K[τ] generates the ideal P, but that is equivalent to saying that F(τp) is irreducible in K[τ]. To conclude, we apply proposition 1.1. For each k-algebra A, we define A(t) := k(t)⊗kA. We also consider the field extension k(∞)=[ m≥1 k(t1 pm). If kis perfect, k(∞)coincides with the perfect closure of k(t), k(t)per. For the sake of brevity, we will write tm=t1 pm. We also define A(m):= A(tm) := A⊗kk(tm) = A(t)⊗k(t)k(tm), A[m]:= A[tm] and 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(because (tm)pm−t= 0). 3 For each prime ideal Pof A(∞)we denote P[∞]:= P∩A[∞],P[m]:= P∩A[m]∈Spec(A[m]) and P(m):= P∩A(m)∈Spec(A(m)). In a similar way, if Qis a prime ideal of A[∞]we denote Q[m]:= Q∩A[m]∈ Spec(A[m]). We have: •P=[ m≥0 P(m),P[∞]=[ m≥0 P[m], (resp. Q=[ m≥0 Q[m]). •P(n)∩A(m)=P(m)and P[n]∩A[m]=P[m]for all n≥m(resp. Q[n]∩ A[m]=Q[m]for all n≥m). The following properties are straightforward: 1. The k-algebras A[m](respectively A(m)) are isomorphic to each other. 2. If Sm=k[tm]− {0}, then A(m)=S−1 mA[m]. 3. Since (Sm)pm⊂S0⊂Sm, we have A(m)=S−1 0A[m]for m≥0. Consequently A(∞)=S−1 0A[∞]. 4. If Ais a domain (integrally closed), then A[m]and A(m)are domains (integrally closed) for all m≥0 or m=∞. 5. If Ais a noetherian k-algebra, then A[m]and A(m)are noetherian rings, for every m≥0. 6. If A=k[X] = k[X1, . . . , Xn], then A[∞]is not noetherian (the ideal generated by the tm,m≥0, is not finitely generated). 7. If I⊂Ais an ideal, then (A/I)(∞)=A(∞)/A(∞)I. 8. If T⊂Ais a multiplicative subset, then (T−1A)(∞)=T−1A(∞). 9. If A=k[X], then A(∞)=k(∞)[X], hence A(∞)is noetherian. Moreover, A(∞)is noetherian for every finitely generated k-algebra A. The main goal of this paper is to characterize whether the ring A(∞)is noetherian (see th. 3.6 and corollary 3.8). Proposition. 1.3 With the above notations, the following properties hold: 4 1. The extensions A[m−1] ⊂A[m]and A(m−1) ⊂A(m)are finite free, and therefore integral and faithfully flat. 2. The corresponding extensions to their quotient fields are purely inseparable. Proof: Straightforward. Corollary. 1.4 A[∞](resp. A(∞)) is integral and faithfully flat over each A[m](resp. over each A(m)). From the properties above, we obtain the following lemmas: Lemma. 1.5 Let P0⊆Pbe prime ideals of A(∞)(resp. of A[∞]). The following conditions are equivalent: (a) P0 ( P (b) There exists an m≥0such that P0 (m) ( P(m)(resp. P0 [m] ( P[m]). (c) For every m≥0,P0 (m) ( P(m)(resp. P0 [m] ( P[m]). Lemma. 1.6 Let Pprime ideal of A(∞)(resp. of A[∞]). The following conditions are equivalent: (a) Pis maximal. (b) P(m)(resp. P[m]) is maximal for some m≥0. (c) P(m)(resp. P[m]) is maximal for every m≥0. Corollary. 1.7 With the notations above, for every prime ideal Pof A(∞) we have ht(P) = ht(P(m)) = ht(P[m])for all m≥0. Moreover, dim(A(∞)) = dim(A(m)). Proof: Since flat ring extensions satisfy the “going down” property, corollary 1.4 implies that ht(P∩A(m))≤ht(P). By corollary 1.4 again, A(∞)is integral over A(m), then ht(P)≤ht(P∩A(m)). The equality ht(P(m)) = ht(P[m]) comes from the fact that A(m)is a localization of A[m]. The last relation is a standard consequence of the “going up” property. 5 Remark. 1.8 Corollary 1.7 remains true if we replace A(m)⊂A(∞)by A[m]⊂A[∞]. Corollary. 1.9 With the notations above, for every Q∈Spec(A(m))there is a unique e Q∈Spec(A(m+1))such that e Qc=Q. Moreover, the ideal e Qis given by e Q={y∈A(m+1) |yp∈Q}. Proof: This is an easy consequence of the fact that (A(m+1))p⊂A(m). Corollary. 1.10 Let us assume that Ais noetherian and for every maximal ideal m of A, the residue field A/ m is algebraic over k. Then for every m≥0 we have: 1. dim(A[∞]) = dim(A[m]) = dim(A[t]) = n+ 1. 2. dim(A(∞)) = dim(A(m)) = dim(A(t)) = n. Proof: The first relation comes from remark 1.8 and the noetherianity hypothesis. The second relation comes from corollary 1.7 and proposition (1.4) of [7]. The following result is a consequence of theorem (1.6) of [7], lemma 1.6 and corollary 1.10. Corollary. 1.11 Let Abe a noetherian, biequidimensional, universally catenarian k-algebra of Krull dimension n, and that for any maximal ideal m of A, the residue field A/ m is algebraic over k. Then every maximal ideal of A(∞)has height n. 2 The biggest perfect subfield of a formal functions field Throughout this section, kwill be a perfect field of characteristic p > 0, A=k[[X]], p ⊂Aa prime ideal, R=A/ p and K=Qt(R). The aim of this section is to prove that the biggest perfect subfield of K,K]=\ e≥0 Kpe, is an algebraic extension of the field of constants, k. This result is proved in prop. 2.5 and it is one of the ingredients in the proof of corollary 3.8. 6 Proposition. 2.1 Under the above hypothesis, it follows that k=R]. Proof: Let m be the maximal ideal of R. It suffices to prove that R]⊆k. If f∈R], then for every e > 0 there exists an fe∈Rsuch that f=fpe e. •Suppose at first that fis not a unit, then feis not a unit for any e > 0, and fe∈ m for every e > 0. Thus, f∈ m pefor every e > 0 and by Krull’s intersection theorem, f∈\ e≥0 m pe=\ r≥0 m r= (0). •If fis unit, then f=f0+e f, with f0∈k⊂R]and e f∈R]and f0is unit. By the above case e f= 0, hence f∈k. Proposition. 2.2 If p = (0), that is R=k[[X]],K=k((X)), then k=K]. Proof: It is a consequence of prop. 2.1 and the fact that Ris a unique factorization domain. In order to treat the general case, let us look at some general lemmas. Lemma. 2.3 (cf. [3] Chap. 5, §15, ex. 8) If Lis a separable algebraic extension of a field Kof characteristic p > 0, then L]is an algebraic extension of K]. Proof: If x∈L], then x=ype ewith ye∈Lfor all e≥0. Since yeis separable over K,K(ye) = K(ype e) = K(x), it follows that ye=xp−e∈K(x) and then x∈Kpe(xpe). Therefore [Kpe(x) : Kpe] = [Kpe(xpe) : Kpe]=[K(x) : K]. Thus xsatisfies the same minimal polynomial over Kpeand over Kfor all e≥0, and the coefficients of this minimal polynomial must be in K]. So x is algebraic over K]. Lemma. 2.4 Every algebraic extension of a perfect field is perfect. 7 Proof: This is obvious because this is true for the finite algebraic extensions. Proposition. 2.5 Let kbe a perfect field of characteristic p > 0,A= k[[X]] = k[[X1, . . . , Xn]], p ⊂Aa prime ideal, R=A/ p and K=Qt(R). Then K]is an algebraic extension of k. Proof: 1Let r= dim(A/ p )≤n. By the normalization lemma for power series rings (cf. [1], 24.5 and 23.7)2, there is a new system of formal coordinates Y1, . . . , Ynof A, such that • p ∩k[[Y1, . . . , Yr]] = {0}, •k[[Y1, . . . , Yr]] ,→A p =Ris a finite extension, and •k((Y1, . . . , Yr)) ,→Kis a separable finite extension. The proposition is then a consequence of proposition 2.2 and lemma 2.33. Remark. 2.6 Actually, under the hypothesis of proposition 2.5, J.M. Giral and the authors have proved that the following stronger properties hold: (1) If Ris integrally closed in K, then K]=k. (2) In the general case, K]is a finite extension of k. 3 Noetherianity of A⊗kk(t)per Throughout this section, kwill be a perfect field of characteristic p > 0, keeping the notations of section 1. 1Due to J. M. Giral. 2The proof of the normalization lemma for power series rings in [1] uses generic linear changes of coordinates and needs the field kto be infinite. This proof can be adapted for an arbitrary perfect coefficient field (infinite or not) by using non linear changes of the form Yi=Xi+Fi(Xp i+1, . . . , Xp n), where the Fiare polynomials with coefficients in F p. 3In particular, if kis algebraically closed, we would have K]=k. 8