2
$\begingroup$

I really need help ! In a previous thread, I have asked for the solution of a general question, without getting answers. Since this question was posted, I have reduced the problem to the following much more specific question :

Let $A$ be an integral integrally closed domain, with fraction field $K$. Assume that $L$ is a finite Galois extension of $K$ of prime degree $p$, and that $\mathfrak p$ is a maximal ideal of $A$ containing $p$. Let $B$ be the integral closure of $A$ in $L$, and $\mathfrak P$ a maximal ideal of $B$ lying above $\mathfrak p$ (that is, ${\mathfrak P} \cap A = \mathfrak p$). We assume furthermore that for every $\sigma \in {\rm Gal}(L/K)$, $\sigma x = x\mod \mathfrak P$ for every $x\in B$.

Is it possible that $[B/{\mathfrak P} : A/{\mathfrak p}] > p$ ?

P.S :

1) the fact that $\sigma x = x\mod \mathfrak P$ implies that $(B/{\mathfrak P})/(A/{\mathfrak p})$ is purely inseparable, and every element not in the ground field has degree $p$.

2) maybe the problem could be simplified assuming that a p-th root of unity is in $K$, or assuming $A$ Noetherian.

$\endgroup$
  • $\begingroup$ It sounds like the problem you are trying to solve is for homework, or if not then may be better-suited to math.stackexchange. $\endgroup$ – user74230 Feb 3 '15 at 7:56
  • 1
    $\begingroup$ @user74230 This problem is not for homework, and I have posted a more general and interesting problem in a previous thread without anyone has been able to answer or vote. This appears to be the reduction of the problem to its minimal form. I may be wrong and this may be a stupid question, but I think you have not understood the difficulty of the question here, and simply voted down without even asking for enlightenment. If you are able to produce easily a counter example, please do it ! $\endgroup$ – MikeTeX Feb 3 '15 at 10:32
  • $\begingroup$ It was the fact that you added the elementary #1 and #2 in the "PS" (which seem too obvious to need to be pointed out, especially #1 and the 2nd part of #2) which made me mistakenly think it is a homework problem (especially since no motivation was given in the question). Since you say it is not homework, I will think about it. $\endgroup$ – user74230 Feb 3 '15 at 15:55
  • $\begingroup$ Of course, this is obvious, but sometimes some paraphrase may help to grasp more quickly the conditions of a problem, especially in this forum where answerers are hurry. Anyway, I removed #1. $\endgroup$ – MikeTeX Feb 3 '15 at 19:37
4
$\begingroup$

The answer is "no" for equicharacteristic cases, and probably also mixed characteristic but I don't know an appropriate mixed-characteristic Bertini theorem for such cases (e.g., recent papers on local mixed-char. Bertini theorem are not sufficient). So let me explain a general method which reaches the end for equicharacteristic cases but gets stuck on lack of Bertini in mixed characteristic.

The punchline is that the Galois-theoretic context is a red herring in the end (in that it is not needed in the solution): if $A$ is an integral closed domain with fraction field $K$ and $B$ its integral closure in a field $L$ of finite degree $d$ over $K$ then we shall prove that all residue field extensions for $B$ over $A$ have degree at most $d$, subject to a caveat for mixed characteristic. The proof will use considerations with noetherian approximation, henselization, and Bertini theorems (standard in equicharacteristic, but not at all in mixed characteristic).

By expressing $L/K$ as a tower of primitive extensions, we can assume $L = K(\alpha)$ where $\alpha$ is a root of a monic irreducible polynomial $f \in K[X]$. Let $\{A_i\}$ be the directed system of finitely generated $\mathbf{Z}$-subalgebras of $A$. The normalization of each $A_i$ is module-finite over $A_i$ (by excellence considerations) and is contained in $A$, so the $A_i$'s that are integrally closed are cofinal within this directed system. Letting $K_i = {\rm{Frac}}(K_i)$, taking $i$ large enough lets us assume $f \in K_i[X]$ for all $i$, so $L_i := K_i[X]/(f)$ is a field in which the integral closure $B_i$ of $A_i$ is module-finite (by excellence considerations) and the direct limit of the $B_i$'s is $B$.

Fix a point $s' \in {\rm{Spec}}(B)$ over $s \in {\rm{Spec}}(A)$, and let $s'_i \in {\rm{Spec}}(B_i)$ and $s_i \in {\rm{Spec}}(A_i)$ be the corresponding images. Then the natural map $k(s'_i) \otimes_{k(s_i)} k(s) \rightarrow k(s')$ is surjective for large $i$ since $k(s')$ is $k(s)$-finite and equality holds in the limit. Thus, if $[k(s'_i):k(s_i)] \le d$ for all $i$ then we will be done. Thus, we may replace $A \rightarrow B$ with $A_i \rightarrow B_i$, so we may assume $A$ is finitely generated over $\mathbf{Z}$. We can assume $A$ is local.

The henselization $A^{\rm{h}}$ is normal noetherian, and $B\otimes_A A^{\rm{h}}$ is the direct product of the (normal noetherian) henselizations of $B$ at its maximal ideals over that of $A$. The fraction fields of those factor rings have degree at most $d$ over the fraction field of $A^{\rm{h}}$, and henselization doesn't change the residue field, so it is harmless to pass to such factor rings. Hence, we may now assume $A$ and $B$ are henselian local. Note that their residue fields are finitely generated over their prime fields.

If $k \rightarrow \kappa$ is the extension of residue fields and $k'/k$ is the maximal separable subextension then by "Hensel's Lemma" for henselian local rings the local finite etale $A$-algebra $A'$ with residue field $k'$ uniquely maps to $B$ over $A$ and must be a normal domain and have fraction field of degree $[k':k]$ over ${\rm{Frac}}(A)$, so consideration of generic fibers over $A$ shows that the map $A' \rightarrow B$ is injective. Hence, we may replace $A$ with $A'$ to reduce to the case that the residual extension is purely inseparable. We may assume the residual degree is $> 1$ or there is nothing to prove, so now the residue characteristic is $p > 0$ and $k$ is imperfect. Since $k$ is finitely generated over $\mathbf{F}_p$, its "constant field" is a finite field $\mathbf{F}$, and that is also algebraically closed in $\kappa$ since $\kappa/k$ is purely inseparable.

Now the problem breaks into two cases, depending on whether the generic characteristic is 0 or $p$. Suppose we are in equicharacteristic $p$, so $A$ is uniquely an $\mathbf{F}$-algebra. In view of our earlier passage to finite type $\mathbf{Z}$-algebras prior to henselizing, we can therefore "spread out" our situation to arrive at the following geometric situation: we have a finite dominant map $f:X \rightarrow Y$ between normal affine varieties over $\mathbf{F}$ (irreducible and reduced) with generic degree $d$, and a geometrically irreducible positive-dimensional proper closed subvariety $X_0 \subset X$ such that $X_0$ is generically purely inseparable over $Y_0 = f(X_0) \subset Y$. We claim that the generic degree of $X_0 \rightarrow Y_0$ is at most $d$. The ambient normal varieties inherit geometric irreducibility over $\mathbf{F}$ from that of $X_0$ and $Y_0$, so extending scalars to $\overline{\mathbf{F}}$ has no effect on the degrees under consideration.

Thus, now we may consider the same problem over an algebraically closed ground field $F$ of characteristic $p$ (in fact an algebraic closure of $\mathbf{F}_p$). Since $X_0$ is a proper closed subvariety of $X$ with positive dimension, the common dimension $\delta$ of $X$ and $Y$ is at least 2. If $\delta=2$ then by normality $X \rightarrow Y$ is flat in codimension 1, hence away from a finite set of closed points in $Y$, so the fiber-degree over $y$ coincides with the fiber-degree $d$ at the generic point of $Y$. Hence, the residue degree $[F(x):F(y)]$ is certainly at most $d$ in such cases. By shrinking $Y$ around $y$ we can arrange that $X_0$ and $Y_0$ are smooth with $X_0 \rightarrow Y_0$ finite flat of degree $[F(x):F(y)]$.

Suppose instead that $\delta>2$ and that the result is known in dimension $\delta-1$. Note that $X \rightarrow Y$ is flat in codimension 1 by normality. By the Bertini theorems (in the general form of Jouanolou's book, for example) a generic hyperplane slice $Y'$ of $Y$ is irreducible and smooth in codimension 1 and inherits $S_2$ from $Y$, so is normal by Serre's criterion. By finiteness of $X \rightarrow Y$ the same holds for its preimage $X'$ in $X$ if the slice is generically chosen, with $X' \rightarrow Y'$ of generic degree $d$ by the flatness in codimension 1. Genericity also ensures that the slices $X'_0 = X' \cap X_0$ and $Y'_0 = Y' \cap Y_0$ are irreducible and reduced of positive dimension when $X_0$ and $Y_0$ are of dimension at least 2, or are finite sets of reduced points when $X_0$ and $Y_0$ are curves. The finite flat $X_0 \rightarrow Y_0$ has constant fiber-degree, so applying that at the generic point of $Y'_0$ then completes the dimension induction when $X_0$ and $Y_0$ have dimension $>1$. If instead $X_0$ and $Y_0$ are curves then the reducedness of $X'_0 = f^{-1}(Y'_0)$ and the normality of $X'$ and $Y'$ implies that $X' \rightarrow Y'$ is etale over $Y'_0$ (see Lemma 1.5 in the book by Freitag & Kiehl on etale cohomology), so that full fiber degree (which is the original residue degree of interest) coincides with the degree $d$ of $X'$ over $Y'$, so we win again. This completes the proof in equicharacteristic $p>0$.

Now consider the mixed characteristic case, so the henselizations are naturally algebras over the finite unramified extension $R$ of $\mathbf{Z}_{(p)}^{\rm{h}}$ corresponding to the finite residue field $\mathbf{F}$. By similar reasoning as at the start of the equicharacteristic case, we can extend scalars to the strict henselization $W$ of $R$, or equivalently of $\mathbf{Z}_{(p)}^{\rm{h}}$. So we can again do "spreading out", but now getting "arithmetic schemes": flat affine normal irreducible $W$-schemes $X$ and $Y$ of finite type (rather than over its residue field). Letting $\delta$ denote the common dimension of $X$ and $Y$, once again $\delta \ge 2$ and the case $\delta=2$ is easy via flatness in codimension 1, so we may assume $\delta > 2$ and that the result is known in dimension $\delta - 1$. We can also assume that the points of interests $x \in X$ and $y \in Y$ in the mod-$p$ fibers are non-generic (or else we can use flatness in codimension 1 to conclude). If there were a mixed-characteristic Bertini theorem for normal flat affine schemes of finite type over $W$ (so algebraically closed residue field) then one should be able to conclude via dimension induction as in the equicharacteristic-$p$ case; lacking a reference for such a result, this is a good place to stop.

$\endgroup$
  • $\begingroup$ Thank you so many for the answer. I have a question: At the end of the fifth paragraph, you say "we may now assume A and B are henselian local". In this case, isn't the result immediate ? for if $v$ and $w$ are the valuations corresponding to $A$ and $B$ resp., then $w$ extends $v$, so $[L : K] \geq e(w/v)f(w/v)$. $\endgroup$ – MikeTeX Feb 4 '15 at 8:29
  • $\begingroup$ @MikeTeX: No. In the noetherian case beyond dimension 1 the notion of henselian local ring has nothing to do with the notion of valuation ring. $\endgroup$ – user74230 Feb 4 '15 at 11:32
  • $\begingroup$ Ooops. Definitely not a homework problem, but probably easy for you (this is a way to thank you again). I intend to work the non galois case later, so I appreciate particularly the fact that your proof does not rely upon the galois framework. $\endgroup$ – MikeTeX Feb 4 '15 at 12:05
  • $\begingroup$ @ user74230 Unfortunately, the result appears to be incorrect (see mathoverflow.net/questions/194818/… ). So, there must be a failure in your proof, but I'm not sufficiently skilled to point it out. $\endgroup$ – MikeTeX Mar 1 '15 at 7:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.