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.

  • $\begingroup$ 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, 2015 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, 2015 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, 2015 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, 2015 at 19:37

1 Answer 1


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.

  • $\begingroup$ 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, 2015 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, 2015 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, 2015 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, 2015 at 7:47

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