Divisibility of the degree of an extension by the degree of its residual field

Let $A$ be an integrally closed domain whose quotient field is $K$, $L$ be a finite Galois extension of $K$, and $B$ be the integral closure of $A$ in $L$. Let $M_A$ be a maximal ideal of $A$, and $M_B$ be a maximal ideal of $B$ that lies above $M_A$ (that is, $M_B\cap A = M_A$). Denote by $F_A$ the field $A/M_A$, and by $F_B$ the field $B/M_B$; so $F_B/F_A$ is an algebraic field extension, and it is normal.

1) Must $F_B/F_A$ be finite ? (this is the case if, say $A$ is Noetherian, and I can show that this is also true if $F_B/F_A$ is separable)

2) The most important question for me : Assuming $F_B/F_A$ finite, must the residual degree $[F_B : F_A]$ divide $[L : K]$ (this is actually the case if $A$ is a Dedekind ring, or $A$ and $B$ are discrete valuation rings).

2 Answers

Let $k$ be the field with $2$ elements. Let $R = k[t_i, x_i; i \in \mathbf{N}]$. Let $\sigma : R \to R$ be the order $2$ automorphism sending $t_i$ to $t_i$ and $x_i$ to $x_i + t_i$. Let $S \subset R$ be the fixed elements under the action. Then $R$ and $S$ are normal domains and $R$ is integral over $S$. In particular, $R$ is the integral closure of $S$ in the fraction field extension and the fraction field extension has degree $2$ by Galois theory.

Let $\mathfrak p \subset R$ be the prime ideal $\mathfrak p = (t_i)$. This lies over the prime ideal $\mathfrak q = S \cap \mathfrak p$. We are going to show that $\kappa(\mathfrak q) \subset \kappa(\mathfrak p)$ is infinite. This will give an example where 1) fails, because we can then consider $A = S_{\mathfrak q} \subset B = R_{\mathfrak p}$. I omit the verification that $B$ is the integral closure of $A$ in the fraction field extension (this relies on the fact that $\mathfrak p$ is the only prime of $R$ lying over $\mathfrak q$). By using finitely many variables you can find an example where 2) fails.

OK, so let us consider an element $f \in R$. We can write $f$ as $$f_0 + \sum t_i f_i + h.o.t.$$ where $f_0, f_i$ depend only on the $x_j$ and all other terms have more $t_k$'s in them. Now if $f \in S$, then $\sigma(f) = f$. Note that $$\sigma(f) = f_0 + \sum t_i(f_i + \partial f_0/\partial x_i) + h.o.t.$$ This means that $\partial f_0/\partial x_i$ is identically zero. In other words, we see that $f_0 \in k[x_i^2]$. Hence we see that $\kappa(\mathfrak q) = k(x_i^2)$. Since $\kappa(\mathfrak p) = k(x_i)$ we are done.

• Beautiful. Can I ask you how did you hit on that? Also, I suggest to replace the local ring $R_{\frak p}$ by $R_{\frak q}$, the ring of fractions of $R$ with respect to the multiplicative set $S-\frak q$: this changes nothing and allows seeing (almost) immediately that $R_{\frak q}$ is the integral closure of $S_{\frak q}$ in the field of fractions of $R$. It would be nice, also, to have such a counter example in characteristic $0$, in the case where, one day, you flash upon upon some other bright idea. – MikeTeX Feb 28 '15 at 19:26
• I also observe that with the same automorphism $\sigma(x_i) = x_i+t_i$, the example is also valid in characteristic $p>0$ : $\sigma$ only becomes an order $p$ automorphism. So, all that is now missing is an example in characteristic $0$. – MikeTeX Feb 28 '15 at 19:56

As pointed out above, the answer of Count Dracula is also valid, mutatis mutandis, if one desires counter-examples where the field extensions are of characteristic $p>2$. It turns out that it can also be adapted to provide an example where the characteristic of the base field extension is $0$, and is $2$ for the residual field extension. Here is how:

Let $R={\mathbb Z}[t_i, x_i, i\in \mathbb N]$, and $L$ the fraction field of $R$. Define the order 2 automorphism $\sigma$ of $R$ by $\sigma(z\in \mathbb Z)=z$, $\sigma(t_i) = t_i$ and $\sigma(x_i) = t_i-x_i$.

Let $S$ be the fixed sub-ring under the action, and $K$ its fraction field. Then $\sigma$ extends to a $K$-automorphism of $L$ in an obvious manner, hence $[L:K]=2$ by Galois theory. $R$ is integrally closed (since factorial), and is integral over $S$ because every $f\in R$ satisfies an integral dependance relation over $S$ of the form $(X-f)(X-\sigma f)=0$. It follows easily that $S$ must be integrally closed too.

We consider the ideals ${\frak P} =(2,t_i)\subset R$, and ${\frak p} = S\cap {\frak P}\subset S$. It is not difficult to see that $\frak P$, and hence $\frak p$, is prime. Now, as in the proof of Count Dracula, we see that $f\in R$ can be written $$f_0 + \sum_i t_i f_i + \hbox{terms multiples of t_j t_k},$$ where $f_0, f_i$ depends only on the $x_j$ (eventually, consider the polynomial $f_0$ lacunary in some variables, and add some null $f_i$ in order for the variables in all the $f_0,f_i$ to be the same). If $f\in S$, then $$\sigma f(\star) = f_0(-\star) + \sum_i t_i(f_i(-\star) + {\partial f_0/\partial x_i}(-\star)) + \hbox{terms multiples of t_j t_k}.$$
Therefore $$\sum_i t_i \partial f_0/\partial x_i(-\star) = f_0(\star)-f_0(-\star) + \sum_i t_i(f_i(\star) - f_i(-\star)).$$ But $-1 = 1 \pmod2$ hence $\partial f_0/\partial x_i = 0 \pmod2$. It follows, as in the proof of Count Dracula, that $A/{\frak p}={\mathbb F_2}(x_i^2)$, and that $B/{\frak P}={\mathbb F_2}(x_i)$ is infinite over $A/\frak p$. Considering finitely many variables, it can be shown similarly that $[L:K]=2$ is not divisible by $[B/{\frak P}:A/\frak p]$.

NOTE: to obtain a more general counter-example, it suffices to consider the order $p$ automorphism $$\sigma(t_i)=t_i,\quad \sigma(z\in {\mathbb Z}[\zeta])=z,\quad \sigma(x_i)= t_i+\zeta x_i,$$ where $\zeta$ is a p-root of unity. The adaptations are straightforward, except, perhaps, the fact that the ideal $(2,t_i)$ must be replaced by the ideal ${\frak P}= (1-\zeta, t_i)$ and not $(p,t_i)$: this is because $(1-\zeta)$ is the only maximal ideal above $(p)$ in ${\mathbb Z}[\zeta]$.