0
$\begingroup$

Let $K$ be a Henselian field and $L/K$ a finite extension. Let $\lambda$ and $\kappa$ be the residue fields of $L$ and $K$ respectively. Is the extension $\lambda / \kappa$ always finite?

I got to this question when reading Algebraic Number Theory by Neukirch , Chapter II.6. Concretely page 150, where he defines the inertia degree and tge ramification index. I do not understand why the inertia degree is assumed to be finite.

$\endgroup$
3
  • $\begingroup$ If $K$ is complete, the answer is yes. In general, it’s easy to see that simple sub-extensions of $\lambda/\kappa$ have degree at most $[L:K]$ (and the extension is algebraic). In particular, if $\lambda/\kappa$ is separable, then it is finite. $\endgroup$
    – Aphelli
    Commented Dec 26, 2022 at 23:30
  • $\begingroup$ I do not see how this answers my question. I am not assuming anything about separability of $\lambda / \kappa$. $\endgroup$
    – Andarrkor
    Commented Dec 26, 2022 at 23:38
  • $\begingroup$ Sorry if that was unclear. It was only intended as a partial argument (hence a comment), and that’s why it’s only a comment. KCd’s argument is indeed the good one. $\endgroup$
    – Aphelli
    Commented Dec 27, 2022 at 10:14

1 Answer 1

2
$\begingroup$

Suppose $\overline{x_1}, \ldots, \overline{x_n}$ in $\lambda$ are linearly independent over $\kappa$, where $x_1, \ldots, x_n$ are in the valuation ring $\mathcal O_L$. We'll show $x_1, \ldots, x_n$ are linearly independent over $K$, so $n \leq [L:K]$. Thus a $\kappa$-linearly independent subset of $\lambda$ has size at most $[L:K]$, so $[\lambda:\kappa] \leq [L:K]$. (I agree with your comment that this has absolutely nothing to do with whether or not $\lambda/\kappa$ is separable.)

Suppose $x_1, \ldots, x_n$ are $K$-linearly dependent, so $c_1x_1 + \cdots + c_nx_n = 0$ where $c_i \in K$ and not all $c_i$ are $0$. Dividing through by the $c_i$ with the maximal absolute value in $L$ (that there is a unique absolute value on $L$ extending that from $K$ comes from the Henselian property), we can assume all $c_i$ are in $\mathcal O_L$ and some $c_i$ is $1$. Then we can reduce the equation $$ c_1x_1 + \cdots + c_nx_n = 0 $$ to the residue field of $L$: $$ \overline{c_1}\,\overline{x_1} + \cdots + \overline{c_1}\,\overline{x_1} = \overline{0} $$ in $\lambda$. By $\kappa$-linear independence, all $\overline{c_i}$ should be $\overline{0}$, but some $\overline{c_i}$ is $\overline{1}$, so we have a contradiction.

P.S. I have never understood the term "inertial degree" for $[\lambda:\kappa]$. When I was a student, I learned to call this number the residue field degree, which makes very good sense. Who came up with the idea of calling it the inertial degree? For example, the term "inertia group" is the first example of the (higher) ramification groups, and those are related to the ramification index, which is not the residue field degree. In particular, the inertial degree is not the size of the inertia group, but rather is in some sense complementary to it. Someone asked about this on MSE here but no good answer was posted there yet.

$\endgroup$
1
  • $\begingroup$ Thanks! I was following Neukirch's terminology, he calls $[ \lambda : \kappa ]$ intertial degree, your assumption was right. $\endgroup$
    – Andarrkor
    Commented Dec 27, 2022 at 10:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .