Let $A$ be a strict henselian DVR, and $\hat A$ is completion of $A$,
is $K(A)^{alg} \longrightarrow {K(\hat A)}^{alg}$ a isomorphism?
where $K(A)$ and $K(\hat A)$ are quotient fields.
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
No, since $K(\hat{A})$ is usually transcendental over $K(A)$.
Let $A$ be the strict henselization of $\mathbb Q[x]$ at $x=0$. Then $A$ is contained in $\mathbb Q(x)^{alg}$, thus $K(A)^{alg}=\mathbb Q(x)^{alg}$, a countable field. But $\hat{A}$ is $\mathbb Q^{alg}[[x]]$ and has a bijection to $\prod_{n=1}^\infty \mathbb Q^{alg}$ and so is uncountable, so $K(\hat{A})^{alg}$ is uncoutnable.
answered May 20, 2012 at 5:57
-
$\begingroup$ Do $K(A)$ and $K(\hat A)$ have isomorphic absolute Galois groups? $\endgroup$– ZhiyuCommented Jun 6, 2018 at 9:35
-
$\begingroup$ @zzy I think always yes. The main thing to check is that every algebraic extension of $K(\hat{A})$ is defined over $K(A)$. To do that, you want to check that if you perturb the coefficients of the defining equation slightly, you get the same Galois extension. This is some famous theorem. $\endgroup$ Commented Jun 6, 2018 at 21:43
-
$\begingroup$ Thank you, what you say looks like Krasner's lemma. Before I found something related in the tilting approach to the proof of Fontaine-Winterberger theorem while he didn't give a proof and I come across this problem while searching on mathoverflow. $\endgroup$– ZhiyuCommented Jun 7, 2018 at 23:29
-
$\begingroup$ @zzy You are right, I have never once in my life remembered the name "Krasner's lemma". $\endgroup$ Commented Jun 8, 2018 at 9:14