2
$\begingroup$

Milne, on his notes about Class Field Theory (available in http://www.jmilne.org/math/CourseNotes/CFT.pdf, on page 47 more precisely), given a local field $K$, constructs its maximal unramified extension contained in its separable closure, $K^{un}$, and, give a prime element $\pi$ of $K$, constructs the extension $K_\pi$, which is the union of totally ramified extensions $K_{\pi,n}$ of degree $n$, using Lubin-Tate theory. Then he argues that $K^{ab}=K^{un}\cdot K_\pi$ (local kronecker-weber), where $K^{ab}$ is the maximal abelian extension.

One of the lemmas he uses to prove this result is the following:

A finite unramified extension $L$ of $K_\pi$ is contained in $K_\pi\cdot K^{un}$.

Here is its proof: $L$ must be of the form $K_\pi\cdot L^{\prime}$, where $L^{\prime}$ is some unramified extension of $K_{\pi,n}$ for some $n$. Also, $L^{\prime}$ must be of the form $K_{\pi,n}\cdot L^{\prime \prime}$, where $L^{\prime \prime} $ is some unramified extension of $K$.

So, my question is: how do you prove the following assertions:

(1) $L$ must be of the form $K_\pi\cdot L^{\prime}$;

(2) $L^{\prime}$ must be of the form $K_{\pi,n}\cdot L^{\prime \prime}$?

Thank you.

$\endgroup$
3
  • $\begingroup$ For (2), I thought of the following: let $L^\prime=K_{\pi,n}[\alpha]$. Then $K[\alpha]$ must be unramified over $K$, since $\bar{L^\prime}=\bar{K}_{\pi,n}[\bar{\alpha}]$ ($\bar{F}$ is the residue field of $F$), and so $\bar{\alpha}$ is algebraic over $\bar{K}_{\pi,n}$, and so algebraic over $\bar{K}$, and so $K[\alpha]$ is unramified over $K$. Hence, we can take $L^{\prime \prime}=K[\alpha]$. $\endgroup$ Oct 5, 2016 at 16:44
  • 1
    $\begingroup$ I think the first one is no harder. The finite extension $L$ of $K_\pi$ is separable, so generated by an element $\zeta$ whose minimal $K_\pi$-equation has only finitely many coefficients. Since $K_\pi$ is union of the chain of subfields $K_{\pi,n}$, one of these must contain all the coefficients of your polynomial. Does that work for you? (These infinite extensions always confuse me.) $\endgroup$
    – Lubin
    Oct 12, 2016 at 1:25
  • $\begingroup$ Thank you for your answer dear @Lubin, that settles my question. $\endgroup$ Oct 12, 2016 at 14:01

1 Answer 1

0
$\begingroup$

This question is answered by the comments above, so the topic can be closed.

$\endgroup$

You must log in to answer this question.

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