# Ramified extension in number theory

Assume that $K$ is a complete field under a discrete valuation with Dedekind ring $A$ and maximal ideal $\mathfrak p$ and $A\diagup\mathfrak p$ is perfect. Let $e$ be a positive integer not divisible by $E$. Let $E$ be a finite extension of $K$, $\pi_0$ a prime element in $\mathfrak p$, and $\beta$ an element of $E$ such that $|\beta|^e=|\pi_0|$. Then there exists an element $\pi$ of order one in $\mathfrak \pi$ s.t. one of the roots of the equation $X^e-\pi=0$ is contained in $K(\beta )$.

I don't see that if $E/K$ is a finite extension then $E/K$ is a totally ramified extension as the proof claims.

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-1: You did not give enough context (even in the linked to page). What fields are you talking about? Discretely valued fields? Local fields? – Pete L. Clark Aug 1 '11 at 22:46

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Judging by the typesetting this is from somewhere in Ch. 2 of Lang's book; but you should include more context. – Dylan Moreland Aug 2 '11 at 5:06

## 2 Answers

In general, we have $ef=n$, where $f$ is the residue class field extension degree, and $e$ is the ramification, and $n$ is the degree. Here, the hypothesis $|\beta|^e=|\pi_0|$ does not suggest (nor imply) that $E/K$ is totally ramified, nor that the subextension $K(\beta)/K$. If you add the hypothesis that $E/K$ is totally ramified, then the residue class field extension degree $f$ is $1$, and $K(\beta)/K$ is totally ramified, so $\beta^e=\eta\cdot \pi_0$ with a unit $\eta$ at first in the integers of the extension... but, since the residue field extension is trivial, we can "correct" $\eta$ by units in the ground field to get $\pi$ in the groundfield so that $X^e-\pi=0$ behaves as you want.

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I checked: this is a Lemma on p. 53 of Lang's Algebraic Number Theory. The proof begins:

"We can write $\beta^e = \pi_0 u$ with $u$ a unit in $B$ [the integral closure of $A$ in $E$]. Since the extension is totally ramified..."

Edit: What I wrote in a previous version was wrong; I hadn't read carefully enough. I am currently of the opinion that "$E/K$ is totally ramified" is simply missing as a hypothesis to this Lemma. My evidence is:

1. It is assumed at the beginning of the proof!
2. The Lemma is used (only) to prove Proposition 12, which has the hypothesis that $E/K$ is totally and tamely ramified.
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Okay. I wasn't sure what extension. Well, I'm still learning some basic stuff on algebraic number theory. – Jaska Aug 2 '11 at 13:52
@paul: As you can now see, you were right all along. (I actually tried to delete my answer, but since it was accepted, I couldn't!) – Pete L. Clark Aug 2 '11 at 13:56
Oh, sorry. You can delete your answer now if you want. – Jaska Aug 2 '11 at 14:04

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@Pete... ah, ok! I put back my "answer"... I remember being surprised the first time I read Lang, that there were no "exercises", but I eventually realized that getting such details straight was the meta-exercise. :) – paul garrett Aug 2 '11 at 14:16
@paul: Hence the old "vengeance." In the 2nd Edition of Lang's Algebra, he put in a section on homological algebra due to pressure, even though he did not think it was worthwhile. The only exercise in the section was "Go to the library, check out a book on Homological Algebra, and prove all the theorems without looking at the proofs in the book." At the time, the only book on homological algebra was Cartan-Eilenberg; rumor has it that either in the next edition of the latter, or in a loose-leaf errata (cont) – Arturo Magidin Aug 2 '11 at 18:25
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