0
$\begingroup$

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.

  • 1
    Judging by the typesetting this is from somewhere in Ch. 2 of Lang's book; but you should include more context.2011-08-02

2 Answers 2

2

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.

1

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.
  • 0
    @arturo... Haha! I'd not heard that rumor about C-E! But, yes, c. 1970, my small gray edition of "Algebra" had that questionable remark. In the late 1970s, I asked Lang about his remark, and he claimed that it was not meant to express disdain for the subject (!?) but to claim that once one knew what to _try_ to prove, it was inevitable (!?). That readers of an Algebra book might not be mature enough, or might want to see examples...? It is also true that, while some mention of categorical ideas appeared in early editions, it was half-hearted...2011-08-02