0
$\begingroup$

I´m currently reading chapter 10 in Washingtons book "Introduction to cyclotomic fields". Crucial for his discussion about Leopoldts Spiegelungssatz is exercise 9.1. of the same book, which consists of 2 questions. I'm not totally sure if my proofs for that exercise are correct, so it would be nice if someone could take a quick look. The first question of that exersice is:

Let $K$ be a number field, $a\in K^{\times}$, and $n\in\mathbb{Z}$. Suppose the extension $L=K(\sqrt[n]{a})$ is unramified. Show that $(a)=I^n$ for some Ideal $I$ of $K$.

I want to look at the Situation locally. Let $\mathfrak{p}$ be a finite prime of $K$, and $\mathfrak{q}$ a prime in $L$ above $\mathfrak{p}$. Let $K_{\mathfrak{p}}, L_{\mathfrak{q}}$ be the completions. In $L_{\mathfrak{q}}$ we have $(\sqrt[n]{a})=(\pi_{\mathfrak{q}}^k)$ for some $k\in\mathbb{Z}$, where $\pi_q$ is the prime element in $L_{\mathbb{q}}$. Since by assumption the extension is unramified we have $\pi_{\mathfrak{q}}=\pi_{\mathfrak{p}}$. Hence $(\sqrt[n]{a})^n=(\pi_{\mathfrak{p}})^{kn}=(a)$. Of course $(a)=(1)$, for all but finite primes. Does this already imply the statement, because of unique prime factorization?

The second question is: Suppose $(a)=I^p$ for a prime number $p$. Show that $K(\sqrt[p]{a})$ is unramified except possibly at the primes above $p$. Actually this holds slightly modified for $p$ not necessarily prime, but Washington only uses this result for the prime case, so this is enough for me.

Now the proof. Suppose $(n,p)=1$. Let $\mathfrak{p}$ be a prime of $K$, not above p. If $\mathfrak{p}|I$, we have $(a)=I^n=(\pi_{\mathfrak{p}})$ in $K_\mathfrak{p}$. Hence $a=u\pi^n$, with $u$ a local unit. If $\mathfrak{p}\nmid I$, $a$ is already a local unit in $K_{\mathfrak{p}}$. In both cases it suffices to adjoin the pth root of a local unit $u$, to obtain the extension $K_\mathfrak{p}(\sqrt[p]{a})$. So the corresponding polynomial to the extension is $x^p-u$. Since $\mathfrak{p}$ is not above $p$ the extension is tamely ramified, but $u$ is a local unit, so the polynomial is not an Eisenstein-polynomial, which implies that the extenson is not totally ramified. Since $p$ is prime, the only option left for the ramification index to be is 1.

$\endgroup$
0
$\begingroup$

Your first proof is correct. For the second proof, you should be more careful - it is not true that the minimal polynomial of any primitive element of a totally ramified extension will be Eisenstein at $ p $. For example, $ \mathbf Q_p(\zeta_p) $ is totally ramified over $ \mathbf Q_p $ for any prime $ p $, but the minimal polynomial is $ X^{p-1} + \ldots + X + 1 $, which is not Eisenstein at $ p $.

Instead, you can argue like so: if $ K $ is a $ \mathfrak q $-adic field for a prime $ \mathfrak q $ and $ a $ is a unit in the ring of integers $ \mathcal O_K $, then $ K(\sqrt[p]{a})/K $ is unramified if $ \mathfrak q $ doesn't divide $ p $. Indeed, if $ \mathfrak q $ doesn't divide $ p $, then the polynomial $ X^p - a $, and by extension, the minimal polynomial $ f(X) $ of $ \sqrt[p]{a} $ is separable modulo $ \mathfrak q $. If it were reducible in the residue field $ \mathcal O_K/\mathfrak q $, by Hensel's lemma this factorization would lift to a factorization in $ K[X] $, contradicting the irreducibility of $ f(X) $. Therefore, $ f $ is irreducible modulo $ \mathfrak q $, and the degree of the residue field extension is at least $ \deg f = [K(\sqrt[n]{a}) : K]$. It follows that the extension must be unramified.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.