Washington: If $K(\sqrt[n]{a})$ is unramified, then $(a)=I^n$ for some ideal $I$.

Question

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.

Answer 1

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.