Given an integer n and a finite extension K of Q , find a polynomial of degree n that is irreducible over K

Question

Given a positive integer n and a finite extension $K$ of $\mathbb{Q}$, can one always find an irreducible polynomial in $K[x]$ of degree n? What if $n$ is prime?

The natural approach is to take a prime ideal $\mathfrak{p}$ in $\mathcal{O}_K$, choose an element $\alpha \in \mathfrak{p} - \mathfrak{p}^2$, and consider the polynomial $x^n - \alpha$. It is irreducible over $\mathcal{O}_K$ by Eisenstein's Criterion. If $\mathcal{O}_K$ is a Schreier domain, then Gauss's Lemma applies and $x^n - \alpha$ is irreducible over $K$. The problem is that $\mathcal{O}_K$ need not be a Schreier domain (and Schreier domains are exactly the domains where Gauss's Lemma works).

Answer 1

The question has been well-answered in the comments. Here's another approach, somewhat like that advocated by Hunter Brooks. Let $p$ be a prime congruent to 1 modulo $n$ and such that $K\cap{\bf Q}(e^{2\pi i/p})=\bf Q$. Then $K(e^{2\pi i/p})$ is a cyclic extension of $K$ of degree a multiple of $n$, so it has a subfield of degree $n$ over $K$.

Answer 2

Here is a more general result.

Theorem: Let $(K,|\ |)$ be a non-Archimedean normed field with completion $\hat{K}$. Let $\mathcal{L}/\hat{K}$ be a finite separable extension of degree $d$. Then there exists a degree $d$ separable field extension $L/K$ such that $L\hat{K} = \mathcal{L}$.

In particular, as long as the completion of $K$ admits a separable field extension of a certain degree $d$, so does $K$ itself, necessarily of the form $K[t]/(P(t))$ by the primitive element theorem. Moreover, as long as $K$ has characteristic zero and carries a nontrivial discrete valuation, it admits finite separable extensions of all finite degrees.

For a proof of this theorem using Krasner's Lemma, see Section 3.5 of

http://alpha.math.uga.edu/~pete/8410Chapter3.pdf

When the norm corresponds to discrete valuation $v$ (e.g. $| \ | = | \ |_{\mathfrak{p}}$ the $\mathfrak{p}$-adic norm for a prime ideal $\mathfrak{p}$ of a number field $K$) one can get away with less: by weak approximation, there exists $\alpha \in K$ with $v(\alpha) = 1$. For any positive integer $n$ prime to the characteristic of $K$, by Eisenstein's Criterion the polynomial $t^n - \alpha \in K[t]$ is (separable and) irreducible even over the completion $\hat{K}$, so is certainly irreducible over $K$.

Answer 3

Your "$x^n-\alpha$" approach is the "ramified" way to go: the extension you get localizes to a totally ramified extension of the local field $K_{\mathfrak{p}}$, which has degree $n$, so the global extension must have degree $n$ too.

One might instead go in the "unramified" extension. Let $\mathfrak{p}$ be a prime ideal of $\mathfrak{O}_K$, and $k=\mathfrak{O}_K/\mathfrak{p}$ be the residue field. Then there is an irreducible polynomial of degree $n$ over the finite field $k$; lifting this to a polynomial over $\mathfrak{O}_K$ gives an irredusible polynomial over $K$.