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

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).

-
I think you are trying to find the polynomial explicitly. If you are just wondering whether it exists, the answer is yes - degree n extensions exist for any n and any number field. Just tack on enough p power roots of unity for each p dividing n and take an appropriate subextension, which exists since the Galois group is abelian. – Hunter Brooks Aug 25 '10 at 3:34
Rather than arguing in $K$ and $\mathcal O_K$, why not first localize at $\mathfrak p$, and then argue in $\mathcal O_{K,\mathfrak p}$ (the localization) and its fraction field. The localization is a DVR, which is good as you could want for applying tools like Eisenstein's criterion and Gauss's Lemma. – Emerton Aug 25 '10 at 3:39
HDK: there is no problem at all. A polynomial that is Eisenstein w.r.t. a prime ideal in O_K is irreducible over K. I never heard of Schreier domains before; that concept is unnecessary. If you look at K as the fraction field of the localization at $\mathfrak p$ then the standard proof that Eisenstein polynomials w.r.t. a prime in a PID goes through. – KConrad Aug 25 '10 at 3:42
By the way, a Schreier domain is not "exactly the domains where Gauss's Lemma holds." There are various meanings of Gauss's Lemma and some of them even work for polynomials over pretty general commutative rings. – KConrad Aug 25 '10 at 3:44
I lost some words at the end of my first comment. Should be: the standard proof that Eisenstein polynomials w.r.t. a prime in a PID are irreducible over the fraction field of the PID goes through. – KConrad Aug 25 '10 at 3:46

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 $K$.

-

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://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 $K$.

-

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 $K$.

-
Robin, is "irredusible" the British spelling, or an error? I know about soluble/solvable and centre/center, but irredusible/irreducible for UK/US spelling is completely unfamiliar to me. – KConrad Aug 25 '10 at 17:12
It must be the accusative, since Robin spells it "irreducible" when it's in the nominative in the line before. – Gerry Myerson Aug 27 '10 at 12:45