# Eisenstein polynomial in Dedekind domain is irreducible in field of fractions

This is a follow-up to this question.

Let $\mathcal{O}$ be a Dedekind domain and $f=X^n+a_{n-1}X^{n-1}+\dots+a_1X+a_0 \in \mathcal{O}[X]$ be a $\mathfrak{p}$-Eisenstein polynomial; that is, $\mathfrak{p}$ is a prime ideal of $\mathcal{O}$ such that $a_i\in \mathfrak{p}$ for all $i$ and $a_0\not\in \mathfrak{p}^2$.

$f$ is irreducible in $\mathcal{O}[X]$ by the Eisenstein criterion. Is it irreducible in $K[X]$, where $K$ is the field of fractions of $\mathcal{O}$?

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If $f(x)$ were reducible in $K[x]$ then $f(x) = g(x)h(x)$ where $g$ and $h$ are nonconstant in $K[x]$. Then without loss of generality we can make $g$ and $h$ monic. Any root of $g$ or $h$ is a root of $f$ and thus is integral over ${\mathcal O}$. A monic polynomial factors over a splitting field as a product of $x - r$ as $r$ runs over the roots, so the coefficients of $g$ and $h$ are integral over ${\mathcal O}$. Those coefficients are in $K$, and ${\mathcal O}$ is integrally closed, so $g$ and $h$ are in ${\mathcal O}[x]$. This contradicts irreducibility of $f(x)$ over ${\mathcal O}$. –  KCd Nov 16 '12 at 22:01

You can easily reduced to the case of a PID: consider $R=\mathcal O_\mathfrak p$. Then $f(X)\in R[X]$ is Eisenstein in the usual sense. Hence it is irreducible in $L[X]$ where $L=\mathrm{Frac}(R)=K$.

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Thanks, that is really slick. I never know when it is useful to localize... –  user46225 Nov 16 '12 at 22:39
@user46225: it is useful to localize at a prime if your task is about a single prime ideal, or more generally if your task can be reduced to checking something one prime at a time (e.g., that the norm of the different is the discriminant). –  KCd Nov 17 '12 at 3:06
@KCd: Thank you for your insight, I'll keep it in mind. –  user46225 Nov 17 '12 at 10:09
Also you can localize away from a finite set of prime ideals and still get a PID (by the approximation lemma) –  edo arad Jul 22 at 12:10