# Polynomial splitting in linear factors modulo a prime ideal

Let $K$ be a number field and let $f(x) \in \mathcal{O}_K[x]$ be an nonconstant irreducible polynomial. Also let $L = K(\alpha)$ be an extension of $K$ containing a root of $f(x)$ and let $P$ be a prime of $\mathcal{O}_K$ that splits completely in $L$.

My question is: Is it true that $f(x) \bmod P$ completely splits into $\deg(f)$ distinct linear factors? (In particular, $f(x) \bmod P$ is separable.)

Looking at this previous question, it seems that the answer should be “yes”. However, I’ve not been able to prove that the $\beta_i$ are indeed distinct.

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Is this what you have in mind?

Suppose $L/E/K$ is a tower of number fields, and $P$ a prime of $K$. Your assumption is that $P$ splits completely in $L$. That is to say, $P$ does not ramify in $L$, and the degree of the residue extensions of the primes of $L$ above $P$ is $1$; the same must therefore hold for $E$. So if Dedekind’s theorem applies, we are in business; correct?

However, say $L=E=\mathbb Q(i)$ and $K=\mathbb Q$. The irreducible $f(x) = x^2 +25$ factors completely over $L$, and has repeated roots in $\mathbb F_5$ (of course!). On the other hand, $(5)$ splits in $L$.

See http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/dedekindf.pdf for Dedekind’s theorem with $K=\mathbb Q$. For a version for arbitrary $K$ see Lang’s ANT page 27.

peter ag gives a counterexample, so let me elaborate on the previous question: Dedekind’s factorization criterion holds for primes that are coprime to the conductor of your subring. The conductor of $\mathcal O_K[\alpha]$ is the largest ideal of $\mathcal O_L$ lying inside it, and Dedekind’s factorization criterion says that for any prime $\mathfrak p$ of $K$ coprime to the conductor ideal, the splitting of $f$ modulo $\mathfrak p$ will mirror the splitting of $\mathfrak p$ in $\mathcal O_L$. Since the conductor is not the zero ideal, it is only divisible by finitely many primes of $\mathcal O_L$, therefore for all but finitely many primes the factorization criterion applies. This was enough to settle the linked question, since in that context we may simply ignore all prime ideals dividing the conductor. However, for the counterexample $K = \mathbf Q$, $L = \mathbf Q(i)$ and $f(x) = x^2 + 25$, the conductor of $\mathbf Z[5i]$ is the principal ideal $(5)$, and therefore for the rational prime $5$ the criterion fails.