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I'm trying to finish an exercise from Daniel Marcus's "Number Fields" book: #30(e) in chapter 3, page 91.

Here's the problem: say $\mathcal{O}_K$ is the ring of integers of a number field $K$ and let $f(x) \in \mathcal{O}_K[x]$ be a nonconstant monic irreducible polynomial. Prove that $f$ splits into linear factors mod $P$ for infinitely many primes $P$ of $\mathcal{O}_K$.

Here's my approach so far: let $L = K(\alpha)$ be a field extension containing a root $\alpha$ of $f$. By an earlier exercise, there are infinitely many primes $P \in Spec(\mathcal{O}_K)$ that split completely in $L$ (this is true for any extension $L/K$). My approach has been to look at the image of $\alpha$ in residue fields of these primes of $L$, then pull those images back to $\mathcal{O}_K/P$, to obtain roots of $f(x)$ mod $P$.

In gory detail, consider such a prime $P$; say it splits into $Q_1 \cdots Q_n$ in $\mathcal{O}_L$ (where $n = \deg(f) = [L:K]$). We have $n$ projections $\pi_i : \mathcal{O}_L \to \mathcal{O}_L/Q_i$.

The inertial degrees are all 1 (since the prime split completely), so we also have natural isomorphisms $\mathcal{O}_K/P \to \mathcal{O}_L/Q_i$, giving us induced maps $\tilde \pi_i : \mathcal{O}_L \to \mathcal{O}_K/P$. These are part of a commutative diagram with the two rings and their residue fields, but I don't know how to write commutative diagrams here...

Okay, so let $\beta_i = \tilde\pi_i(\alpha) \in \mathcal{O}_K/P$. It's easy to see that $\beta_i$ is then a root of $f(x)$ mod $P$.

My question is, why must these $\beta_i$'s be "distinct"? (I'm putting quote marks because, a priori, there's no reason why $f(x)$ should be separable over $\mathcal{O}_K/P$, so it's plausible that some of the $\beta_i$ actually are equal. But then hopefully $f(x)$ actually has them as repeated roots.)

Thanks a lot. (Incidentally, this method does work fine for a simple example: take $f(x) = x^2+1$; then you can directly show that $f$ splits mod $5$, for example, by going up to the Gaussian integers and looking at the image of $i$ mod the ideals $(2 \pm i)$. In one case, you get $2$ and in the other you get $-2$, which are the roots mod 5.)

EDIT: Never mind, I figured it out in the cold light of day. I was basically reproducing the theorem that the factorization of $f(x)$ mod $P$ gives the factorization of $P$ in $\mathcal{O}_L$. Given that theorem, this problem follows immediately. So the approach above does work: the last step would be to show that $Q_i = P\mathcal{O}_L + (\alpha-b_i)$, where $b_i \in \mathcal{O}_K$ is any lift of $\beta_i$. Since the $Q_i$ are distinct ideals, the $\beta_i$ can't be the same.

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