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The question doesn't of course make sense as written in the title. Here is what I really mean:

Given a global field $k$ and an irreducible polynomial $P \in k[x]$

Is it true that $P$ is reducible at almost all places?

I would guess that Hensel's lemma plus an approximation theorem will give an affirmative answer.

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    QiaochuYuan: you want not $1/n$ but $1/m$, where $m$ is the order of the Galois group of$P$(so $n \le m \le n!$). Hurkyl: what you say is correct for Galois group $S_3$ (since$1/6$of its elements are the identity,$1/3$are of order 3, and$1/2$of order 2).2012-04-21

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In fact the situation is even worse than Michalis' answer suggests: a globally irreducible polynomial can be (in fact usually is) locally reducible almost everywhere. For instance, consider the polynomial $x^4 - 8x^2 + 36$. This is the minimal polynomial of $\sqrt{5} + i$, so it is irreducible. But mod $p$ for any prime $p \notin \{2, 5\}$, at least one of $-1$, $5$ and $-5$ is a quadratic residue; hence the polynomial is not irreducible over $\mathbb{Q}_p$ for any $p > 5$.

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    Nice example, but in fact the OP was conjecturing this behaviour (or maybe he mistyped and meant "is it true that P is irreducible at almost all places"?). Here's another easy example: $X^4-X^2+1$ is the 12th cyclotomic polynomial and hence irreducible. It is easy to see that it is reducible for every prime!2012-04-22
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Take the irreducible polynomial $X^2+1$ over $\mathbb{Q}$. If $p\equiv 3\mod 4$ this polynomial is irreducible in $\mathbb Q_3$ by Hensels lemma. If $p\equiv 1\mod 4$ it is reducible, so $X^2+1$ is reducible / irreducible for half of the places. In general you can use Chebotarev's theorem to determine for how many primes a polynomial will split.

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    Chebotarev tells you that the frequency with which a polynomial remains irreducible mod a prime is equal to the frequency of elements of the Galois group with cycle type a full $n$-cycle. If the Galois group has no such elements (which cannot occur when $n = 2, 3$ but can occur for $n \ge 4$), as in David Loeffler's example...2012-04-20