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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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What do you mean that $P$ is reducible at 'one place'? –  Beni Bogosel Apr 19 '12 at 9:40
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IIRC, for a number field, Chebotarev's density theorem implies that the asymptotic density of primes $p$ for which $P$ is irreducible modulo $p$ is $1/n$, where $n$ is the degree of $P$. I'm guessing this is relevant to the question asked.... –  Hurkyl Apr 19 '12 at 9:46
    
@Hurkyl: that is false. $\frac{1}{n}$ is the asymptotic density of primes $p$ for which $P$ splits. –  Qiaochu Yuan Apr 20 '12 at 14:42
    
That doesn't sound right. IIRC, the probabilities are the same as what you'd expect for picking a random polynomial modulo $p$. For cubic polynomials, you get $1/3$ irreducible, $1/6$ totally split, and $1/2$ that factor into a linear-quadratic. Maybe I'm assuming a Galois group of $S_n$ or something? –  Hurkyl Apr 20 '12 at 15:48
    
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). –  David Loeffler Apr 21 '12 at 1:21

2 Answers 2

up vote 7 down vote accepted

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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In fact here is the worst possible example: $x^4 - 5x^2 + 16$, whose splitting field is $\mathbb{Q}(\sqrt{13}, \sqrt{-3})$. This polynomial is reducible in $\mathbb{Q}_p$ for every $p$. I found this by using the construction in this lovely paper: ams.org/journals/proc/2005-133-11/S0002-9939-05-07855-X/… –  David Loeffler Apr 20 '12 at 22:10
    
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! –  Michalis Apr 22 '12 at 0:37

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... –  Qiaochu Yuan Apr 20 '12 at 14:40

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