We know that $f(x)=x^4+1$ is a polynomial irreducible over $\mathbb{Q}$ but reducible over $\mathbb{F}_p$ for every prime $p$.

My question is:

Is it true that if $f(x)$ has a linear factor over $\mathbb{F}_p$ for every prime $p$, then $f(x)$ has a linear factor over $\mathbb{Q}$?

Edit: Thanks for @Jyrki Lahtonen's answer, I want to do some modifications:

Is it true that if $f(x)$ has a linear factor over $\mathbb{F}_p$ for every prime $p$, then $f(x)$ is reducible over $\mathbb{Q}$?

Thanks in advance!

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    $\begingroup$ For a possible counterexample, do you want $f$ to be irreducible over $\mathbb{Q}$ also? $\endgroup$ – ronno Dec 16 '13 at 10:42
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    $\begingroup$ @ronno I know why you ask this now, thank you! $\endgroup$ – Next Dec 16 '13 at 10:50
  • $\begingroup$ I assume that $f \in \mathbb{Z}[x]$? +1 for the new question. $\endgroup$ – Martin Brandenburg Dec 16 '13 at 11:15
  • $\begingroup$ @Martin Brandenburg Yep.. $\endgroup$ – Next Dec 16 '13 at 11:46

For your follow-up question, only linear polynomials don't work :

Suppose $G$ is a subgroup of $S_n$ such that $G$ acts transitively on $\{1,\ldots, n\}$, and let $H_i^j = \{\sigma \in G \mid \sigma(i)=j \}$.

If $\tau(i)=j$ then $H_i^k = H_j^k \tau$, and $H_k^j = \tau H_k^i$. Since $G$ is transitive, every $H_i^j$ has the same cardinal. Since every element of $G$ is in $n$ such $H_i^j$, we have $|H_i^j| = |G|/n$, and in particular, elements in $G$ have on average $\sum |H_i^i|/|G| = 1$ fixed point. Since the identity element has $n$ fixed points, if $n>1$ there must be some elements in $G$ without fixed points.

So if you have an irreducible polynomial of degree $n$ over $\Bbb Q$ its Galois group is transitive on its $n$ roots, so if $n>1$, it has some elements without fixed points. Then Cebotarev's theorem says that there are infinitely many primes $p$ for which the polynomial doesn't have a linear factor over $\Bbb F_p$.

So if $P$ has linear factors for all primes, then it is reducible or of degree $1$.

  • $\begingroup$ If $\sigma$ is an element of the Galois group without fixed points among the roots, then Chebotarev's theorem tells us that the Frobenius element is in the conjugacy class of $\sigma$ for infinitely many primes $p$. But why does this imply that the polynomial has no root in $\Bbb{F}_p$ for those primes $p$? $\endgroup$ – Jyrki Lahtonen Dec 16 '13 at 12:28
  • $\begingroup$ I mean, $f$ has a root in $\Bbb{F}_p$, iff the splitting field of $f$ has an intermediate field such that one of the ideals $\mathfrak{P}$ lying above $p$ in that intermediate field has $f({\mathfrak{P}}|p)=1$, but Chebotarev works at the level of the splitting field, right? $\endgroup$ – Jyrki Lahtonen Dec 16 '13 at 12:38
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    $\begingroup$ @JyrkiLahtonen : the Frobenius element of $P$ is (up to reordering of the roots) how the Frobenius automorphism acts on the roots of $P$ in $\overline{\Bbb F_p}$. The permutation has a fixpoint if and only if one of the roots is in $\Bbb F_p$, if and only if $P$ has a linear factor mod $p$. More generally, having a $t$-cycle means that we have $t$ roots in $\Bbb F_{p^t}$, and so an irreducible factor of degree $t$ in $P$ mod $p$ : the length of the orbits of the Frobenius elements of $P$ mod $p$ are the degrees of the irreducible factors of $P$ mod $p$ $\endgroup$ – mercio Dec 16 '13 at 12:38
  • $\begingroup$ Ok. For some reason I forgot about that interpretation of the Frobenius automorphism, and was trying to unravel it from the wrong end. In spite of the fact that the automorphism described by Qiaochu here and by Jacobson in BAI is the Frobenius automorphism by construction. Very well done!! $\endgroup$ – Jyrki Lahtonen Dec 16 '13 at 13:05

No. Consider $$ f(x)=(x^2+1)(x^2+2)(x^2-2). $$ Modulo any prime at least one of the numbers $-1$, $-2$, $2$ is a quadratic residue. Therefore $f(x)$ has a linear factor modulo $p$ for all primes $p$.

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    $\begingroup$ Thanks! Is it true that if $f(x)$ has a linear factor over $\mathbb{F}_p$ for every prime $p$, then $f(x)$ is reducible over $\mathbb{Q}$? $\endgroup$ – Next Dec 16 '13 at 10:48
  • $\begingroup$ I don't know. A very good question! $\endgroup$ – Jyrki Lahtonen Dec 16 '13 at 10:49

A degree $5$ counterexample is cited in this question: $(x^2 + 31)(x^3 + x + 1)$.

The reason can be understood by studying the cubic. The splitting field of a cubic always contains $\sqrt{\Delta}$, where $\Delta$ is the discriminant of the polynomial. Consequently, there are three ways a cubic can split in a finite field:

  • It has three roots
  • It is irreducible (and $\Delta$ must be square, because the splitting field is the cubic extension)
  • It has one root (and $\Delta$ is nonsquare)

For the given cubic, $\Delta = -31$; consequently if the cubic factor has no roots in the finite field, then the quadratic factor does.

  • $\begingroup$ (I assume characteristic greater than $3$; things become weird in characteristic 2 or 3, so those need to be checked separately) $\endgroup$ – Hurkyl Sep 4 '16 at 23:39

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