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 $\mathbb{Q}$?

Question

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!

For a possible counterexample, do you want $f$ to be irreducible over $\mathbb{Q}$ also? — ronno, Dec 16 '13 at 10:42
I assume that $f \in \mathbb{Z}[x]$? +1 for the new question. — Martin Brandenburg, Dec 16 '13 at 11:15

score 13 · Accepted Answer · edited Mar 27 '16 at 14:45

13

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 $1$.

edited Mar 27 '16 at 14:45

user26857

52,094

answered Dec 16 '13 at 12:14

mercio

50,180

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 $p$? – Jyrki Lahtonen Dec 16 '13 at 12:28
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? – Jyrki Lahtonen Dec 16 '13 at 12:38
3

@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 $p$ – mercio Dec 16 '13 at 12:38
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!! – Jyrki Lahtonen Dec 16 '13 at 13:05
@mercio Hi. Can you please confirm if by Chebotarev's theorem you mean this: https://en.wikipedia.org/wiki/Chebotarev%27s_density_theorem Thanks. – caffeinemachine Nov 26 '20 at 17:54
@caffeinemachine yes, that's the one – mercio Nov 28 '20 at 22:54

score 11 · Answer 2 · answered Dec 16 '13 at 10:46

11

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 $p$.

answered Dec 16 '13 at 10:46

Jyrki Lahtonen

133,153

2

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 $\mathbb{Q}$? – lsr314 Dec 16 '13 at 10:48
I don't know. A very good question! – Jyrki Lahtonen Dec 16 '13 at 10:49

score 3 · Answer 3 · edited Apr 13 '17 at 12:19

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.

(I assume characteristic greater than $3$; things become weird in characteristic 2 or 3, so those need to be checked separately) — , Sep 04 '16 at 23:39

Rosie F · Answer 4 · 2021-03-25T09:25:23.927

The idea used by user14972 in their answer can be used to find other examples:

$f(x)=(x^2+3)(x^3+2)$. How many zeroes the cubic factor has in $\Bbb F_p$ depends on which quadratic form of discriminant $D=-108=-3\cdot6^2$, if any, represents $p$; if 3 zeroes, $a^2+27b^2$; if no zeroes, $4a^2+2ab+7b^2$; if one zero, neither quadratic form. If that cubic factor has no zeroes, then $-108$ is a square $c^2$; $p>3$ so 6 has an inverse, and $(6^{-1}c)^2+3=0$.

Similarly, $f(x)=(x^2+3)(x^3+3)$, $D=-243=-3\cdot9^2$.

user14972's example, and the two above, use the characteristic cubic of quadratic forms whose discriminants have some value $D$ where $h(D)=3$, i.e. where quadratic forms of discriminant $D$ are in three classes. But it is not necessary to use such concepts of quadratic form theory.

If $p=3$ or $p=1\mod 6$, then $p$ has the form $a^2+3b^2$, so $0=(a^2+3b^2)b^{-2}=(ab^{-1})^2+3$. So, if $x^2+3$ has no zeroes in $\Bbb F_p$, then $p=2$ or $p=5\mod6$, so everything is a cube.[1] So we may take $f(x)=(x^2+3)(x^3+a)$ where $a$ is any integer except a cube (so that $f$ has no linear factor in $\Bbb Q$).

[Brandl][2] gives the example $f(x)=(x^2+x+1)(x^3-2)$. The following proof that $f$ has a zero in each $\Bbb F_p$ is adapted from [Brandl]. If $p\ne 1\mod 3$, then the order of the multiplicative group of $\Bbb F_p$ is not divisible by 3. Hence 2 is a cube modulo $p$.[1] If $p=1\mod 3$, then there exists a primitive cube root of unity $x$ modulo $p$. [This follows from Sylow's first theorem.] This is obviously a root of the polynomial $x^2+x+1$. [This is because $0=x^3-1=(x-1)(x^2+x+1)$ but $x\ne 1$.]

[1] Proof. If $p=2$ or 3, $a^3=a$. If $p=6k+5$, then, by Fermat's little theorem, $a=a^p=a^{6k+5}$, so $a=r^3$ where $r=a^{4k+3}$. Alternatively, if any element of $\Bbb F_p$ were not a cube, then, by the pigeonhole principle, at least two unequal elements $a, b$ must have equal cubes. Then $0=a^3-b^3=(a-b)(a^2+ab+b^2)$; $a-b\ne 0$ so $a^2+ab+b^2=0$. The only element whose cube is 0 is 0, so $b\ne 0$, so $b$ has an inverse. So $(a^2+ab+b^2)b^{-2}=0$ so $c^2+c+1=0$ where $c=ab^{-1}$. So $0=(c^2+c+1)(c-1)=c^3-1$ so $c^3=1$. $a\ne b$ so $c\ne 1$, so $c$'s order is 3, so $3\mid p-1$, contradiction.

[2] Rolf Brandl, Daniela Bubboloni and Ingrid Hupp. Polynomials with roots mod $p$ for all primes $p$. J. Group Theory 4 (2001), p.233--239.

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 $\mathbb{Q}$?

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