# nth powers modulo all primes

Let $a \in \mathbb{Z}$, $n \in \mathbb{N}^*$ be integers, and set $P=X^n - a$. Let us consider the three following statements :

1) $P$ has a root in $\mathbb{Z}$ (i.e. $a$ is an nth power)

2) $P$ has a root in $\mathbb{Z}/d\mathbb{Z}$ for all $d \ge 1$ (i.e. $a$ is an nth power mod $d$ for all $d$)

3) $P$ has a root in $\mathbb{Z}/p\mathbb{Z}$ for all prime $p$ (i.e. $a$ is an nth power mod $p$ for all prime $p$)

I would like to prove that they are equivalent, but I still can't prove $3 \Rightarrow 1$.

Here's what I already know :

• Obviously we have $1 \Rightarrow 2 \Rightarrow 3$

• Proving $2 \Rightarrow 1$ is not very difficult ($2 \Rightarrow$ all $p$-adic valuations of a are multiples of $n$)

• I think I can prove $3 \Rightarrow 1$ in the case $n = 2$. But my proof seems quite involved (and frankly a bit of an overkill) for this question : Roughly, if $a$ is not a square, I use the quadratic reciprocity law and Dirichlet's theorem to prove that there exist a prime $p$ for which the Legendre symbol of $a$ mod $p$ is $-1$.

So I guess my questions are : Can we prove $3 \Rightarrow 1$ ? Is there an "elementary" proof ? Thanks in advance for any comment or answer.

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How are you using quadratic reciprocity? This is an $n$th power. –  quanta May 6 '11 at 16:08
@ quanta : I was using quadratic reciprocity in the case $n = 2$. That is why I couldn't tackle the general case (I don't know much about higher order explicit reciprocity laws). Seeing the answer below, I realize now that my attempt was misguided. –  Joel Cohen May 6 '11 at 16:19
–  Qiaochu Yuan May 6 '11 at 16:59
Yes, this completely answers my question. Quite surprising and interesting... –  Joel Cohen May 6 '11 at 17:25

These statements are not equivalent. I claim that 3), but not 1) or 2), hold for the polynomial $x^8 - 16$ (see the Wikipedia article on the Grunwald-Wang theorem). The fact that 1) doesn't hold is obvious. Note that $16$ is an eighth power in $\mathbb{Z}/2\mathbb{Z}$, but not in $\mathbb{Z}/32\mathbb{Z}$, so 2) doesn't hold. Factor the polynomial as

$$(x^2 - 2)(x^2 + 2)(x^2 - 2x + 2)(x^2 + 2x + 2).$$

For any odd prime $p$, it's not hard to see that either $2, -2$, or $-1$ is a quadratic residue $\bmod p$. Depending on which of these is a quadratic residue, one of the quadratic fators above splits. Hence $x^8 - 16$ has a root $\bmod p$ for all odd primes $p$ and 3) holds.

Edit: Nevertheless, here is a positive result.

Proposition: Let $P(x) \in \mathbb{Z}[x]$ be a monic irreducible polynomial of degree $n > 1$. Then there exists a prime $p$ such that $P(x)$ has no roots $\bmod p$.

Proof. Given a prime $p$ not dividing the discriminant of $P$, consider the action of the Frobenius map $x \mapsto x^p$ on the roots of $P(x)$ in $\overline{ \mathbb{F}_p }$. Any orbit of this action gives an irreducible factor of $P(x)$, and conversely: hence the cycle type of the Frobenius map determines the factorization of $P(x)$ into irreducible factors $\bmod p$. Now, it is known (see the Wikipedia article) that any cycle type of a Frobenius map agrees with the cycle type of some element of the Galois group $G$ of $P$. Conversely, if the Galois group $G$ has an element of a particular cycle type, there is a prime $p$ such that the Frobenius element at $p$ has that cycle type by the Frobenius density theorem.

So it remains to show that $G$ has an element with no fixed points. Since $P$ is irreducible, $G$ acts transitively on its roots. By Burnside's lemma, it follows that

$$\sum_{g \in G} |\text{Fix}(g)| = |G|.$$

But the identity element has $n > 1$ fixed points, so it must be the case that $|\text{Fix}(g)| = 0$ for some $g$ or else the LHS would be at least $(n-1) + |G|$.

The upshot is that the equivalence between 2) and 3) holds whenever $x^n - a$ would be irreducible if $a$ weren't an $n^{th}$ power, which holds (for example) when $n = 2, 3$.

Note that it is not true that $P(x)$ remains irreducible modulo some prime: this is equivalent to the claim that the Galois group contains a $n$-cycle, which is always true when $n = 2, 3$ and false in general for higher values of $n$. However, it is true for a generic polynomial in the sense that the generic polynomial of degree $n$ has Galois group $S_n$.

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My mistake. Hensel's Lemma does not hold if $p$ divides $n$. Oh well, thank you for your counterexample. –  Joel Cohen May 6 '11 at 17:16
Very interesting addition to your original answer. So it seems one cannot avoid using some sort of density principle. –  Joel Cohen May 6 '11 at 19:33
The fact that the argument goes without considering the ramified primes separately bothers me a bit. Let say $p$ is a prime that divides the discriminant of $P(x)$, then $P(x)$ will have multiple roots mod $p$. What does it mean "any cycle type of a Frobenius map agrees with the cycle type of some element of the Galois group G of P" in this case? –  Jiangwei Xue May 7 '11 at 8:07
@Jiangwei: whoops. That first sentence should be "given a prime $p$ not dividing the discriminant of $P$." –  Qiaochu Yuan May 7 '11 at 8:19

Also, the simplest example of a polynomial that is irreducible over $\mathbb Q$ but reducible modulo every prime is the traditional fave, $x^4+1$.

Edit: as suggested by a comment (thx! @Pierre-Yves G.), one proof of this begins by noting that $x^4+1$ is the 8th cyclotomic polynomial. Since $\mathbb F_{p^2}^\times$ is cyclic, there will be an 8th root of unity in $\mathbb F_{p^2}$ exactly when $8|p^2-1$, which would mean that $x^4+1$ factors into at worst quadratic factors over $\mathbb F_p$. Since $\mathbb Z/8^\times$ is a 2,2 group, $p^2=1$ mod $8$ for every odd prime $p$. ($x^4+1=(x+1)^2$ at $p=2$.)

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Dear Paul: I think it would be nice if you could add a proof of your statement (or at least a link to such a proof). –  Pierre-Yves Gaillard Oct 13 '11 at 1:50
This is a nice example. Thanks. –  Joel Cohen Oct 13 '11 at 21:18