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This is a qual problem from Princeton's website and I'm wondering if there's an easy way to solve it:

For which $p$ is $3$ a cube root in $\mathbb{Q}_p$?

The case $p=3$ for which $X^3-3$ is not separable modulo $p$ can easily be ruled out by checking that $3$ is not a cube modulo $9$. Is there an approach to this that does not use cubic reciprocity? If not, then I'd appreciate it if someone would show how it's done using cubic reciprocity. I haven't seen good concrete examples of it anywhere.

EDIT: I should have been more explicit here. What I really meant to ask was how would one find all the primes $p\neq 3$ s.t. $x^3\equiv 3\,(\textrm{mod }p)$ has a solution? I know how to work with the quadratic case using quadratic reciprocity, but I'm not sure what should be done in the cubic case.

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Isn't it the case that $3$ is a cube in $\mathbb{Q}_p$ if and only if it is a cube in $\mathbb{Z}_p$, and that this holds if and only if $x^3\equiv 3\pmod{p^k}$ has a solution for all $k$, which in turn is equivalent (by Hensel's Lemma) to checking that $x^3\equiv 3\pmod{p}$ has a solution (provided $p\neq 3$)? – Arturo Magidin Jan 11 '12 at 21:36
Yes, that's obviously true. I don't know if I'm just interpreting the problem wrong. But at least if a problem asks for something like "What primes are inert in $\mathbb{Q}(\sqrt{n})$?" the expectation would be to write down some congruence conditions that classify them. It might be that I'm just interpreting the problem in the wrong way... – pki Jan 11 '12 at 21:42
Another way to rephrase my question would then be: Being confronted with this question, would you expect that you have to write down congruence conditions? :) – pki Jan 11 '12 at 21:44
Yes, the point is that the question "for which primes does $x^3\equiv 3\pmod{p}$ have a solution" is a question that will give you an expression in terms of congruence classes of primes. So, yes, I would expect the answer to be given in terms of congruence conditions. – Arturo Magidin Jan 11 '12 at 21:47
The Galois group of $\mathbb{Q}(3^{1/3})$ is $S_3$, which is solvable but not abelian. Therefore, one expects that a congruence condition on $p$ will not be sufficient, but congruence conditions on $p$ and on the primes lying above $p$ in the intermediate number field $\mathbb{Q}(\sqrt{-3})$ will. Indeed, this is what happens in Arturo Magidin's answer. See… for a similar example. – David Speyer Jan 11 '12 at 22:26
up vote 4 down vote accepted

As noted in the comments, the question comes down to:

For which primes $p\gt 3$ is $3$ a cubic residue modulo $p$?

This is answered in detail in Franz Lemmermeyer's Reciprocity Laws, Chapter 7 ("Cubic Reciprocity").

If $p\equiv 2\pmod{3}$, then the order of the units modulo $p$ is prime to $3$, so every element is a cube; thus, $3$ is a cube modulo $p$ for all primes $p\equiv 2\pmod{3}$.

If $p\equiv 1\pmod{3}$, then one can write $4p = L^2 + 27M^2$ for integers $L$ and $M$, and $3$ is a cubic residue modulo $p$ if and only if $M\equiv 0\pmod{3}$ (Proposition 7.2 in Lemmermeyer).

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For odd primes $q \equiv 2 \pmod 3,$ the cubing map is a bijection, 3 is always a cube $\pmod q.$

For odd primes $p \equiv 1 \pmod 3,$ by cubic reciprocity, 3 is a cube $\pmod p$ if and only if there is an integer representation
$$ p = x^2 + x y + 61 y^2, $$ or $4p=u^2 + 243 v^2.$ In this form this is Exercise 4.15(d) on page 91 of Cox. Also Exercise 23 on page 135 of Ireland and Rosen. The result is due to Jacobi (1827).

For more information when cubic reciprocity is not quite good enough, see Representation of primes by the principal form of discriminant $-D$ when the classnumber $h(-D)$ is 3 by Richard H. Hudson and Kenneth S. Williams, Acta Arithmetica (1991) volume 57 pages 131-153.

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