Suppose $(a,3q)=1$ and $a\equiv 1\pmod 3$. Are there infinitely many primes $p\equiv a\pmod {3q}$ such that $3$ is a cubic nonresidue modulo $p$?

Or, an equivalent formulation using quadratic forms: by cubic reciprocity the problem is equivalent to asking whether there are infinitely many primes $p\equiv a\pmod {3q}$ that cannot be expressed in the form $4p=a^2+243 b^2$ for integers $a$ and $b$. Since 243 is not convenient, we know that there are no integers $a_1,\ldots,a_k, N$ such that $4p=a^2+243b^2$ is solvable if and only if $p\equiv a_i \pmod N$ for some $i$. Can we conclude a stronger result, that there is no residue class $a \pmod {3q}$ with $a\equiv 1\pmod 3$ containing only primes expressible in the form $4p=a^2+243b^2$?


The answer is yes; this follows from the Chebotarev density theorem.

To avoid technicalities, I only consider here the case when $3$ does not divide $q$. Let $K_1 = \mathbf{Q}(\zeta_q)$ and $K_2=\mathbf{Q}(\zeta_3, \sqrt[3]{3})$. These are both Galois extensions of $\mathbf{Q}$. We claim that these are linearly disjoint.

To see this, notice that $L:=K_1 K_2 = \mathbf{Q}(\zeta_{3q}, \sqrt[3]{3})$. Now $\sqrt[3]{3} \not \in \mathbf{Q}(\zeta_{3q})$. Otherwise $\mathbf{Q}(\sqrt[3]{3}) \subset \mathbf{Q}(\zeta_{3q})$; since the latter field is abelian over $\mathbf{Q}$, this would imply that $\mathbf{Q}(\sqrt[3]{3})$ is Galois over $\mathbf{Q}$, an absurdity. Since $\sqrt[3]{3}$ is not in $\mathbf{Q}(\zeta_{3q})$, it follows that $x^3-3$ is irreducible over $\mathbf{Q}(\zeta_{3q})$, and so $[L:\mathbf{Q}] = 3 \cdot [\mathbf{Q}(\zeta_{3q}):\mathbf{Q}] = 3 \varphi(3q) = 6 \phi(q) = [K_2:\mathbf{\mathbf{Q}}] [K_1:\mathbf{Q}]$. So we have linear disjointness and the Galois group of $L/\mathbf{Q}$ is just the direct product of the Galois groups corresponding to $K_1$ and $K_2$.

But now life is good: Let $\sigma_1$ be the automorphism of $K_1/\mathbf{Q}$ identified with $a\bmod{q}$, under the usual identification of $\mathrm{Gal}(\mathbf{Q}(\zeta_q)/\mathbf{Q})$ with $(\mathbf{Z}/q\mathbf{Z})^{\times}$, and let $\sigma_2$ be the automorphism of $K_2/\mathbf{Q}$ keeping $\zeta_3$ fixed but sending $\sqrt[3]{3}$ to one of the other roots of $x^3-3$. Let $\sigma$ be an automorphism of $L$ that restricts to $\sigma_1$ on $K_1$ and $\sigma_2$ on $K_2$. Note that there are $2$ possible choices of $\sigma$, and that these define a conjugacy class in the Galois group of $L$ over $\mathbf{Q}$.

By Chebotarev, there are infinitely many (unramified) primes $p$ whose Frobenius, in $L$, is the conjugacy class of $\sigma$. Then $p \equiv 1 \pmod{3}$ (since $\sigma_2$ restricts to the identity on $\mathbf{Q}(\zeta_3)$) and $p\equiv a\pmod{q}$, which implies that $p\equiv a\pmod{3q}$; on the other hand, since the Frobenius restricted to $K_2$ is not trivial, $p$ does not split completely in $K_2$, and hence $3$ is not a cube mod $p$.

In fact, the proportion of primes $p$ with $p \equiv a\pmod{3q}$ and $3$ not a cube is, by this argument, $2/[L:\mathbf{Q}] = \frac{1}{3\phi(q)}$. Since the proportion of primes $p$ with $p \equiv a\pmod{3q}$ is $\frac{1}{\phi(3q)} = \frac{1}{2\phi(q)}$, there's a $2/3$ chance that a prime in the given residue class will not have $3$ as a cubic residue. (Of course, this is what one would naively guess.)

A similar approach will work if $3\mid q$; here one can let $q'$ be the $3$-free part of $q$ and then argue with $K_1 = \mathbf{Q}(\zeta_{q'})$ and $K_2 = \mathbf{Q}(\zeta_{3 q/q'}, \sqrt[3]{3})$.

  • 1
    $\begingroup$ Nice, but I do not understand why you chek that $2$, rather than $3$, be a cubic non-residue. $\endgroup$ – Filippo Alberto Edoardo Jul 17 '14 at 8:07
  • 1
    $\begingroup$ Thanks Filippo. That's what I get for not reading the question carefully! Edited. $\endgroup$ – so-called friend Don Jul 17 '14 at 14:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.