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 $4p=a^2+243b^2$?


1 Answer 1


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 $K_2 = \mathbf{Q}(\zeta_{3 q/q'}, \sqrt[3]{3})$.

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    $\begingroup$ Nice, but I do not understand why you chek that $2$, rather than $3$, be a cubic non-residue. $\endgroup$ Jul 17, 2014 at 8:07
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    $\begingroup$ Thanks Filippo. That's what I get for not reading the question carefully! Edited. $\endgroup$ Jul 17, 2014 at 14:30

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