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Forgive me if this is a silly question. We know for $K = \mathbb{Q}(i)$ that the primes $p \in \mathbb{Z}$ which are inert in $\mathcal{O}_{K}=\mathbb{Z}[i]$ are exactly the primes which are $3$ modulo $4$. Hence, the inert primes lie in the arithmetic progression $\{4k + 3 : k \in \mathbb{Z}\}$. Of course the same can be said for split primes in this case.

Is the same true for other quadratic extensions? That is, are there $D \in \mathbb{Z}$ square-free such that the primes inert (or split) in $\mathcal{O}_{\mathbb{Q}(\sqrt{D})}$ lie in a (non-trivial) arithmetic progression?

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  • $\begingroup$ @Mathmo123 Hmmm okay. The symbol $\left( \frac{D}{p} \right)$ is determined by $D$ being a square modulo $p$ or not. So the only relation I can see to get would be quadratic. Where does $4D$ come in? $\endgroup$ – Freddie Nov 22 '19 at 1:11
  • $\begingroup$ @user714237 Thanks for the correction. I wrote that comment way too quickly! I've corrected it in an answer. $\endgroup$ – Mathmo123 Nov 23 '19 at 16:38
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In the case of quadratic extensions, we can be very explicit. By the Kummer-Dedekind theorem, a prime $p\nmid 2D$ splits in $\mathbb Q(\sqrt{D})$ if and only if $$X^2 - D$$ has a root modulo $p$, i.e. if and only if $\left(\frac Dp\right) =1$. (If $p\mid D$ and $D$ is squarefree, then the extension is ramified. If $p = 2$, one needs to be a bit more careful if $D\equiv 1 \pmod 4$.)

If $D$ is an odd prime, then, by quadratic reciprocity, $D$ is a square modulo $p$ if and only if $$\left(\frac pD\right)(-1)^{\frac{p-1}2\frac{D-1}2}=1.$$ If $D \equiv 1\pmod 4$, this is equivalent to $\left(\frac pD\right) =1$, which gives a congruence condition modulo $D$. If not, then $$\left(\frac pD\right)(-1)^{\frac{p-1}2\frac{D-1}2}=\left(\frac pD\right)(-1)^{\frac{p-1}2} = \left(\frac pD\right)\left(\frac p4\right) =\left(\frac p{4D}\right),$$ and we obtain a congruence condition modulo $4D$.

More generally, without loss of generality, we can assume that $D$ is squarefree and write $$D = \mathrm{sign}(D)2^a q_1\cdots q_n$$ where the $q_i$ are distinct primes and $a = 0$ or $1$. By quadratic reciprocity, $\left(\frac Dp\right) = 1$ if and only if \begin{align} 1 = \left(\frac Dp\right) &= \left(\frac{\mathrm{sign}(D)}p\right)\left(\frac{2^a}p\right)\prod_i\left(\frac{q_i}p\right)\\ &=(-1)^{\frac{p-1}2\frac{D-1}2}(-1)^{\frac{p^2-1}8a}\prod_i\left(\frac{p}{q_i}\right)(-1)^{\frac{p-1}2\frac{q_i-1}2}\\ &=\left(\frac pD\right)(-1)^{M\frac{D-1}2}, \end{align} where $M$ is some explicit constant. This gives a congruence condition modulo $D$ or $4D$ as before.


One can also ask the same question for an arbitrary extension $K/\mathbb Q$: is there are congruence criterion that determines how $p$ splits in $K$. As @user714237 points out, the answer is yes if $K/\mathbb Q$ is an abelian extension, i.e. a Galois extension with abelian Galois group. By the Kronecker-Weber theorem, any abelian extension is a subfield of $\mathbb Q(\zeta_N)$ for some $N$, and we obtain a congruence condition modulo $N$.

If $K/\mathbb Q$ is not abelian, then the splitting properties of $p$ will not be given by a congruence criterion alone. One of the goals of the Langlands program is to show that splitting properties in general can be given in terms of the Hecke eigenvalues of modular forms and their generalisations (see this question and its answers for an explicit example). However, this question is wide open in general.

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  • $\begingroup$ Very nice answer, and thank you for the followup question. I know that $\left( \frac{p}{D} \right) = 1$ exactly when $p \equiv n^{2} \mod D$ for some $n$. So if $a_{1}, \ldots, a_{\ell}$ are the squares modulo $D$, then $p$ lies in the progression $\{kD + gcd(a_{1}, |a_{2}-a_{1}|, \ldots, |a_{\ell} - a_{\ell-1}|\}_{k}$. Is this what you are implying? $\endgroup$ – Freddie Nov 23 '19 at 19:52
  • $\begingroup$ I don't think arithmetic progressions are the best way to think of these questions. But if you want to think in those terms, I would phrase it as $p$ splits if and only if $p$ lies in one of the progressions $a_i + kD$. $\endgroup$ – Mathmo123 Nov 23 '19 at 20:03

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