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Let $p$ be a fixed prime. How do I prove that there are infinitely many imaginary quadratic fields in which $p$ splits completely?

I know the statement of the Chebotarev density theorem and I think it should be applicable here, though I do not see immediately how. Can you please help me out?

dekimashita
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    Not obvious to me that Tchebotarev is exactly the point... maybe think about Dirichlet's theorem on primes in arithmetic progressions...? – paul garrett Oct 26 '21 at 22:13
  • Yes, I actually thought so too since Chebotarev applies for a fixed number field and varying $p$ whereas here we varying number fields and have a fixed $p$. – dekimashita Oct 26 '21 at 22:17
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    Given a quadratic field $\mathbb Q(\sqrt{-D})$, do you know how to compute whether (an odd prime) $p$ splits completely in terms of the polynomial $x^2 + D$? If so, can you find a criterion for $p$ to split completely in terms of $D$? – Mathmo123 Oct 27 '21 at 04:51

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The Dedekind-Kummer theorem tells us in quadratic number fields $\mathbb Q(\sqrt d)$, $p$ splits iff $\left(\frac \Delta p \right) = 1$, that is using Kronecker symbol, $\Delta$ is a quadratic residue for odd $p$ or $\pm 1 \pmod 8$ for $p = 2$.

Recall $\Delta$ is the discriminant, which is $4d$ for $d \equiv 2,3 \pmod 4$ or $d$ for $d \equiv 1 \pmod 4$. Then it should be easy to reason about the $d$ such that $p$ splits.

qwr
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For a fixed prime p and varying quadratic number fields $K = \mathbf Q (\sqrt D)$, it seems easier to look at the completions of K at the prime ideals of $O_K$ above p. Here p is required to split, say (p)=P.Q , with $P\neq Q$, so the local degrees $n_P, n_Q$ over $\mathbf Q_p$ must both be $1$. Equivalently, $K_P = K_Q = \mathbf Q_p$, and $D$ is a square in $\mathbf Q_p$. Using the well-known structure of $\mathbf Q_p ^*$ mod squares, this last condition means that : 1) If $p \neq 2$ and $D= p^n u$, where $u$ is a p-adic unit, then $n$ is even and the image of $u$ is a non null square in the residual field $\mathbf F_p$ ; 2) If $p=2$ and $D= 2^n u$, then $n$ is even and $u \equiv 1$ mod $8$ .

nguyen quang do
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