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Given a finite set $S$ of primes, is it possible to find an imaginary quadratic field $K$ such that all primes in $S$ are split completely in $K$?

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up vote 3 down vote accepted

Sure. Let me just assume WLOG that $S$ contains $2$. Let $D$ be squarefree. If $D \equiv 1 \bmod 4$, then $\mathbb{Q}(\sqrt{D})$ has ring of integers $\mathbb{Z}[x]/(x^2 - x - \frac{D-1}{4})$. Then $2$ splits if and only if $D \equiv 1 \bmod 8$. An odd prime $p$ splits if $D \equiv 1 \bmod p$, so it suffices to find a (negative) squarefree $D \equiv 1 \bmod 8$ such that $D \equiv 1 \bmod p$ for all odd primes $p \in S$. In fact a prime with this property exists by Dirichlet's theorem.

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2 splits completely in $\mathbb{Q}(\sqrt{d})$ for $d$ squarefree and 1 mod 8, right? – user29743 Aug 1 '12 at 16:53
Oops. Let me fix that. – Qiaochu Yuan Aug 1 '12 at 16:54
I think with that addition the answer to OP's question is yes and you can do it explicitly with CRT + quadratic reciprocity, I am a little lazy on working out the details and would make a mistake anyway! – user29743 Aug 1 '12 at 16:55
Also, you want $D \equiv 1 mod 4$. – David Speyer Aug 1 '12 at 16:55
@countinghaus: I am not sure things are so easy. I think I know how to prove that there are infinitely many primes $D$ congruent to $1 \bmod 4$ such that $\left( \frac{D}{p} \right) = 1$ if $p$ is prime, but here we need $p$ to be a product of primes and the Jacobi symbol no longer reflects quadratic residue-ness. – Qiaochu Yuan Aug 1 '12 at 17:27

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