Let $K_q$ denote the unique quadratic subextension of the ray class field over $\mathbb{Q}$ of conductor $q\times\infty$. Then $K_q$ should be $\mathbb{Q}(\sqrt{q})$ if $q$ if 1 mod 4 and $\mathbb{Q}(\sqrt{-q})$ if $q$ is 3 mod 4 if I'm not mistaken. I've shown if $p$ splits in $K_q$ then $p$ is a square mod $q$ (without QR) and I need to show the converse without using quadratic reciprocity. (I need to generalize it to a setting where I haven't proven the exact quadratic reciprocity law I believe exists). Here $q$ and $p$ are both primes and we can assume $q$ is odd.

The question in the title would show this for $q=3$.

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    $\begingroup$ Hi Christine. There is a quadratic reciprocity law in every number field $K$, formulated as a product of Hilbert symbols is always $1$: if $a$ and $b$ are in $K^\times$ then $\prod_v (a,b)_v = 1$ where $v$ runs over all places of $K$. Maybe that will be enough for what you need. $\endgroup$ – KConrad Feb 19 '16 at 0:51
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    $\begingroup$ The question in the title is different from the one in the text. $\endgroup$ – Franz Lemmermeyer Feb 19 '16 at 5:34
  • $\begingroup$ Christine, The question here is what your already know in this context and are able to use. The ray class field is a cyclotomic field. If you know about the splitting of primes in cyclotomic fields and the Frobenius element in the Galois group, then you can apply those ideas. $\endgroup$ – user88166 Feb 25 '16 at 19:48
  • $\begingroup$ I figured it out based on Keith Conrad's suggestion. Thanks everyone! $\endgroup$ – Christine McMeekin Apr 20 '16 at 19:01

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