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Say I want to know which primes ramify, split, or stay inert in the field $\mathbb{Q}(\sqrt{a})$, where $a$ is a squarefree integer. The ramified primes are easy to find - they are just primes that divide the discriminant. For other primes, we know the splitting behavior can be classified by what the prime is modulo $N$, where $N = a$ for $a \equiv 1 \bmod 4$ and $N = 4a$ for $a \equiv 2,3 \bmod 4$. My question is: is there an easy way to prove this fact?

The way this is presented that I've seen is $\mathbb{Q}(\sqrt{a})$ can be embedded in $\mathbb{Q}(\zeta_N)$ for $N$ as given above, and this $N$ is the smallest $N$ possible. And we know that for a subfield $L$ of $\mathbb{Q}(\zeta_N)$ corresponding to a subgroup $H$ of $Gal(\mathbb{Q}(\zeta_N)/\mathbb{Q}) \cong (\mathbb{Z}/N\mathbb{Z})$, a prime $p$ splits completely in $L$ if and only if $p$ mod $N \in H$.

This feels somewhat unnatural to me... so I wonder if there's an easy way to see this, maybe not involving cyclotomic fields?

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    $\begingroup$ This is basically quadratic reciprocity. $\endgroup$ Jan 23, 2018 at 7:10

2 Answers 2

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$p$ splits in $\Bbb Q(\sqrt a)$ iff $a$ is a quadratic residue modulo $p$.

Suppose that $a>0$ and $a\equiv1\pmod 4$. Then the Legendre symbol $\left(\frac{a}p\right)$ equals the Jacobi symbol $\left(\frac pa\right)$ and so only depends on $p$ modulo $a$.

Suppose that $a>0$ and $a\equiv3\pmod 4$. Then $\left(\frac{a}p\right)$ equals $(-1)^{(p-1)/2}\left(\frac pa\right)$ and so only depends on $p$ modulo $4a$.

And so on, with the other cases (with $a$ even, $a<0$ or both). In each case quadratic reciprocity tells you that $\left(\frac{a}p\right)$ will depend only on $p$ modulo $4a$ (or possibly just modulo $a$).

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This is explained rather well in Samuel's Algebraic Theory of Numbers in Chapter 5.4. If you don't have access to that book (it might be online) then I could give a quick sketch of the ideas involved. Also Marcus' Number Fields probably does this explicitly too.

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