3
$\begingroup$

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?

$\endgroup$
1
  • 4
    $\begingroup$ This is basically quadratic reciprocity. $\endgroup$ – Angina Seng Jan 23 '18 at 7:10
1
$\begingroup$

$p$ 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 $a$).

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.