Suppose that $p$ is unramified in a number field $K$ of degree $n$. Apparently, Stickelberger proved that $\big( \frac{Disc(K)}{p}\big) = (-1)^{n - g}$, where $g$ is the number of prime ideal factors over $(p)$ in $K$.
Is there a convenient reference for this fact?
Thank you! -Frank
First reduce the question to the local case--This can be done since after base change to $\mathbb{Q}_{p}$, $K \otimes_{\mathbb{Q}} \mathbb{Q}_{p}=K_1 \times...\times K_{g}$, the trace form is the orthogonal sum of the local trace forms so the discriminant is the product of the local discriminants, and then use that the Legendre symbol is multiplicative. Now, since $p$ is unramified the extension $K_{i}/\mathbb{Q}_{p}$ is a cyclic extension of degree $f_{i}$, notation as usual, with Galois group generated by $\sigma$. We want to show that $\left ( \frac{disc(K_i)}{p}\right)= (-1)^{f_i -1}$. This is done in the exact same way one proves that a cubic extension is Galois iff its discriminant is a square; just notice that $(-1)^{f_i -1}$ is the the signature of $\sigma$ under the regular representation of $Gal(K_{i}/\mathbb{Q}_{p})$.
Remark: If you, like Conway, like the notation $p=-1$ for the infinite prime then the above proof works at that prime too. Just read Kronecker symbol instead of Legendre, and what one gets is the usual "sign of discriminant is $(-1 )^{s}$ "
