TITLE: Proof of Stickelberger’s Theorem
QUESTION [18 upvotes]: I am having some trouble in understanding the proof of Stickelberger’s  Theorem,
$\textbf{Theorem :}$ If $K$ is an algebraic number field then $\Delta_K$, the discriminant of $K$, satisfies $$\Delta_K\equiv 0,1\pmod{4}$$
$\textbf{Proof :}$ Let $\lbrace a_1,\ldots ,a_n\rbrace\subseteq\mathfrak{O}_K$ be an integral basis for $K$ and $\sigma_1,\ldots\sigma_n :K\to \mathbb{C}$ be all the embeddings of $K$. Then we have by definition, $$\sqrt {\Delta_K}=\det([\sigma_i(a_j)])$$ and this can be written as $$\sqrt{\Delta_K}=\sum_{\pi\in A_n}\prod_{i=1}^n\sigma_i\left(a_{\pi (i)}\right)-\sum_{\pi\not\in A_n}\prod_{i=1}^n\sigma_i\left(a_{\pi (i)}\right):=P-N$$
Now for each embedding $\sigma_i$ we have, $$\sigma_i(P+N)=P+N,\hspace{5mm}  \sigma_i(PN)=PN$$ and hence $P+N, PN\in\mathbb{Q}$.
Hence we have $P+N,PN\in\mathbb{Z}$, because $P$ and $N$ are both algebraic integers. Now using the identity $$(P-N)^2=(P+N)^2-4PN$$ it follows that $\Delta_K\equiv0,1\pmod{4}.$
$\underline{\textbf{My questions}}:$ 
$(1)$ How can we apply $\sigma_i$ to $P+N$ and $PN$, I mean how does it follow that $P+N, PN\in K$ ?
$(2)$ Why is $\sigma_i(P+N)=P+N$ and $\sigma_i(PN)=PN$ ?
$(3)$ How does it follow that $P+N, PN\in\mathbb{Q}$ ?

REPLY [12 votes]: Maybe it's a bit late but you can also give a proof by only looking at ramification of $2$ in $K$. The theorem follows by proving that $\Delta_K\equiv 0\bmod 2$ implies $\Delta_K\equiv 0\bmod 4$, and that $\Delta_K\equiv 1\bmod 2$ implies $\Delta_K\equiv 1\bmod 4$.
Assume that $2|\Delta_K$. Then there exists a prime $\mathfrak{p}$ of $K$ lying over $2$ with $e=e(\mathfrak{p}/2)\geq 2$. If $e=2$ then we have wild ramification and hence $\mathfrak{p}^2|\mathfrak{D}_K$. Taking norms we conclude that $4|\Delta_K$. If $e\geq 3$ then we get $\mathfrak{p}^2|\mathfrak{p}^{e-1}|\mathfrak{D}_K$ so that similarly we obtain $4|\Delta_K$.
If $\Delta_K$ is odd, write $\Delta_K=u^2d$ for $d$ square-free. If $d=1$ then $\overline{\Delta}_K$ is square in $(\mathbb{Z}/4\mathbb{Z})^{*}$ so that $\Delta_K\equiv 1\bmod 4$. If $d\neq 1$ then $\mathbb{Q}(\sqrt{\Delta_K})=\mathbb{Q}(\sqrt{d})\neq \mathbb{Q}$ and if $N$ is the normal closure of $K/\mathbb{Q}$ then $\mathbb{Q}(\sqrt{\Delta_K})\subset N$. As $\Delta_K$ is odd we have that $2$ is unramified in $K$, hence also in $N$, hence also in $\mathbb{Q}(\sqrt{d})$, which implies $d\equiv 1\bmod 4$. As $u$ is odd we conclude that $\Delta_K\equiv 1\bmod 4$ as well.