# Proof of Stickelberger’s Theorem

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 $P+N, PN\in\mathbb{Q}$ ?

First of all, a short remark: I cannot answer any of the three questions you asked at the end of your post. (For example, I also have the impression that it is not immediately clear why $P+N,PN \in K$.)
Similar as in the proof you outlined, we can also write $$\Delta_K=(\sum_{\pi \in X} \prod_{i=1}^n \sigma_{\pi(i)}a_i-\sum_{\pi \in Y} \prod_{i=1}^n \sigma_{\pi(i)}a_i)^2=(P-N)^2$$ where $X$ is the set of even permutations and $Y$ is the set of odd permuations in $S_n$. Note that we can assume $K \subset \mathbb{C}$. Let $L$ be a field with $K \subseteq L \subseteq \mathbb{C}$ and such that $L/ \mathbb{Q}$ is finite and Galois. (It is a well known fact in Galois Theory that a suitable $L$ exists). I will only explain the "critical step" of the proof, namely that $P+N, PN \in \mathbb{Q}$. (The rest should then be clear.) First of all, I show the following statement:
($\star$)\phantom{aaaaaaa}$$\phi(P+N)=P+N and \phi(PN)=PN for all \phi \in \text{Gal}(L/\mathbb{Q}) Let \phi \in \text{Gal}(L/\mathbb{Q}). For all the embeddings \sigma_i, we have \sigma_iK \subseteq L. Proof: Extend every \sigma_i to an embedding \overline{\sigma_i}:L \to \mathbb{C}. It follows from the normality of L that \overline{\sigma_i}L=L. A fortiori, we have \sigma_iK\subseteq L. Therefore, for every \sigma_i, we can build the composition \phi \circ \sigma_i and this is an embedding K \to \mathbb{C}. Now, it is not hard to see that the association \sigma_i \mapsto \phi \circ \sigma_i gives us a bijection$$\{\sigma_1,...,\sigma_n\} \to \{\sigma_1,...,\sigma_n\}$$But this means that we can find a permutation \tau \in S_n such that for every i \in \{1,...,n\}$$\phi\circ \sigma_i=\sigma_{\tau(i)}$$Distinguish two cases concerning \tau. Case 1: \tau is even. Then, \tau X=X and we have$$\begin{align} \phi(\sum_{\pi \in X} \prod_{i=1}^n \sigma_{\pi(i)}a_i) & = \sum_{\pi \in X} \prod_{i=1}^n \phi\circ\sigma_{\pi(i)}a_i \\ & = \sum_{\pi \in X} \prod_{i=1}^n \sigma_{\tau\pi(i)}a_i \\ & = \sum_{\pi \in \tau X} \prod_{i=1}^n \sigma_{\pi(i)}a_i \\ & =\sum_{\pi \in X} \prod_{i=1}^n \sigma_{\pi(i)}a_i \end{align}$This shows$\phi(P)=P$. In a similar fashion, you can show that$\phi(N)=N$(the key point is here that$Y=\tau Y$.) Case 2:$\tau$is odd. Here we have$\tau X=Y$and$\tau Y=X$. It follows that$\phi(P)=N$and$\phi(N)=P$(the proof is similar to the one in case 1: simply write down the formulas for$\phi(P)$and$\phi(N)$and during the calculations, make use of$\tau X=Y$and$\tau Y=X$). In every case,$\phi(P+N)=P+N$and$\phi(PN)=PN$and this shows ($\star$). Now,$(\star)$says exactly that$P+N$and$PN$are in the fixed field of the Galois group. But this means$P+N, PN \in \mathbb{Q}$. 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.