Questions about the proof of Stickelberger's theorem on discriminants

Question

I was going through the proof of Stickelberger's theorem about discriminants in the book 'Algebraic Number Theory' by Richard A. Mollin, and I am having some problems in understanding the proof. I will state the theorem and the proof, and I will be highly grateful if anyone can answer my questions. I have also asked this question in MSE but have not got any answers.

$\textbf{Theorem :}$ 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. From the above how does it follow that $P+N$, $PN\in\mathbb{Q}$ ?

Answer 1

A bit too long for a comment, so I write it as an answer.

$(1)$ Let $L$ be the Galois closure of $K$ over $\mathbb{Q}$. We can apply $\sigma\in Gal(L,\mathbb{Q})$ on $P+N$ and $PN$, which lie in $L$.

$(2)$ Since either $\sigma(P)=P$ and $\sigma(N)=N$ or $\sigma(P)=N$ and $\sigma(N)=P$, it always follows that $\sigma(P+N)=P+N$ and $\sigma(PN)=PN$, for $\sigma\in Gal(L/\mathbb{Q})$, where $L$ is the Galois closure of $K$ over $\mathbb{Q}$.

$(3)$ So $P+N$ and $PN$ are fixed under elements of $Gal(L/\mathbb{Q})$, hence in $\mathbb{Q}$ by definition.

EDIT: Look here for more details (page 12): web.mit.edu/~holden1/www/math/ant.pdf

Answer 2

This is just a clarification of Dietrich Burde's remarks taken from Schur's original proof. Let $K$ be generated by a root $\beta$ of a monic polynomial $f \in {\mathbb Z}[x]$. Then each $\alpha_i$ is a rational polynomial in $\beta$, and each $\sigma_j$ permutes the roots of $f$; thus we may apply the $\sigma_j$ to $P$ and $N$. Since $P+N$ and $PN$ are symmetric functions of these roots (even permutations fix $P$ and $N$, odd permutations switch them), these expressions are rational algebraic integers, i.e., ordinary integers.