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

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I sympathize with the question, but please tell us the source of the proof! This is not about shaming the author (many books have worse blunders), but about knowing the context and possibly the standing assumptions that might explain how things are being identified with others. – darij grinberg May 18 '13 at 15:55
This proof originates with Schur ( – Chandan Singh Dalawat May 19 '13 at 1:30

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):

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Can you please tell how do you prove your first and second claim ? – Pritam Majumder May 18 '13 at 18:11
I am not convinced of (1). (The Holden proof sidesteps this completely by only using that $P$ and $N$ lie in $K^{\mathrm{gal}}$. – darij grinberg May 18 '13 at 22:04
@darij: sorry, you are right. – Dietrich Burde May 18 '13 at 22:55

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