11
$\begingroup$

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 $P+N$, $PN\in\mathbb{Q}$ ?
$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ May 18, 2013 at 15:55
  • 1
    $\begingroup$ This proof originates with Schur (eudml.org/doc/168085). $\endgroup$ May 19, 2013 at 1:30

2 Answers 2

9
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ Can you please tell how do you prove your first and second claim ? $\endgroup$ May 18, 2013 at 18:11
  • $\begingroup$ I am not convinced of (1). (The Holden proof sidesteps this completely by only using that $P$ and $N$ lie in $K^{\mathrm{gal}}$. $\endgroup$ May 18, 2013 at 22:04
  • $\begingroup$ @darij: sorry, you are right. $\endgroup$ May 18, 2013 at 22:55
  • $\begingroup$ Just a note for myself: given any $\tau \in Gal(L / \Bbb Q)$, the set $\{\tau \circ \sigma_i\}_i$ is a permutation of the $\{\sigma_i\}$, so that we have the dichotomy in (2). $\endgroup$
    – Alphonse
    Jun 22, 2021 at 19:33
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.