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


2 Answers 2


First of all a short remark: I cannot really answer any of the three questions you asked at the end of your post. (For example, I have also the impression that it is not immediately clear why $P+N,PN \in K$.)

Nevertheless, here is my reconstruction of the proof you asked for. In some details, it is different from the proof you outlined. Instead, my proof will be in accordance with the answer to your question on mathoverflow (In fact, my proof hopefully answers the questions you asked in the comments to the mathoverflow-answer.)

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 $\phi \in \text{Gal}(L/\mathbb{Q})$. It follows from the normality of $L$ that $\sigma_iK \subseteq L$ for all the embeddings $\sigma_i$. Therefore, all the $\sigma_i a_j$ are in $L$ and we can apply $\phi$ to those elements (and thereby to $P$ and $N$). Moreover, 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.

  • $\begingroup$ This is a very nice argument! I just used in in my algebraic number theory course. Do you have an attribution for it, by any chance? $\endgroup$ Nov 29, 2022 at 1:18
  • $\begingroup$ Thanks! I didn't see it anywhere else, I came up with the proof myself. Of course it still might be somewhere, I haven't looked extensively. The way I stumbled upon it is I was once was playing around with what kind of ramification patterns can occur in an S3-extension of Q based on the value of the discriminant, and I noticed that using Stickelberger never gave me more information compared to the type of arguments in the above proof. This made me try to prove it using such arguments, and luckily, it worked :) $\endgroup$
    – Tim.ev
    Nov 30, 2022 at 14:27

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