I am having trouble understanding part of the proof of Theorem 28 from Marcus's Number Fields. Let $L$ be a normal extension of $K$ (both number fields), let $R$ and $S$ be their respective integer rings, and let $Q$ be a prime ideal of $S$ lying over $P$. Let $L_H$ denote the fixed field of a subgroup $H$, and more generally for a set $X$ we write $X_H=X\cap L_H$. Let $G=\mathrm{Gal}(L/K)$ and $E=\{\sigma\in G\mid \sigma(\alpha)\equiv\alpha\text{ for all }\alpha\in S\}$.

The claim is that $f(Q, Q_E)$, the inertia degree of $Q$ over $Q_E$, is $1$. Equivalently $S/Q$ is the trivial extension of $S_E/Q_E$. It is sufficient to show the Galois group of $S/Q$ over $S_E/Q_E$ is trivial. To do this, we can show that for each $\theta\in S/Q$ the polynomial $(x-\theta)^m$ has coefficients in $S_E/Q_E$ for some $m\geq 1$.

The line I'm having trouble with: "Fix any $\alpha\in S$ corresponding to $\theta\in S/Q$. Then the polynomial $g(x)=\prod_{\sigma\in E}(x-\sigma\alpha)$ has coefficients in $S_E$."

Why is this true? I know that the coefficients will be sums of products of the form $\sigma\alpha$, which are sums and products of conjugates of $\alpha$, and so they will lie in $S$. However, $\alpha\in S$ means that $\sigma(\alpha)\equiv\alpha\mod Q$ for all $\sigma\in E$, not $\sigma(\alpha)=\alpha$, right? What am I missing?

  • $\begingroup$ Just from the definition of $g(x)$ and $S_E$. Since you know coefficients are in $S$, take any element in $E$ and apply to the coefficients of $g$. By the definition of $g$, the coefficients remain intact. $\endgroup$ Jan 14, 2020 at 20:13
  • $\begingroup$ Note also for any finite group $E$, a multiplication by an element $\tau\in E$, is a bijection on $E$. $\endgroup$ Jan 14, 2020 at 20:19
  • $\begingroup$ @SungjinKim I don't follow. $\endgroup$
    – ponchan
    Jan 16, 2020 at 20:50
  • $\begingroup$ Again, all I end up with is congrunce $\mathrm{mod}$ Q. $\endgroup$
    – ponchan
    Jan 16, 2020 at 20:55
  • $\begingroup$ Please see my answer. $\endgroup$ Jan 16, 2020 at 21:37

1 Answer 1


We can prove a slightly more general claim.


Let $\alpha\in S$, $H$ is a subgroup of the Galois group $G$, and $S_H=S \cap L_H$. Then $g(x)=\prod_{\sigma\in H} (x-\sigma \alpha)$ has coefficients in $S_H$.


Let $\tau \in H$. Then the multiplication by $\tau$ is a bijection on $H$. i.e. $x\mapsto \tau x$ is a bijection on $H$. Write the coefficients of $g$ as $g(x)=x^k+a_{k-1}x^{k-1}+\cdots +a_1 x + a_0$. Then we see that $a_i\in S$ for all $i\leq k-1$. This is what you also had so far. Now, we apply $\tau$ to the coefficients, then we have $$ \tau g(x)=x^k+\tau a_{k-1} x^{k-1}+ \cdots + \tau a_1 x + \tau a_0. $$ For each $i$, $a_i$ is a sum of products of $\sigma \alpha$ with $\alpha\in H$.

By the multiplicative and additive properties of $\tau$, we see that $\tau a_i$ is a sum of products of $\tau \sigma \alpha$ with $\sigma\in H$. Then $\tau g(x) = \prod_{\sigma\in H} (x-\tau\sigma \alpha)$. Since $x\mapsto \tau x$ is a bijection, $\tau\sigma$ ranges over all elements of $H$.Therefore, we have $\tau g(x) = \prod_{\sigma \in H} (x-\sigma \alpha)=g(x)$. That means the coefficients $a_i$'s are not changed by $\tau$. So, for each $i$, we have $a_i\in L_H$. Thereby, proving that $a_i\in S\cap L_H$ as desired.


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