# Inertia and Decomposition fields

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?

• 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. Jan 14, 2020 at 20:13
• Note also for any finite group $E$, a multiplication by an element $\tau\in E$, is a bijection on $E$. Jan 14, 2020 at 20:19
• @SungjinKim I don't follow. Jan 16, 2020 at 20:50
• Again, all I end up with is congrunce $\mathrm{mod}$ Q. Jan 16, 2020 at 20:55

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.