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?