Decomposition group and Galois group I have a specific question about Serre's proof in his book, Local Fields, Proposition 20 in Chapter 1

The homomorphism from the decomposition group D, $\epsilon: D \to G(\overline{L}/\overline{K})$ is surjective.
He says to choose $\overline{a}$ to be a generator of the largest separable extension of $\overline{K}$ in $\overline{L}$. Then use approximation lemma to find a representative $a$ belonging to all the prime ideals $s(\beta)$ for $s \notin D$. Then look at $P(x) = \prod (x - s(a))$. The nonzero roots of $\overline{P}(x)$ all have the form $\overline{s(a)}$, with $s \in D$.

I don't understand this. It seems that if we are choosing $a$ to belong to $s(\beta)$ for $s \notin D$, then the roots should also be for $s \notin D$.
 A: I'll first explain the application of the approximation lemma in case it's unclear to anyone who happens to read this:

Approximation Lemma. Let $k$ be a positive integer. For every $i$, $1 \leq i \leq k$, let $\mathfrak{p}_i$ be distinct prime ideals of $A$, $x_i$ elements of $K$, and $n_i$ integers. Then there exists an $x \in K$ such that $v_{\mathfrak{p}_i}(x - x_i)\geq  n_i$ for all $i$, and $v_{\mathfrak{q}}(x)\geq 0$ for $\mathfrak{q} \neq \mathfrak{p}_1\dots \mathfrak{p}_k$.

With the notation as in the statement of the lemma, we let $x_1$ be any lift of $\overline{a}$, $\mathfrak{p}_1=\beta$, and $n_1=1$, and then take $\mathfrak{p}_2\dots\mathfrak{p}_k$ to be the prime ideals $s(\beta)$, for each $s\not\in D$, with $n_i=1$ and $x_i=0$ for $2\leq i\leq k$. Then the lemma gives us an element $a\in \mathscr{O}_L$ such that $a-x_1\in\beta$ (so that $a$ is a lift of $\overline{a}$ by our choice of $x_1$) and $a\in s(\beta)$ for each $s\not\in D$.
Then the nonzero roots of $P(x)$ are of the form $s(a)$ for $s\in D$. To see this, take $s\not\in D$, so that $s^{-1}\not\in D$. But from the application of the approximation lemma, we have $a\in s^{-1}(\beta)$, and hence $s(a)\in \beta$, which means $\overline{s(a)}=0$. The point is that we chose the lift $a$ of the generator $\overline{a}$ to have the property that whenever you conjugate it by an element not in the decomposition group, it lands in $\beta$. Because of this, the only way for one of the roots $\overline{s(a)}$ of $\overline{P}(x)$ to be nonzero is if $s\in D$.
