I couldn't figure out a proof of the following statement while I'm reading the book "Fields and Galois Theory" by J. Milne.

Let $A$ to be a UFD and let $P$ be a prime ideal of $A$, and let $Q=\operatorname{Frac}(A)$ and $Q'=\operatorname{Frac}(A/P)$. Assume $f(x)\in A[x]$ is a monic polynomial without multiple roots, such that its reduction mod $P$ $f'(x)\in (A/P)[x]$ is also without multiple roots. Then the Galois group $G'$ of $f'$ over $Q'$ is a subgroup of the Galois group $G$ of $f$ over $Q$ as permutation groups.

Any ideas?

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    $\begingroup$ You should consider accepting some of the answers to your previous questions. It marks them as completed and is a good way to reward the people who answer them. $\endgroup$ – Alex Becker Feb 22 '12 at 0:11
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    $\begingroup$ @Alex, looks like CC took your advice - and accepted an answer I posted over two months ago. $\endgroup$ – Gerry Myerson Feb 22 '12 at 0:41
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    $\begingroup$ This is Lemma 4.28 in the notes that you mentioned. Did you look at the references given there? van der Waerden is a little ancient, but I remember it being readable. $\endgroup$ – Dylan Moreland Feb 22 '12 at 4:19

I assume that you have elementary knowledge of commutative algebra; integral dependence, the lying-over theorem, localizations of rings by multiplicative subsets, etc.(for example, see Atiyah-MacDonald). I will prove a slightly more general theorem than yours. Since a UFD is integrally closed, your theorem will follow immediately.

Theorem Let $A$ be an integrally closed domain and let $P$ be a prime ideal of $A$. Let $K$ be the field of fractions of A. Let $\tilde{K}$ be the field of fractions of $A/P$. Let $f(X) \in A[X]$ be a monic polynomial without multiple roots. Let $\tilde{f}(X) \in (A/P)[X]$ be the reduction of $f(X)$ mod $P$. Suppose $\tilde{f}(X)$ is also wihout multiple roots. Let $L$ be the splitting field of $f(X)$ over $K$. Let $G$ be the Galois group of $L/K$. Let $\tilde{L}$ be the splitting field of $\tilde{f}(X)$ over $\tilde{K}$. Let $\tilde{G}$ be the Galois group of $\tilde{L}/\tilde{K}$. Then $\tilde{G}$ is isomorphic to a subgroup of $G$.

To prove this theorem, we need some notations.

Definition Let $A$ be a ring. Let $G$ be a group. Let $Aut(A)$ be the automorphism group of $A$. Supppose there exists a homomorphism $\psi:G \rightarrow Aut(A)$. We say $G$ acts on $A$. For $\sigma \in G$ and $x \in A$, we denote $\psi(\sigma)(x)$ by $\sigma.x$. We denote the set $\{x \in A;\sigma.x = x$ for all $\sigma \in G\}$ by $A^G$. Then $A^G$ is a subring of $A$.

The proof of the theorem: Let $\alpha_1, ..., \alpha_n$ be the roots of $f(X)$ in $L$. Let $B = A[\alpha_1, ..., \alpha_n]$. Since $f(X)$ is monic, $B$ is integral over A. Let $S = A - P$. Let $A_S$ and $B_S$ be localizaions of $A$ and $B$ respectively. Then $B_S = A_S[\alpha_1, ..., \alpha_n]$. Since A is integrally closed, $A_S$ is also integrally closed. By replacing $A$ and $P$ by $A_S$ and $PA_S$ respectively, we can assume $P$ is a maximal ideal of $A$. By the lying-over theorem, there exits a prime ideal $M$ of $B$ such that $P = A \cap M$. Since $B$ is integral over $A$ and $P$ is a maximal ideal of $A$, $M$ is a maximal ideal of $B$.

Obviously $G$ acts on $B$. $A \subset B^G \subset K$ and $B^G$ is integral over $A$. Since $A$ is integrally closed, $B^G = A$.

Let $\pi:B → B/M$ be the canonical map. Then $B/M = (A/P)[\pi(\alpha_1), ..., \pi(\alpha_n)] = \tilde{K}(\pi(\alpha_1), ..., \pi(\alpha_n))$. Since $\pi(\alpha_1), ..., \pi(\alpha_n)$ are all the roots of $\tilde{f}(X)$, $B/M$ can be identified with $\tilde{L}$. Since $\tilde{f}(X)$ is wihout multiple roots, $B/M$ is separable over $A/P$. Hence $B/M$ has a primitive element $\theta$ over $A/P$. Let y be an element of B such that $\pi(y) = \theta$. Let $H = \{\sigma \in G; \sigma(M) = M\}$. Each $\sigma \in H$ induces $\tilde{\sigma} \in \tilde{G}$. Hence we get a homomorphism $\psi: H \rightarrow \tilde{G}$. It suffices to prove that $\psi$ is an isomorphism. For each $\sigma \in G - H, \sigma(M) \neq M$. Hence, by the Chinese remainder theorem, there exists $x \in B$ such that $x \equiv y$ $(mod M)$, $x \equiv 0$ $(mod$ $\sigma(M))$ for each $\sigma \in G - H$

Let $F(X) = \prod_{\sigma \in G}(X - \sigma(x)) \in B[X]$. Since each coefficient of $F(X)$ is invariant by $G, F(X) \in (B^G)[X] = A[X]$. Let $\tilde{F}(X) = \prod_{\sigma \in G}(X - \pi(\sigma(x))$. Then $\tilde{F}(X) \in (A/P)[X]$. Let $\lambda$ be any element of $\tilde{G}$. Since $\pi(x)$ is a root of $\tilde{F}(X)$, $\lambda(\pi(x))$ is also a root of $\tilde{F}(X)$. Hence there exists $\tau \in G$ such that $\lambda(\pi(x)) = \pi(\tau(x))$. For each $\sigma \in G - H$, $\sigma(x) \in M$. Hence $\pi(\sigma(x)) = 0$. On the other hand, $\pi(\tau(x)) = \tilde{\tau}(\pi(x)) = \tilde{\tau}(\theta) \neq 0$ Hence $\tau \in H$. Since $\lambda(\theta) = \pi(\tau(x)) = \tilde{\tau}(\pi(x)) = \tilde{\tau}(\theta)$, $\lambda = \tilde{\tau}$. Therefore $\psi: H \rightarrow \tilde{G}$ is surjective.

It remains to show that $\psi$ is injective. Suppose $\psi$ is not injective. Then there exists $\sigma \in H$ such that $\sigma \neq 1$ and $\psi(σ) = 1$. Hence there exists a root $\alpha$ of $f(X)$ such that $\sigma(\alpha) \neq \alpha$. Since $\sigma(\alpha)$ $\equiv$ $\alpha$ $(mod M)$, $\tilde{f}(X)$ has a multiple root $\pi(\alpha)$. This is a contradiction. QED

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