The Stacks project

Proof. Choose $a \in A$ lifting $b$ and set $P = \prod _{\sigma \in G} (T - \sigma (a))$. $\square$

Proof. Since $I^ G = R^ G \cap I$ it is clear that the map is injective. Lemma 15.110.1 shows that Algebra, Lemma 10.46.11 applies. $\square$

Proof. Observe that the polynomial is indeed monic and has coefficients in $R^ G$. We can write $x = f_1 b_1 + \ldots + f_ m b_ m$ with $f_ j \in J$ and $b_ j \in R$. Thus, arguing by induction on $m$, we may assume that $x = y - fb$ with $f \in J$, $b \in R$, and $y \in JR$ such that the result holds for $y$. Then we see that

where we have

A computation we omit shows that $a_ i \in R^ G$ (hint: the given expression is symmetric). Thus the polynomial of the statement of the lemma for $x$ is congruent modulo $J$ to the polynomial for $y$ and this proves the induction step. $\square$

Proof. Part (1) follow from Lemma 15.110.1 with $I = JR$. If $a$ is as in part (2), then $a$ is the image of $x \in R^ G \cap JR$. Hence $(T - x)^ n = \prod _{\sigma \in G} (T - \sigma (x))$ is congruent to $T^ n$ modulo $J$ by Lemma 15.110.3. This proves part (2). To see the final statement we may apply Algebra, Lemma 10.46.11. $\square$

Proof. Choose a surjection $E \to A$ where $E$ is a polynomial algebra over $R^ G$. Then $(E \otimes _{R^ G} R)^ G = E$ because $E$ is free as an $R^ G$-module. Denote $J = \mathop{\mathrm{Ker}}(E \to A)$. Since tensor product is right exact we see that $A \otimes _{R^ G} R$ is the quotient of $E \otimes _{R^ G} R$ by the ideal generated by $J$. In this way we see that our lemma is a special case of Lemma 15.110.4. $\square$

Proof. To see the first statement consider the exact sequence $0 \to R^ G \to R \to \bigoplus _{\sigma \in G} R$ where the second map sends $x$ to $(\sigma (x) - x)_{\sigma \in G}$. Tensoring with $A$ the sequence remains exact if $R^ G \to A$ is flat. Thus $A$ is the $G$-invariants in $(A \otimes _{R^ G} R)^ G$.

The second statement follows from Lemma 15.110.6 and Algebra, Lemma 10.46.11. $\square$

Proof. The extension $R^ G \subset R$ is integral because every $x \in R$ is a root of the monic polynomial $\prod _{\sigma \in G}(T - \sigma (x))$ in $R^ G[T]$. Thus there are no inclusion relations among the primes lying over a given prime $\mathfrak p$ (Algebra, Lemma 10.36.20). If the lemma is wrong, then we can choose $x \in \mathfrak q'$, $x \not\in \sigma (\mathfrak q)$ for all $\sigma \in G$. See Algebra, Lemma 10.15.2. Then $y = \prod _{\sigma \in G} \sigma (x)$ is in $R^ G$ and in $\mathfrak p = R^ G \cap \mathfrak q'$. On the other hand, $x \not\in \sigma (\mathfrak q)$ for all $\sigma $ means $\sigma (x) \not\in \mathfrak q$ for all $\sigma $. Hence $y \not\in \mathfrak q$ as $\mathfrak q$ is a prime ideal. This is impossible as $y \in \mathfrak p \subset \mathfrak q$. $\square$

Proof. With $A = (R^ G)_\mathfrak p$ and $B = A \otimes _{R^ G} R$ we see that $A = B^ G$ as localization is flat, see Lemma 15.110.7. Observe that $\mathfrak pA$ and $\mathfrak qB$ are prime ideals, $D$ is the stabilizer of $\mathfrak qB$, and $\kappa (\mathfrak p) = \kappa (\mathfrak pA)$ and $\kappa (\mathfrak q) = \kappa (\mathfrak qB)$. Thus we may replace $R$ by $B$ and assume that $\mathfrak p$ is a maximal ideal. Since $R^ G \subset R$ is an integral ring extension, we find that the maximal ideals of $R$ are exactly the primes lying over $\mathfrak p$ (follows from Algebra, Lemmas 10.36.20 and 10.36.22). By Lemma 15.110.8 there are finitely many of them $\mathfrak q = \mathfrak q_1, \mathfrak q_2, \ldots , \mathfrak q_ m$ and they form a single orbit for $G$. By the Chinese remainder theorem (Algebra, Lemma 10.15.4) the map $R \to \prod _{j = 1, \ldots , m} R/\mathfrak q_ j$ is surjective.

First we prove that the extension is normal. Pick an element $\alpha \in \kappa (\mathfrak q)$. We have to show that the minimal polynomial $P$ of $\alpha $ over $\kappa (\mathfrak p)$ splits completely. By the above we can choose $a \in \mathfrak q_2 \cap \ldots \cap \mathfrak q_ m$ mapping to $\alpha $ in $\kappa (\mathfrak q)$. Consider the polynomial $Q = \prod _{\sigma \in G} (T - \sigma (a))$ in $R^ G[T]$. The image of $Q$ in $R[T]$ splits completely into linear factors, hence the same is true for its image in $\kappa (\mathfrak q)[T]$. Since $P$ divides the image of $Q$ in $\kappa (\mathfrak p)[T]$ we conclude that $P$ splits completely into linear factors over $\kappa (\mathfrak q)$ as desired.

Since $\kappa (\mathfrak q)/\kappa (\mathfrak p)$ is normal we may assume $\kappa (\mathfrak q) = \kappa _1 \otimes _{\kappa (\mathfrak p)} \kappa _2$ with $\kappa _1/\kappa (\mathfrak p)$ purely inseparable and $\kappa _2/\kappa (\mathfrak p)$ Galois, see Fields, Lemma 9.27.3. Pick $\alpha \in \kappa _2$ which generates $\kappa _2$ over $\kappa (\mathfrak p)$ if it is finite and a subfield of degree $> |G|$ if it is infinite (to get a contradiction). This is possible by Fields, Lemma 9.19.1. Pick $a$, $P$, and $Q$ as in the previous paragraph. If $\alpha ' \in \kappa _2$ is a Galois conjugate of $\alpha $ over $\kappa (\mathfrak p)$, then the fact that $P$ divides the image of $P$ in $\kappa (\mathfrak p)[T]$ shows there exists a $\sigma \in G$ such that $\sigma (a)$ maps to $\alpha '$. By our choice of $a$ (vanishing at other maximal ideals) this implies $\sigma \in D$ and that the image of $\sigma $ in $\text{Aut}(\kappa (\mathfrak q)/\kappa (\mathfrak p))$ maps $\alpha $ to $\alpha '$. Hence the surjectivity or the desired absurdity in case $\alpha $ has degree $> |G|$ over $\kappa (\mathfrak p)$. $\square$

Proof. Proof of (1). Consider pairs $(M, \sigma )$ where $K \subset M \subset L$ is a subfield such that $M/K$ is Galois, $\sigma \in \text{Gal}(M/K)$ with $\sigma (\mathfrak q \cap M) = \mathfrak q' \cap M$. We say $(M', \sigma ') \geq (M, \sigma )$ if and only if $M \subset M'$ and $\sigma '|_ M = \sigma $. Observe that $(K, \text{id}_ K)$ is such a pair as $A = K \cap B$ since $A$ is a normal domain. The collection of these pairs satisfies the hypotheses of Zorn's lemma, hence there exists a maximal pair $(M, \sigma )$. If $M \not= L$, then we can find $M \subset M' \subset L$ with $M'/M$ nontrivial and finite and $M'/K$ Galois (Fields, Lemma 9.16.5). Choose $\sigma ' \in \text{Gal}(M'/K)$ whose restriction to $M$ is $\sigma $ (Fields, Lemma 9.22.2). Then the primes $\sigma '(\mathfrak q \cap M')$ and $\mathfrak q' \cap M'$ restrict to the same prime of $B \cap M$. Since $B \cap M = (B \cap M')^{\text{Gal}(M'/M)}$ we can use Lemma 15.110.8 to find $\tau \in \text{Gal}(M'/M)$ with $\tau (\sigma '(\mathfrak q \cap M')) = \mathfrak q' \cap M'$. Hence $(M', \tau \circ \sigma ') > (M, \sigma )$ contradicting the maximality of $(M, \sigma )$.

Part (2) is proved in exactly the same manner as part (1). We write out the details. Pick $\overline{\sigma } \in \text{Aut}(\kappa (\mathfrak q)/\kappa (\mathfrak p))$. Consider pairs $(M, \sigma )$ where $K \subset M \subset L$ is a subfield such that $M/K$ is Galois, $\sigma \in \text{Gal}(M/K)$ with $\sigma (\mathfrak q \cap M) = \mathfrak q \cap M$ and

commutes. We say $(M', \sigma ') \geq (M, \sigma )$ if and only if $M \subset M'$ and $\sigma '|_ M = \sigma $. As above $(K, \text{id}_ K)$ is such a pair. The collection of these pairs satisfies the hypotheses of Zorn's lemma, hence there exists a maximal pair $(M, \sigma )$. If $M \not= L$, then we can find $M \subset M' \subset L$ with $M'/M$ finite and $M'/K$ Galois (Fields, Lemma 9.16.5). Choose $\sigma ' \in \text{Gal}(M'/K)$ whose restriction to $M$ is $\sigma $ (Fields, Lemma 9.22.2). Then the primes $\sigma '(\mathfrak q \cap M')$ and $\mathfrak q \cap M'$ restrict to the same prime of $B \cap M$. Adjusting the choice of $\sigma '$ as in the first paragraph, we may assume that $\sigma '(\mathfrak q \cap M') = \mathfrak q \cap M'$. Then $\sigma '$ and $\overline{\sigma }$ define maps $\kappa (\mathfrak q \cap M') \to \kappa (\mathfrak q)$ which agree on $\kappa (\mathfrak q \cap M)$. Since $B \cap M = (B \cap M')^{\text{Gal}(M'/M)}$ we can use Lemma 15.110.9 to find $\tau \in \text{Gal}(M'/M)$ with $\tau (\mathfrak q \cap M') = \mathfrak q \cap M'$ such that $\tau \circ \sigma $ and $\overline{\sigma }$ induce the same map on $\kappa (\mathfrak q \cap M')$. There is a small detail here in that the lemma first guarantees that $\kappa (\mathfrak q \cap M')/\kappa (\mathfrak q \cap M)$ is normal, which then tells us that the difference between the maps is an automorphism of this extension (Fields, Lemma 9.15.10), to which we can apply the lemma to get $\tau $. Hence $(M', \tau \circ \sigma ') > (M, \sigma )$ contradicting the maximality of $(M, \sigma )$. $\square$

Proof. Let $\sigma \in D_\mathfrak q$. Pick $\tau \in H$ mapping to $\sigma $. This is possible by Fields, Lemma 9.22.2. Then $\tau (\mathfrak r)$ and $\mathfrak r$ both lie over $\mathfrak q$. Hence by Lemma 15.110.10 there exists a $\sigma ' \in \text{Gal}(M/L)$ with $\sigma '(\tau (\mathfrak r)) = \mathfrak r$. Hence $\sigma '\tau \in D_\mathfrak r$ maps to $\sigma $. The case of inertia groups is proved in exactly the same way using surjectivity onto automorphism groups. $\square$