Lemma 15.110.1. Let $\varphi : A \to B$ be a surjection of rings. Let $G$ be a finite group of order $n$ acting on $\varphi : A \to B$. If $b \in B^ G$, then there exists a monic polynomial $P \in A^ G[T]$ which maps to $(T - b)^ n$ in $B^ G[T]$.

## 15.110 Group actions and integral closure

This section is in some sense a continuation of Algebra, Section 10.38. More material of a similar kind can be found in Fundamental Groups, Section 58.12

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

Lemma 15.110.2. Let $R$ be a ring. Let $G$ be a finite group acting on $R$. Let $I \subset R$ be an ideal such that $\sigma (I) \subset I$ for all $\sigma \in G$. Then $R^ G/I^ G \subset (R/I)^ G$ is an integral extension of rings which induces homeomorphisms on spectra and purely inseparable extensions of residue fields.

**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.
$\square$

Lemma 15.110.3. Let $R$ be a ring. Let $G$ be a finite group of order $n$ acting on $R$. Let $A$ be an $R^ G$-algebra.

for $b \in (A \otimes _{R^ G} R)^ G$ there exists a monic polynomial $P \in A[T]$ whose image in $(A \otimes _{R^ G} R)^ G[T]$ is $(T - b)^ n$,

for $a \in A$ mapping to zero in $(A \otimes _{R^ G} R)^ G$ we have $(T - a)^{n^2} = T^{n^2}$ in $A[T]$.

**Proof.**
Write $A$ as the quotient of a polynomial algebra $P$ over $R^ G$. Then $(P \otimes _{R^ G} R)^ G = P$ because $P$ is free as an $R^ G$-module. Hence part (1) follows from Lemma 15.110.1.

Let $J = \mathop{\mathrm{Ker}}(P \to A)$. Lift $a$ as in (2) to an element $f \in P$. Then $f \otimes 1$ maps to zero in $A \otimes _{R^ G} R$. Hence $f \otimes 1$ is in $(J')^ G$ where $J' \subset P \otimes _{R^ G} R$ is the image of the map $J \otimes _{R^ G} R \to P \otimes _{R^ G} R$. Apply Lemma 15.110.1 to $f \otimes 1$ and the surjective ring map

which defines $A'$. We obtain $P \in (\text{Sym}^*_{R^ G}(J) \otimes _{R^ G} R)^ G[T]$ mapping to $(T - f \otimes 1)^ n$ in $A'[T]$. Apply part (1) to see that there exists a $P' \in \text{Sym}^*_{R^ G}(J)[T, T']$ whose image is $(T' - P)^ n$. Since $\text{Sym}_{R^ G}^*(P)$ is still free over $R^ G$ we conclude that $P'$ maps to $(T' - (T - f)^ n)^ n$ in $\text{Sym}_{R^ G}^*(P)$. On the other hand, tracing through the construction of the polynomials $P$ and $P'$ in Lemma 15.110.1 we see that $P'$ is congruent to $(T' - T^ n)^ n$ modulo the irrelevant ideal of the graded ring $\text{Sym}^*_{R^ G}(J)$. It follows that

in $A[T', T]$. Setting $T' = 0$ for example we obtain the statement of the lemma. $\square$

Lemma 15.110.4. Let $R$ be a ring. Let $G$ be a finite group acting on $R$. Let $R^ G \to A$ be a ring map. The map

is an isomorphism if $R^ G \to A$ is flat. In general the map is integral, induces a homeomorphism on spectra, and induces purely inseparable residue field extensions.

**Proof.**
The first statement follows from Lemma 15.110.3 and Algebra, Lemma 10.46.11. To see the second consider the exact sequence $0 \to R^ G \to R \to \bigoplus _{\sigma \in G} R$ where the second map sends $x$ to $(\sigma (x) - x)$. Tensoring with $A$ the sequence remains exact if $R^ G \to A$ is flat.
$\square$

Lemma 15.110.5. Let $G$ be a finite group acting on a ring $R$. For any two primes $\mathfrak q, \mathfrak q' \subset R$ lying over the same prime in $R^ G$ there exists a $\sigma \in G$ with $\sigma (\mathfrak q) = \mathfrak q'$.

**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 $y \in \mathfrak p \subset \mathfrak q$.
$\square$

Lemma 15.110.6. Let $G$ be a finite group acting on a ring $R$. Let $\mathfrak q \subset R$ be a prime lying over $\mathfrak p \subset R^ G$. Then $\kappa (\mathfrak q)/\kappa (\mathfrak p)$ is an algebraic normal extension and the map

is surjective^{1}.

**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 15.110.4. 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 15.110.5 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 $R \to \prod _{j = 1, \ldots , m} R/\sigma (\mathfrak q_ j)$ 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. $\alpha \in \kappa _2$ 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 $, then the above 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 $\kappa (\mathfrak p)$. $\square$

Lemma 15.110.7. Let $A$ be a normal domain with fraction field $K$. Let $L/K$ be a (possibly infinite) Galois extension. Let $G = \text{Gal}(L/K)$ and let $B$ be the integral closure of $A$ in $L$.

For any two primes $\mathfrak q, \mathfrak q' \subset B$ lying over the same prime in $A$ there exists a $\sigma \in G$ with $\sigma (\mathfrak q) = \mathfrak q'$.

Let $\mathfrak q \subset B$ be a prime lying over $\mathfrak p \subset A$. Then $\kappa (\mathfrak q)/\kappa (\mathfrak p)$ is an algebraic normal extension and the map

\[ D = \{ \sigma \in G \mid \sigma (\mathfrak q) = \mathfrak q\} \longrightarrow \text{Aut}(\kappa (\mathfrak q)/\kappa (\mathfrak p)) \]is surjective.

**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 15.110.5 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 15.110.6 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 $(M, \sigma )$. $\square$

Lemma 15.110.8. Let $A$ be a normal domain with fraction field $K$. Let $M/L/K$ be a tower of (possibly infinite) Galois extensions of $K$. Let $H = \text{Gal}(M/K)$ and $G = \text{Gal}(L/K)$ and let $C$ and $B$ be the integral closure of $A$ in $M$ and $L$. Let $\mathfrak r \subset C$ and $\mathfrak q = B \cap \mathfrak r$. Set $D_\mathfrak r = \{ \tau \in H \mid \tau (\mathfrak r) = \mathfrak r\} $ and $I_\mathfrak r = \{ \tau \in D_\mathfrak r \mid \tau \bmod \mathfrak r = \text{id}_{\kappa (\mathfrak r)}\} $ and similarly for $D_\mathfrak q$ and $I_\mathfrak q$. Under the map $H \to G$ the induced maps $D_\mathfrak r \to D_\mathfrak q$ and $I_\mathfrak r \to I_\mathfrak q$ are surjective.

**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 15.110.7 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.
$\square$

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