The Stacks project

Proof. Assume the lemma is not true to get a contradiction. Then there exists a monic irreducible polynomial $P(T) \in \kappa (\mathfrak p)[T]$ of degree $d > 1$. After replacing $P$ by $a^ d P(a^{-1}T)$ for suitable $a \in A$ (to clear denominators) we may assume that $P$ is the image of a monic polynomial $Q$ in $A[T]$. Observe that $Q$ is irreducible in $K[T]$. Namely a factorization over $K$ leads to a factorization over $A$ by Algebra, Lemma 10.38.5 which we could reduce modulo $\mathfrak p$ to get a factorization of $P$. As $K$ is separably closed, $Q$ is not a separable polynomial (Fields, Definition 9.12.2). Then the characteristic of $K$ is $p > 0$ and $Q$ has vanishing linear term (Fields, Definition 9.12.2). However, then we can replace $Q$ by $Q + a T$ where $a \in \mathfrak p$ is nonzero to get a contradiction. $\square$

Proof. Let $(A, \mathfrak m, \kappa )$ be normal local with separably closed fraction field $K$. If $A = K$, then we are done. If not, then the residue field $\kappa $ is algebraically closed by Lemma 58.12.1 and it suffices to check that $A$ is henselian. Let $f \in A[T]$ be monic and let $a_0 \in \kappa $ be a root of multiplicity $1$ of the reduction $\overline{f} \in \kappa [T]$. Let $f = \prod f_ i$ be the factorization in $K[T]$. By Algebra, Lemma 10.38.5 we have $f_ i \in A[T]$. Thus $a_0$ is a root of $f_ i$ for some $i$. After replacing $f$ by $f_ i$ we may assume $f$ is irreducible. Then, since the derivative $f'$ cannot be zero in $A[T]$ as $a_0$ is a single root, we conclude that $f$ is linear due to the fact that $K$ is separably algebraically closed. Thus $A$ is henselian, see Algebra, Definition 10.153.1. $\square$

Proof. Since $\mathfrak q$ is the inverse image of $\mathfrak q'$ and since $\kappa (\mathfrak q) \subset \kappa (\mathfrak q')$, we get $I_{\mathfrak q'} \subset I_\mathfrak q$. Conversely, if $\sigma \in I_\mathfrak q$, the $\sigma $ acts trivially on the fibre ring $A \otimes _{R^ G} \kappa (\mathfrak q)$. Thus $\sigma $ fixes all the primes lying over $\mathfrak q$ and induces the identity on their residue fields. $\square$

Proof. The strategy of the proof is to use étale localization to reduce to the case where $R \to R^ I$ is a local isomorphism at $R^ I \cap \mathfrak p$. Let $R^ G \to A$ be an étale ring map. We claim that if the result holds for the action of $G$ on $A \otimes _{R^ G} R$ and some prime $\mathfrak q'$ of $A \otimes _{R^ G} R$ lying over $\mathfrak q$, then the result is true.

To check this, note that since $R^ G \to A$ is flat we have $A = (A \otimes _{R^ G} R)^ G$, see More on Algebra, Lemma 15.110.7. By Lemma 58.12.3 the group $I$ does not change. Then a second application of More on Algebra, Lemma 15.110.7 shows that $A \otimes _{R^ G} R^ I = (A \otimes _{R^ G} R)^ I$ (because $R^ I \to A \otimes _{R^ G} R^ I$ is flat). Thus

is cartesian and the horizontal arrows are étale. Thus if the left vertical arrow is étale in some open neighbourhood $W$ of $(A \otimes _{R^ G} R)^ I \cap \mathfrak q'$, then the right vertical arrow is étale at the points of the (open) image of $W$ in $\mathop{\mathrm{Spec}}(R^ I)$, see Descent, Lemma 35.14.5. In particular the morphism $\mathop{\mathrm{Spec}}(R^ I) \to \mathop{\mathrm{Spec}}(R^ G)$ is étale at $R^ I \cap \mathfrak q$.

Let $\mathfrak p = R^ G \cap \mathfrak q$. By More on Algebra, Lemma 15.110.8 the fibre of $\mathop{\mathrm{Spec}}(R) \to \mathop{\mathrm{Spec}}(R^ G)$ over $\mathfrak p$ is finite. Moreover the residue field extensions at these points are algebraic, normal, with finite automorphism groups by More on Algebra, Lemma 15.110.9. Thus we may apply More on Morphisms, Lemma 37.42.1 to the integral ring map $R^ G \to R$ and the prime $\mathfrak p$. Combined with the claim above we reduce to the case where $R = A_1 \times \ldots \times A_ n$ with each $A_ i$ having a single prime $\mathfrak q_ i$ lying over $\mathfrak p$ such that the residue field extensions $\kappa (\mathfrak q_ i)/\kappa (\mathfrak p)$ are purely inseparable. Of course $\mathfrak q$ is one of these primes, say $\mathfrak q = \mathfrak q_1$.

It may not be the case that $G$ permutes the factors $A_ i$ (this would be true if the spectrum of $A_ i$ were connected, for example if $R^ G$ was local). This we can fix as follows; we suggest the reader think this through for themselves, perhaps using idempotents instead of topology. Recall that the product decomposition gives a corresponding disjoint union decomposition of $\mathop{\mathrm{Spec}}(R)$ by open and closed subsets $U_ i$. Since $G$ is finite, we can refine this covering by a finite disjoint union decomposition $\mathop{\mathrm{Spec}}(R) = \coprod _{j \in J} W_ j$ by open and closed subsets $W_ j$, such that for all $j \in J$ there exists a $j' \in J$ with $\sigma (W_ j) = W_{j'}$. The union of the $W_ j$ not meeting $\{ \mathfrak q_1, \ldots , \mathfrak q_ n\} $ is a closed subset not meeting the fibre over $\mathfrak p$ hence maps to a closed subset of $\mathop{\mathrm{Spec}}(R^ G)$ not meeting $\mathfrak p$ as $\mathop{\mathrm{Spec}}(R) \to \mathop{\mathrm{Spec}}(R^ G)$ is closed. Hence after replacing $R^ G$ by a principal localization (permissible by the claim) we may assume each $W_ j$ meets one of the points $\mathfrak q_ i$. Then we set $U_ i = W_ j$ if $\mathfrak q_ i \in W_ j$. The corresponding product decomposition $R = A_1 \times \ldots \times A_ n$ is one where $G$ permutes the factors $A_ i$.

Thus we may assume we have a product decomposition $R = A_1 \times \ldots \times A_ n$ compatible with $G$-action, where each $A_ i$ has a single prime $\mathfrak q_ i$ lying over $\mathfrak p$ and the field extensions $\kappa (\mathfrak q_ i)/\kappa (\mathfrak p)$ are purely inseparable. Write $A' = A_2 \times \ldots \times A_ n$ so that

Since $\mathfrak q = \mathfrak q_1$ we find that every $\sigma \in I$ preserves the product decomposition above. Hence

Observe that $I = D = \{ \sigma \in G \mid \sigma (\mathfrak q) = \mathfrak q\} $ because $\kappa (\mathfrak q)/\kappa (\mathfrak p)$ is purely inseparable. Since the action of $G$ on primes over $\mathfrak p$ is transitive (More on Algebra, Lemma 15.110.8) we conclude that, the index of $I$ in $G$ is $n$ and we can write $G = eI \amalg \sigma _2I \amalg \ldots \amalg \sigma _ nI$ so that $A_ i = \sigma _ i(A_1)$ for $i = 2, \ldots , n$. It follows that

Thus the map $R^ G \to R^ I$ is étale at $R^ I \cap \mathfrak q$ and the proof is complete. $\square$

Proof. We can write $L$ as the filtered colimit of finite Galois extensions of $K$. Hence it suffices to prove this lemma in case $L/K$ is a finite Galois extension, see Algebra, Lemma 10.154.3. Since $A = B^ G$ as $A$ is integrally closed in $K = L^ G$ the result follows from Lemma 58.12.4. $\square$