The Stacks project

Lemma 58.12.4. Let $G$ be a finite group acting on a ring $R$. Let $\mathfrak q \subset R$ be a prime. Set

\[ I = \{ \sigma \in G \mid \sigma (\mathfrak q) = \mathfrak q \text{ and } \sigma \bmod \mathfrak q = \text{id}_\mathfrak q\} \]

Then $R^ G \to R^ I$ is étale at $R^ I \cap \mathfrak q$.

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

\[ \xymatrix{ \mathop{\mathrm{Spec}}((A \otimes _{R^ G} R)^ I) \ar[d] \ar[r] & \mathop{\mathrm{Spec}}(R^ I) \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] & \mathop{\mathrm{Spec}}(R^ G) } \]

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

\[ R = A_1 \times A' \]

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

\[ R^ I = (A_1)^ I \times (A')^ I \]

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

\[ R^ G = (A_1)^ I. \]

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

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BST. Beware of the difference between the letter 'O' and the digit '0'.