Timeline for Are quotients of affine schemes by finite groups faithfully flat?

6 events

when toggle format	what		by	license	comment
Sep 17, 2021 at 22:25	comment	added	user267839		alright I see, thank you!
Sep 17, 2021 at 21:18	comment	added	Piotr Achinger		P.S. The fiber at zero is the fiber over the prime $(x^2,xy,y^2)R^G = (x,y)R\cap R^G$, not over the zero ideal.
Sep 17, 2021 at 21:16	comment	added	Piotr Achinger		Say $A\to B$ is a finite and flat homomorphism of Noetherian rings. Therefore $B$ is a finitely generated flat $A$-module and hence projective. If $A$ is a domain, $M$ a finitely generated projective $A$-module, and $P$ is a prime of $A$, let $K={\rm Frac}(A)$ and $F={\rm Frac}(A/P)$. Then $\dim_K (M\otimes_A K) = \dim_F (M\otimes_A F)$, both equal to the rank of $M$ when treated as a vector bundle on ${\rm Spec}(A)$.
Sep 17, 2021 at 21:03	comment	added	user267839		could you clarify why the observations that the map $R^G \to R$ has degree $2$ (ie $[ \operatorname{Frac}(R): \operatorname{Frac}(R^G) ]=2$) and that the fiber at $0$ has lenght $3$ (ie the ${Frac}(R)$-module $R \otimes_{R^G} \operatorname {Frac}(R)$ has length $3$ or equivalently has dimension $3$ as ${Frac}(R)$-vector space) implies that $R^G \to R$ is not flat?
Jun 5, 2014 at 2:38	vote	accept	jacob
Jun 5, 2014 at 2:09	history	answered	Piotr Achinger	CC BY-SA 3.0