Let $f:X\rightarrow Y$ be a proper morphism of Noetherian schemes, $\mathcal{F}$ a coherent $O_X$-module and $y\in Y.$ Denote by $k(y)$ the residue field of $Y$ at $y$, $X_y=X\times_Y\operatorname{Spec} k(y)$ and $\mathcal{F}_y=p^*\mathcal{F}$ where $p$ is the projection $X\times_Y\operatorname{Spec}k(y)\rightarrow X$.
How is the "canonical" homomorphism $(R^if_*\mathcal{F})_y\otimes_{O_{Y,y}}k(y)\rightarrow H^i(X_y,\mathcal{F}_y)$ defined ?
How is the “canonical” homomorphism $(R^pf_*\mathcal{F})_y\otimes_{O_{Y,y}}k(y)\rightarrow H^p(X_y,\mathcal{F}_y)$ defined
algebraic-geometryschemessheaf-cohomology
Related Solutions
$\newcommand{\spec}[1]{\mathrm{Spec}\,(#1)}$ I'm new at this, so it is more thoughts put into words than an answer.
By definition of a fiber product (in any locally small category for that matter), we have : $$ X_y(-) = X(-) \times_{Y(-)} \spec{k(y)}(-)$$ (where $Z(-)$ is the functor of points $\hom_{k-\mathbf{Sch}}(-,Z)$ of the $k$-scheme $Z$).
Explicitely, for a $k$-scheme $Z$, $X_y(Z) = \{ (\varphi,\psi) \in X(Z) \times \spec{k(y)}(Z) : f \circ \varphi = i_y \circ \psi \}$ where $i_y$ is the (inclusion) morphism $\spec{k(y)} \to Y$. But then, the functor of points of $\spec{k(y)}$ admits a nice description : any morphism $Z \to \spec {k(y)}$ is topologically the constant map to the single point of $\spec {k(y)}$, and the map of structuring sheaves is just a morphism $k(y) \to \mathcal O_Z(Z)$ of $k$-algebra.
Now take $Z$ to be $\spec A$ for a $k$-algebra $A$, we have : $$ \spec{k(y)} (A) \cong \hom_{k}(k(y), A). $$ So an element of $X_y(A)$ is the data of a morphism $\spec A \to X$ and a structure of $k(y)$-algebra on $A$ extending the structure of $k$-algebra, such that
- topologically, $\spec A$ factors through $f^{-1}(\{y\})$,
- for every $x \in f^{-1}(\{y\})$, the structure of $k(y)$-algebra of $A$ factors through the residue field $k(x)$ of $x$.
If now $A$ is a $k(y)$-algebra and you're interesting in the $A$-points of $X_y$ as $k(y)$-schemes$^{(1)}$, then the structure of $k(y)$-algebra is imposed but the conditions remain. So I would say that your assumption
Using the universal property of the fiber product, is this equivalent to the $A$-points of $X$ where $A$ is a commutative $k$-algebra via $k \hookrightarrow k(y)$?
was not restrictive enough.
(1) This is not clear in the OP what kind of $A$-points you are looking for.
There are multiple ways of attacking this:
Method 1:
Show that if $V\subseteq U$ is another affine containing $x$, then the maps $\text{Spec}(\mathcal{O}_{X,x})\to V$ and $\text{Spec}(\mathcal{O}_{X,x})\to U$ are the same (this isn't that hard) (EDIT: As Asal Beag Dubh points out below, this means that $\text{Spec}(\mathcal{O}_{X,x})\to U$ factors as $\text{Spec}(\mathcal{O}_{X,x})\to V\to U$). Then, for any two affines $W,U$ pass to some affine open $V\subseteq W\cap U$.
Method 2:
Let $Z\subseteq X$ be the subset of $X$ consisting of points which generalize $x$. Consider the topological map $i:Z\hookrightarrow X$. Define $\mathcal{O}_Z:=i^{-1}\mathcal{O}_X$. Shown then that $(Z,\mathcal{O}_Z)$ is a scheme, and that for any choice of affine $x\in U$ one has that $Z\to X$ is isomorphic to $\mathcal{O}_{X,x}$ (i.e. that $Z\cong \mathcal{O}_{X,x}$ in a way compatible with these mappings).
Best Answer
This is an application of a more general fact. Let $X,Y,Y'$ be noetherian schemes, let $f:X\to Y$ be of finite type and separated, let $u:Y'\to Y$ be arbitrary, and let $X'=X\times_YY'$ be the base change, organized in the following commutative diagram:
$$\require{AMScd} \begin{CD} X' @>{v}>> X\\ @VV{g}V @VV{f}V \\ Y' @>{u}>> Y \end{CD}$$
Let $\def\cF{\mathcal{F}}\cF$ be a quasi-coherent sheaf on $X$. I claim that there is a natural map $$u^*R^if_*(\cF)\to R^ig_*(v^*\cF).$$
Proof: The question is local on $Y$ and $Y'$, so we may assume that $Y=\operatorname{Spec} A$ and $Y'=\operatorname{Spec} A'$. So it suffices to demonstrate a map $H^i(X,\cF)\otimes_A A'\to H^i(X',\cF')$ where $\cF'=v^*\cF$. But as $X$ is separated and $\cF$ is quasi-coherent, we may use the Cech cohomology to compute $H^i(X,\cF)$. So let $\mathfrak{U}$ be a cover of $X$ by open affines $U_i$ and let $C(\mathfrak{U},\cF)$ be the Cech complex. Since affine morphisms are stable under base change, $v$ is also affine, and the Cech complex associated to the open cover $v^{-1}(U_i)$ and the sheaf $\cF'$ is exactly $C(\mathfrak{U},\cF)\otimes_A A'$, which computes $H^i(X',\cF')$ for the same reasons as before. This construction gives us our natural map: it's the induced map on cohomology from the chain map $C(\mathfrak{U},\cF)\to C(\mathfrak{U},\cF)\otimes_A A'$ sending $c\mapsto c\otimes 1$. $\blacksquare$
To apply this to your specific case, let $Y'=\operatorname{Spec} k(y)$ and $X'=X_y$ so the diagram is a base-change diagram along the natural inclusion $y\to Y$. Since your canonical homomorphism can be computed Zariski-locally in a neighborhood of $y$, we may assume $Y$ is affine so that $R^if_*(\cF)$ is the sheaf associated to $H^i(X,\cF)$. Then pullback along $\operatorname{Spec} k(y)\to Y$ is exactly applying $-\otimes_{\mathcal{O}_{Y,y}} k(y)$, so we've identified the LHS of your formula with the more general fact. The RHS is essentially the same: since $\operatorname{Spec} k(y)$ is affine, $R^ig_*(-)=H^i(X_y,-)$.