# Hartshorne Corollary III.9.4

I am reading the section of Flatness from Hartshorne. I have a doubt in the proof of the corollary of the following proposition :

$\textbf{Proposition 9.3}$ Let $f:X\longrightarrow Y$ be a separated morphism of finite type of noetherian schemes and let $\mathcal{F}$ be a quasi-coherent sheaf on $X$. Let $u:Y'\longrightarrow Y$ be a flat morphism of noetherian schemes.

Then for all $i\geq 0$, there are natural isomorphisms

$u^*R^if_*(\mathcal{F})\cong R^ig_*(v^*\mathcal{F})$. I understood the proof of this.

I have a doubt in the corollary to this.

$\textbf{Corollary 9.4}$ Let $f:X\longrightarrow Y$ and $\mathcal{F}$ be as in (9.3) and assume that $Y$ is affine. For any point $y\in Y$, the $X_y$ be the fibre over $y$, and let $\mathcal{F}_y$ be the induced sheaf. On the other hand, let $k(y)$ denote the constant sheaf $k(y)$ on the closed subset $\bar{\{y\}}$ of $Y$. Then for all $i\geq 0$ there are natural isomorphisms. $H^i(X_y,\mathcal{F}_y)\cong H^i(X,\mathcal{F}\otimes k(y))$

They begin the proof by saying : Let $Y'\subset Y$ be the reduced induced subscheme structure on $\bar{\{y\}}$., and let $X'=X\times_Y Y'$, which is a closed subscheme of $X$. Then both sides of the desired isomorphism depend only on the sheaf $\mathcal{F}'=\mathcal{F}\otimes k(y)$ on $X'$. Thus we can replace $X,Y,\mathcal{F}$ by $X',Y',\mathcal{F}'$.

Why can we assume this? Why can we replace as said above? Clearly, on the right side $\mathcal{F}\otimes k(y)$ is $\mathcal{F}'$. But why is it true on the left?

Any help will be appreciated!

@poorna In order to apply Prop. 9.3 in the proof of Cor. 9.4 one needs a flat base change

$$u:A \longrightarrow k(y).$$

When $y \in Y$ is not closed then $u$ is not flat: For the proof consider $u$ as the composition $A \longrightarrow A_{p_y}$ and $A_{p_y} \longrightarrow k(y), y = p_y$. The former morphism is flat being a localisation. Hence we show that the latter is not flat. Flatness of the latter were equivalent to the field $k(y)$ being a free $A_{p_y}$-module of rank 1 (Matsumura, Hideyuki: Commutative ring theory. Theor. 7.10). Apparently, that does not hold because the ideal $p_y \subset A$ is not maximal.

Therefore Hartshorne introduces the intermediate ring $A' := A/p_y$. It is an integral domain. If $Y' := Spec \ A'$ then $y \in Y'$ is the generic point with residue field $k(y) = A'_{(0)}$ a localisation. Hence

$$u':A' \longrightarrow \ k(y)$$

is flat. By definition

$$X_y := X \times_Y Spec \ k(y) = (X \times_{Y} Y') \times_{Y'} Spec \ k(y) = X' \times_{Y'} Spec \ k(y)$$

$$\mathscr F_y := \mathscr F \otimes_{A} k(y) = (\mathscr F \otimes_A A') \otimes_{A'} k(y) = \mathscr F' \otimes_{A'} k(y).$$

Therefore I differ from Hartshorne's notation and propose to set $\mathscr F' := \mathscr F \otimes_A A'$. Was this your point, too?

• I have a more trivial doubt. They talk of $\mathcal{F}\otimes k(y)$ as a sheaf on $X$. Does that mean $\mathcal{F}\otimes f^*k(y)$ Nov 11, 2014 at 5:21
• The second doubt is this. With your reduction, we can apply Prop 9.3 to conclude that $H^i(X_y,\mathcal{F}_y)\cong H^i(X',\mathcal{F}')\otimes k(y)$. But why will this mean that $H^i(X_y,\mathcal{F}_y)\cong H^i(X,\mathcal{F})\otimes k(y)$ Nov 11, 2014 at 8:35
• @poorna 1. You are right: $\mathscr F \otimes k(y)$ means $\mathscr F \otimes_{\mathscr O_X} f^*k(y)$. 2. $H^i(X_y, \mathscr F_y) = H^i(X',\mathscr F') \otimes_{A'} k(y)) = H^i(X', \mathscr F' \otimes_{A'} k(y))$ because $k(y)$ is a flat A'-module (Projection formula). Now $H^i(X', \mathscr F' \otimes_{A'} k(y)) = H^i(X, \mathscr F' \otimes_{A'} k(y))$ because $supp (\mathscr F' \otimes_{A'} k(y)) \subset X' \subset X$. Furthermore, $H^i(X, \mathscr F' \otimes_{A'} k(y)) = H^i(X, \mathscr F \otimes_{A} k(y))$. I wonder why Hartshorne seems a bit short; does he not need these considerations? Nov 11, 2014 at 14:12

In the statement of the corollary by $$\mathcal{F} \otimes k(y)$$ Hartshorne means $$F \otimes \rho_* \pi^* k(y)$$ where $$\pi: X \times_Y Y' \rightarrow Y'$$ and $$\rho: X \times_Y Y' \rightarrow X$$ are the natural projections. This is a quasi-coherent sheaf on $$X$$. As such, and since $$f$$ is separated, we can compute $$H^i(X,F \otimes \rho_* \pi^* k(y))$$ by Cech cohomology. Hence the data that are relevant for this computation are the sections of $$F \otimes \rho_* \pi^* k(y)$$ over an affine open cover $$\operatorname{Spec}C_i$$ of $$X$$. These sections coincide with the sections of the quasi-coherent sheaf $$\rho^* \mathcal{F} \otimes \pi^* k(y)$$ of $$X'= X \times_Y Y'$$ on the open affine cover $$\operatorname{Spec}C_i \times_Y Y'$$ (this is what $$\mathcal{F}'=\mathcal{F} \otimes k(y)$$ means inside Hartshorne's proof). Hence the cohomology groups $$H^i(X,\rho_* \pi^* k(y))$$ will remain invariant upon the replacement of $$X,Y,\mathcal{F}$$ with $$X',Y',\mathcal{F}'$$. An even simpler argument shows that $$H^i(X_y,\mathcal{F}_y)$$ will also remain invariant. The final piece is that $$Y'$$ is required to have the reduced induced structure so that $$Y'=\operatorname{Spec}A$$ is an integral scheme, $$y$$ its generic point and $$\operatorname{Spec} k(y) \rightarrow Y'$$ is flat because $$k(y)$$ is the field of fractions of $$A$$, i.e. the localization of $$A$$ at the zero ideal (and localization is flat).