The question I have is from the proof of the lemma above, but it is actually a more general statement about quasi-coherent sheaves on an affine scheme. Suppose $X= \text{Spec }A$ for some ring $A$, and $\mathscr{F}$ is a quasi-coherent sheaf on $X$. Then for some open affine cover of $X$, the restriction sheaf is isomorphic to a sheaf of a module over the corresponding ring. In particular, if $\text{Spec }B$ is in the cover, then $\mathscr{F}|_{\text{Spec } B} \cong \widetilde{M}$ for a $B$-module $M$. This part is by definition.

Now $\text{Spec }B$ is covered by distinguished open sets of the form $D(g)$ for $g\in A$, and for any such open set the inclusion $D(g)\subseteq \text{Spec }B$ is induced by the ring map $B\to A_g$. Thus $\mathscr{F}|_{D(g)} \cong (M\otimes_B A_g)^{\tilde{}}$.

He deduces the last sentence from a previous proposition that deals with properties of the sheaves of modules. The two properties that seem important for this deduction are the following: For a ring map $A \to B$ inducing the map of spectra $f:\text{Spec }B \to \text{Spec }A$,

(1) If $M$ and $N$ are $A$-modules, then $(M\otimes N)^{\tilde{}} \cong \widetilde{M} \otimes_{\mathcal{O}_{\text{Spec }A}} \widetilde{N}$.

(2) For any $A$-module $M$, $f^*(\widetilde{M})\cong (M\otimes_{A} B)^{\tilde{}}$.

I cannot seem to make the connection. So any help with his last statement would be great. Thanks.


I am clarifying further the crucial point raised in the comments to @hwong557's answer.

Let $(f,f^\#): (X,\mathcal{O}_X) \rightarrow (Y,\mathcal{O}_Y)$ be a morphism of locally ringed spaces. Suppose that $f(X)$ is open in $Y$ and that $f$ induces an isomorphism $\big(X,\mathcal{O}_X\big) \rightarrow \big(f(X),\mathcal{O}_Y|_{f(X)}\big)$ of locally ringed spaces. Hence, we have an isomorphism of sheaves $\mathcal{O}_Y|_{f(X)} \cong f_*\mathcal{O}_X$. Moreover, for every open set $U$ of $X$ we have an isomorphism of rings $\mathcal{O}_Y(f(U)) = \mathcal{O}_Y|_{f(X)}(f(U)) \cong f_*\mathcal{O}_X(f(U)) = \mathcal{O}_X(U)$.

Now let $\mathscr{G}$ be an $\mathcal{O}_Y$-module. Let us look at the sections of the presheaf that gives rise to $f^* \mathscr{G}$. Let $U$ be open in $X$. Then the sections of that presheaf over $U$ are $f^{-1}\mathscr{G}(U) \otimes_{f^{-1} \mathcal{O}_Y(U)} \mathcal{O}_X(U)$. But $f^{-1} \mathcal{O}_Y(U) = \lim_{V' \supset f(U)} \mathcal{O}_Y(V') =\mathcal{O}_Y(f(U))$, and we already know that this latter ring is isomorphic to $\mathcal{O}_X(U)$. Hence, $f^{-1}\mathscr{G}(U) \otimes_{f^{-1} \mathcal{O}_Y(U)} \mathcal{O}_X(U) \cong f^{-1}\mathscr{G}(U)$. This now shows that $f^* \mathscr{G}$ is the same as $f^{-1}\mathscr{G}$, and the latter is by definition $\mathscr{G}|_X$.


Let $\phi: Spec (A_g) \to Spec (B)$ be the inclusion map. Then:

$$\mathcal F\mid_{D(g)} = (\mathcal F\mid_{V})\mid_{D(g)} = (\tilde M) \mid_{D(g)} = \phi^*(\tilde M) = (M \otimes_B A_g)^\tilde{} $$

  • $\begingroup$ Why is it clear that $(\tilde{M})|_{D(g)} = \phi^{*}(\tilde{M})$? $\endgroup$
    – user194928
    Aug 9 '16 at 0:17
  • $\begingroup$ It's not clear, but if you compute all the components of the definition of $\phi^*$ as on p. 109–110, it'll work out. $\endgroup$
    – Mr. Chip
    Aug 9 '16 at 0:19
  • $\begingroup$ Right, I agree with Mr. Chip. It's not immediately clear. Another way (may or may not more or less painful) is to use the uniqueness of adjoint. That is, show $\tilde M|_{D(g)}$ satisfies the isomorphism of homs on page 110. I think Hartshorne gives as exercise the uniqueness of adjoint somewhere in the sheaf section. $\endgroup$
    – hwong557
    Aug 9 '16 at 0:25
  • $\begingroup$ It seems rather daunting but if you really just need to look at the situation of the inclusion $\phi: U \hookrightarrow X$ and $\mathcal F$ is a $\mathcal O_X$-mod, then $\phi^* \mathcal F = \mathcal F|_{U}$. It's easier here to grind through the definitions because $\phi$ is an open map. $\endgroup$
    – hwong557
    Aug 9 '16 at 0:46

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