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Hartshorne, Algebraic Geometry, Proposition II.5.2, reads (in part):

Let $A$ be a ring and let $X = \operatorname{Spec} A$. Also let $A \to B$ be a ring homomorphism and $f : \operatorname{Spec} B \to \operatorname{Spec} A$ be the corresponding map of spectra. Then:

[...]

(d) for any $B$-module $N$ we have $f_\ast(\widetilde{N}) \cong \widetilde{_{A}N}$, where $_A N$ means $N$ considered as an $A$-module;

(e) for any $A$-module $M$ we have $f^\ast(\widetilde{M}) \cong \widetilde{M \otimes_A B}$.

The corresponding proof in the text reads:

The last statements... follow directly from the definitions.

I don't get it. For instance,

$$(f_\ast\widetilde{N})(U) = \left\{s : f^{-1}(U) \to \bigsqcup_{f(\mathfrak{p}) \in U} N_\mathfrak{p} \; \middle\vert \; s(\mathfrak{p}) \in N_\mathfrak{p}, s = \frac{n}{b} \text{ locally}\right\}$$

while

$$\widetilde{_{A}N}(U) = \left\{t : U \to \bigsqcup_{\mathfrak{q} \in U} ({}_AN)_\mathfrak{q} \; \middle\vert \; s(\mathfrak{q}) \in ({}_AN)_\mathfrak{q}, s = \frac{n}{a} \text{ locally}\right\}$$

I can't see the connection. What am I missing?

Daniel McLaury
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    You have to use the adjunctions, not the constructions. Of course Hartshorne's book doesn't tell you that. Don't read it when you really want to understand the details of the foundations of AG. There are many better books. – Martin Brandenburg Oct 19 '13 at 10:58
  • Thanks. The proof supplied is incredibly misleading, then! – Daniel McLaury Oct 19 '13 at 10:59
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    Which proof? ;) – Martin Brandenburg Oct 19 '13 at 11:31
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    Dear Daniel, The statement for $f_$ is fairly simple: take an open subset in Spec $A$ of the form Spec $A_a$; then its preimage in Spec $B$ is Spec $B_a$. By construction the sections of $\widetilde N$ over Spec $B_a$ are $N_a$, and so by definition of pushforward, the sections of $f_\widetilde{N}$ over Spec $A_a$ are $N_a$. Thus this pushforward is the sheaf that associates to Spec $A_a$ the module $N_a$; and that is precisely the sheaf associated to $N$ when it is regarded as an $A$-module. As Martin says, the case of $f^$ can be understood via its adjunction with $f_$. Regards, – Matt E Oct 19 '13 at 12:43

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