# Hartshorne's proof that $\mathcal{O}_{\operatorname{Spec} A}(D(f)) \cong A_f$, Prop II.2.2(b)

Hartshorne, "Algebraic Geometry," Proposition II.2.2(b) on page 71 reads (roughly):

$\mathcal{O}_{\operatorname{Spec} A}(D(f)) \cong A_f$

The relevant section of the proof reads (after some simplification):

We define $\psi : A_f \to \mathcal{O}_{\operatorname{Spec} A}(D(f))$ by sending $a/f^n$ to the section which assigns to each $\mathfrak{p}$ the image of $a/f^n$ in $A_\mathfrak{p}$. First we show that $\psi$ is injective. If $\psi(a/f^n) = 0$, then for every $\mathfrak{p} \in D(f)$ we have $a/f^n = 0$ in $A_\mathfrak{p}$, so by definition there is some $h \not \in \mathfrak{p}$ such that $h a = 0$ in $A$. Let $\mathfrak{a}$ be the annihilator of $a$ in $A$. Then $h \in \mathfrak{a}$ and $h \not \in \mathfrak{p}$, so $\mathfrak{a} \not \subseteq \mathfrak{p}$. We conclude that $V(\mathfrak{a}) \cap D(f) = \emptyset$. Therefore $f \in \sqrt{\mathfrak{a}}$, so $f^\ell \in \mathfrak{a}$ for some $\ell$, so $f^\ell a = 0$. Since $f$ is a unit in $A_f$, this says that $a = 0$ in $A_f$. The hard part is to show that $\psi$ is surjective...

I follow the argument fine, but it strikes me as particularly ingenious, despite Hartshorne's implicit assertion that it's the "easy part" of the argument. (To be fair, the proof of surjectivity takes a whole page.) In particular, the fact that we're checking a commutative algebra result to apply to a problem in algebraic geometry by applying algebraic geometry to the commutative algebra problem kind of blows my mind at the moment.

Is this really just a very clever argument, or is there some perspective from which it's straightforward? Alternatively, is there some other way of seeing this result?

If $S\subset A$ is a multiplicative system, and $x\in S^{-1}A$ is such that $x=0$ in every $A_P$ with $P\cap S=\emptyset$, then $x=0$ since it is $0$ in all localizations of $S^{-1}A$.
$D(f)$ is the collection of points where $f$ doesn't vanish. So it stands to reason that if a function $a$ vanishes at all those points, then its product with $f$ should be zero. And since $f$ is invertible on $D(f)$, $a$ is annihilated by a unit, and so must be algebraically zero there.