# The fundamental theorem of $\OO_{\spec A}$ (Wednesday, September 08)

:::{.remark}
Recall that we defined a first version of *affine schemes*, namely pairs $(\spec A, \OO_{\spec A})$ where for $U \subseteq \spec A$ open we have $s\in \OO_{\spec A}(U)$ locally represented by $\ro{s}{V} = a/f$ for $V \subseteq U$ where $a, f\in A$ and $V(f) \intersect V = \emptyset$, so $f$ doesn't vanish on $V$.
Note that the $D(f)$ form a topological basis for $\spec A$, and the gluing condition is difficult, i.e. $\OO_{\spec A}(U)$ may be hard to compute, even given an open cover $\mcv\covers U$.
We proved that $\OO_{\spec A, \mfp} = A_\mfp$ last time, and today we're showing

- $\OO_{\spec A}(D(f)) = A_{f}$,
- $\Globsec{\Spec A; \OO_{\spec A}} \cong A$.

:::

## Proof of the fundamental theorem

:::{.proof title="of b and c"}
$b\implies c$:
Take $f=1\in A$, then $\OO(\Spec A) = \OO(D(1)) = A$ using (b), so the only hard part is showing (b).

To prove (b), by definition of $\OO$ there is a ring morphism
\[
\psi: A_{f} &\to \OO(D(f)) \\
{a\over f^n} &\mapsto {a\over f^n}
.\]

Note that this is just a careful statement, since the morphisms on stalks $\psi_{\mfp}: A_{f} \to A_{\mfp}$ by not be injective in general.
The proof will follow if $\psi$ is both injective and surjective.
:::

:::{.claim}
$\psi$ is bijective.
:::

:::{.proof title="of injectivity"}
Suppose $\psi(s) = 0$, we then want to show $s=0$.
Write $s = a/f^n$, then for all $\mfp \in D(f)$ we know $a/f^n = 0 \in A_{\mfp}$.
So for each $\mfp$ there is some $h_{\mfp} \not\in\mfp$ where 
\[
h_{\mfp}(a\cdot 1 - f^n\cdot 0) = 0 && \text{in } A
\]
in $A$.
Consider the ideal $\mfa \da \Ann(a) \da \ts{b\in A \st ab=0 \in A} \ni h_{\mfp}$.
So take the closed subset $V(\mfa)$, which does not contain $\mfp$ since $\mfa \not\subseteq\mfp$.
Now iterating over all $\mfp \in D(f)$, we get $V(\mfa) \intersect D(f) = \emptyset$.
So $V(\mfa) \subseteq V(f) = D(f)^c$, thus $f\in \sqrt{\mfa}$ and $f^m a = 0$ for some $m$.
Then $f^m(a\cdot 1 - f^n\cdot 0) = 0$ in $A$, so $a/f^n = 0$ in $A_{f}$.

:::

:::{.proof title="of surjectivity"}
**Step 1**: Expressing $s\in \OO(D(f))$ nicely locally.

By definition of $\OO_{D(f)}$, there exist $V_i$ with $\ro{s}{V_i} = a_i/g_i$ for $a_i, g_i\in A$.
We'd like $g_i = h_i^{m_i}$ for some $m_i$, so $g$ is a power of $h_i$, but this may not be true a priori.
Fix $V_i = D(h_i)$, then $a_i / g_i\in \OO(D(h_i))$ implies that $g_i\not\in \mfp$ for any $\mfp \in D(h_i)$.
This implies that $D(h_i) \subseteq D(g_i)$, and taking complements yields $V(h_i) \contains V(g_i)$, and $h_i \in \sqrt{\gens{g_i}}$ and $h_i^{n} = g_i$.
Writing $g_i = h_i^n/c$ we have $a_i/g_i = ca_i / h_i^n$.
Note that $D(h_i) = D(h_i^n)$.
Now replace $a_i$ with $ca_i$ and $g_i$ with $h_i$ to get 
\[
\ro{s}{D(h_i)} = a_i / h_i
.\]

**Step 2**: Quasicompactness of $D(f)$.

Note that $\ts{D(h_i)}_{i\in I} \covers D(f)$, so take a finite subcover $\ts{D(h_i)}_{i\leq m}$.

Proof of quasicompactness: since $D(f) \contains \Union_{i\in I} D(h_i)$, we get 
\[
V(f) \subseteq \Intersect_{i\in I} V(h_i) = V\qty{ \sum h_i}
.\]
So $f^u \in \sum h_i$, and up to reordering we can conclude $f^u = \sum_{i\leq m} b_i h_i$ for some $b_i \in A$.
Then $D(f) \subseteq \Union_{i\leq m} D(h_i)$.

:::{.remark}
Since we can write $\spec A = D(1)$, it is quasicompact.
:::

**Step 3**: Showing surjectivity.

Next time.

:::