# Sections of the structure sheaf (Friday, September 03)

## Sections of puncturing at zero

:::{.remark}
Last time: we defined $\spec A$ as a topological space and $\OO_{\spec A}$, a sheaf of rings on $\spec A$ which evidently satisfied the gluing condition:
\[
\OO_{\spec A}(U) \da \ts{s: U\to \Disjoint_{p\in U} A_p \st s(p) \in A_{p} \, \forall p \text{ and } s \text{ is locally a fraction}}
.\]

:::

:::{.example title="?"}
Set $X\da \AA^1\slice k \da \spec k[t]$ for $k=\bar k$.
Take a point $\gens t = \gens{t-0} \in \spec k[t]$ corresponding to $0\in X$, then 
\[
\OO_{X}(X\smts{0}) = k[t, t\inv]
= \ts{{f(t) \over t^\ell } \st f\in k[t], \ell \geq 0 }
.\]
Generally for $p = \gens{t-a_1, \cdots, t-a_m}$ we get $s_p \in k[t]\localize{\ts{t-a_i}_{1\leq i\leq m} }$.
Note that for $p = \gens 0$, we get $s_p \in k\rff{t}$.
:::

:::{.claim}

A section $s$ is determined by $s_{p}$ for $p = \gens{0}$, so there is an injective map
\[
\OO_{\spec k[t]}(\spec k[t]\smts{0}) &\to k(t) \\
s &\mapsto s_{\gens 0}
.\]
:::

:::{.proof title="?"}
Note that $\gens 0$ is in every open set, so locally near $p$ there exists a $P\in V$ and $a,f$ with $f\not\in Q$ for all $Q$ and $s_Q = a/f$ for all $Q\in V$.
Since $\gens 0 \in V$, we have $s_{\gens 0} = a/f \in k(t)$ and $s_p = a/f\in A_p$.
Since $A_p \subseteq k(t)$, we get $s_p = s_{\gens 0}$ under this inclusion.
:::

:::{.claim}
\[
\OO_{\spec k[t]}(\spec k[t]\smts{0}) = k[t, t\inv] = k[t]\localize{\ts{t^\ell}_{\ell\geq 0}}
.\]
:::

:::{.proof title="?"}
We showed that the LHS is a subset of $k(t)$, so which subsets can be written as things that are locally fractions on the complement of zero.

$\supseteq$: 
This can clearly be done in $k[t, t\inv]$ since every element is locally the fraction $f/t^k$.

$\subseteq$:
Suppose $f/g$ with $f,g$ coprime (this is a PID!) with a pole away from zero, so $g\in Q$ for some $Q\neq \gens{0}$.
But then $f/g$ isn't in $A_Q$.
:::

:::{.remark}
Note that $X \da \mspec k[t] \subseteq X' \da \spec k[t]$ as the set of closed points, and restricting $\OO_{X'}$ to $X$ yields the sheaf of regular functions.
Having the extra generic point was useful!
:::

:::{.exercise title="?"}
Show that the maximal ideals $m\normal A$ correspond precisely to closed points of $X=\spec A$.
:::

:::{.example title="of a function that is locally but not globally a fraction"}
Take $A \da k[x,y,z,w]/\gens{xy-zw}$, which is the cone over a smooth quadric surface and $X\da \spec A$.
Then take $U = \spec(A) \sm V(y, w) = V(y)^c \intersect V(w)^c$ and consider the section
\[
s(p) \da
\begin{cases}
x/w & p\in V(w)^c 
\\
z/y & p\in V(y)^c.
\end{cases}
\]

For $p\in U$, it makes sense to consider $x/w$ and $z/y$.
Are they equal?
The answer is yes because $xy-zw = 0$.
Check that this can't be a global fraction, which is a consequence of this random open set not being the complement of localizing at a prime ideal.
:::

## Distinguished opens

:::{.definition title="Distinguished open sets"}
Given $f\in A$, the **distinguished open** $D(f)$ corresponding to $f$ is defined as
\[
D(f) = V(\gens f)^c \da \ts{p\in \spec(A) \st f\in p}^c = \ts{p\in \spec A \st f\not\in p}
,\]

i.e. the points of $\spec(A)$ where $f$ doesn't vanish.
:::

:::{.remark}
The sets $\ts{D(f) \st f\in A}$ form a basis for the topology on $\spec (A)$.
This follows from writing $V(I)^c = \Union_{f\in I} D(f)$.
:::

## The fundamental theorem of $\OO_{\spec A}$ (Hartshorne Proposition 2.2)

:::{.theorem title="Hartshorne Prop 2.2"}
Let $A\in \CRing$ be unital with $1\neq 0$ unless $A=0$ and consider $(\spec A, \OO)$.
Then

a. For any $p\in \spec A$, the stalk $\OO_p \cong A_{p}$.
b. For any $f\in A$, $\OO(D(f)) = A_{f}$.
c. Taking $f=1$, $\Gamma(\spec A, \OO) = A$.

:::

:::{.remark}
Note that (b) gives the values of $\OO$ on a basis of opens, which determines the sheaf.
:::

:::{.proof title="of a"}
Define a map
\[
f_p: \OO_p &\to A_p \\
(U, s) &\mapsto s(p)
.\]
This is well-defined since $p\in W$ for any $W \subseteq U \intersect V$ for equivalent germs $(U, s) \sim (V, t)$.

Surjectivity: given $x=a/g \in A_p$, we want $(U, s)\in \OO_p$ such that $f_p(U, s) = a/g$, so just take $U = D(g)$ and $s=a/g$ (which makes sense!) and evidently maps to $a/g$.

Injectivity: supposing $f_p(U, s) = 0$ in $A_p$, we want $(U, s) = 0$.
If $s(p) = 0$, then there exists some $h\in P$ with $h\cdot s(p) = 0$.
Since $s(p)$ is locally a fraction, we can find $p\in V \subseteq U$ with $s=a/g$ on $V$ with $g\neq 0$ on $V$, so consider $V \intersect D(h)$.
The claim is that $s$ is 0 here, which follows from $h\cdot (a/g) = 0$.
:::