# Wednesday, September 22

:::{.remark}
Today: how to build more schemes by gluing known ones together. 
Let $(X_1, \OO_{X_1})$ and $(X_2, \OO_{X_2}) \in \Sch$, i.e. locally ringed spaces locally isomorphic to $(\spec R, \OO_{\spec R})$.
Let $U_i \subseteq X_i$, and suppose we're given an isomorphism in $\LRS$:
\[
\phi: 
(U_1, \ro{\OO_{X_1}}{U_1}) \to 
(U_2, \ro{\OO_{X_2}}{U_2}) 
.\]

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Define a locally ringed space as follows: set $X = X_1 \disjoint X_2/ x\sim \phi(x)$, and define
\[
\OO_X(U) \da 
\ts{ 
s_1 \in \OO_{X_1}(X_1 \intersect U), s_2\in \OO_{X_2}(X_2 \intersect U) \st \ro{s_1}{U_1 \intersect U} = \phi^\#(U_2 \intersect U)\qty{ \ro{s_2}{U_2 \intersect U} }
} 
.\]

:::

:::{.example title="?"}
Consider $I \da (0, 1) \subseteq \RR$ and take $X_1 = X_2 \da (I, C^\infty(\wait; \RR))$.
Using these to cover the circle, we can obtain $(S^1, C^\infty(\wait; \RR))$, using smooth functions that agree on the overlap (here a disjoint union of smaller intervals).
:::

:::{.example title="A non-affine scheme"}
Let $X_1 = X_2 \da \AA^1\slice k \da \spec k[x]$, and set $U_1 \subseteq X_1 \da D(x)$.
Then take the clear isomorphism 
\[
(U_1, \OO_{X_1}\ro{}{U_1}) \mapsvia{\id} 
(U_2, \OO_{X_2}\ro{}{U_2})
,\]
since they're the same open subset of the same affine variety.
Gluing yields the following:

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:::

:::{.exercise title="?"}
Prove this is not an affine scheme.
Use that regular functions are determined by their values on a Zariski open.
:::

:::{.claim}
For $X = \spec R$, recall that $D(f) \da \ts{ p\in \spec R \st p\not\ni f}$.
Then $(D(f), \ro{ \OO_X }{D(f)})$ is an affine scheme.
:::

:::{.proof title="?"}
Consider $R\localize{f} \da \ts{r/f^k \st r\in R, k\geq 0}/\sim$.
There's a map $R\to R\localize{f}$ which induces a map $\spec R\localize{f} \to \spec R$, and we claim it's the inclusion of $D(f)$.

:::{.claim}
$\spec R\localize f = D(f)$ as spaces.
:::

This uses that primes in the localization are primes in $R$ not intersecting the inverted set.
So $\spec R\localize f = \ts{p\in \spec R \st p \intersect \gens{f} = \emptyset}$, then use that $p \intersect \ts{f^k} = \emptyset \iff f\not\in p$, since prime ideals are radical.
We now want to show $\OO_{\spec R\localize f} = \ro{\OO_{\spec R}}{D(f)}$.
Check that 
\[
\OO_{\spec R\localize f} = \ts{s:U\to \Disjoint_{p\in U} (R\localize f)\localize p}
\]
and 
\[
\ro{\OO_{\spec R}}{D(f)}(U) = \ts{s: U \to \Disjoint_{p\in U} R\localize{p}}
,\]
but $(R_f)_p = R_p$ since $f\not \in p$.
This uses that $(R\localize{S_1})\localize{S_2} = R\localize{S_2}$ when $S_1 \subseteq S_2$.
The first are functions of the form $(a/f^k)/(b/f^\ell) = f^\ell a/f^k b$, so anything locally a fraction in $R_f$ is locally a fraction in $R$.
:::

:::{.example title="?"}
Let $X_1 = \AA^1\slice k$ with $U_1 \da D(x) \subseteq X_2$ and $X_2 = \AA^1\slice k$ with $U_2 = D(y)$.
Then 
\[
(U_1, \OO_{X_1}\ro{}{U_1}) \cong (k[x, x\inv], \OO_{\spec k[x, x\inv]}) 
\mapsvia{(\phi, \phi^\#)} 
(U_2, \OO_{X_2}\ro{}{U_2}) \cong (k[y, y\inv], \OO_{\spec k[y, y\inv]})
.\]

Then $(\phi, \phi^\#)$ is an isomorphism in $\LRS$ is given by a unique isomorphism
\[
\iota: k[x, x\inv] &\to k[y, y\inv] \\
y &\mapsto x \\
y\inv &\mapsto x\inv
.\]
Note that there is another isomorphism:
\[
\iota': k[x, x\inv] &\to k[y, y\inv] \\
y &\mapsto x\inv \\
y\inv &\mapsto x
.\]
So glue instead by the morphism $(\phi, \phi^\#)$ induced by $\iota'$.
We'll then define **projective space** as
\[
\PP^1\slice k \da \AA^1\slice k \Disjoint_{(\phi, \phi^\#)} \AA^1\slice k
.\]
Note that this works over any ring!

What does this do to closed points?
The closed points of $\spec k[x, x\inv]$ are $\ts{(x-c) \st c\neq 0}$ if $k=\bar k$, which corresponded to the closed points $c\in \AA^1\slice k$ as a variety.
Under $x\mapsto y\inv$, we have $(x-c)\mapsto (y\inv - c) = (y-c\inv)$, thus the variety point $c$ gets sent to $c\inv$.

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:::