# Wednesday, September 15

## asdsadas

:::{.remark}
Last time: for $\Aff\Var$ we considered $X = V(I) \subseteq \AA^n\slice k$, and for $\Aff\Sch$ we considered $\spec k[X]$ where $k[X] \da\kxn / I(X)$.
Both had the Zariski topology, and $X = \mspec k[X] \subseteq \spec k[X]$.
We had structure sheaves $\OO_{\spec k[X]}$, and for $X$, we have $U' \da U \intersect \mspec k[X]$.
On $\mspec k[X]$, we have the notion of a regular function, and $\OO_X(U') = \OO_{\spec k[X]}(U')$.
The previous setup only worked for rings finitely generated over a field, whereas for schemes, we can take things like $\spec \ZZ$, so they're much more versatile (e.g. for number theory).

:::

## adssads 

:::{.example title="?"}
Compare $\AA^2\slice k \in \Aff\Var$ to $\spec k[x, y]$.
Note that $\gens 0 \in \spec k[x, y]$, and taking its closure yields
\[
\cl(\gens{0}) 
&= \Intersect_{V(I)\contains \gens 0 } V(I) \\
&= \Intersect_{V(I)\ni 0 } V(I) \\
&= V(0) \\
&= \spec k[x, y]
,\]
so $0$ is a dense point!


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But there are prime ideals of height $>1$.
For example, any irreducible subvariety of $\AA^2$ yields a generic point.

\todo[inline]{Krull's dimension theorem?}

:::

:::{.exercise title="?"}
Try to draw $\spec \ZZ$ and $\spec \ZZ[t]$.
:::

:::{.remark}
We'll now try a naive definition of schemes, which we'll find won't quite work.
:::

:::{.definition title="A wrong definition of a scheme!"}
A **scheme** is a ringed space $(X, \OO_X)$ which is locally an affine scheme, i.e. there exists an open cover $\mcu \covers X$ such that there is a collection of rings $A_i$ with
\[
(U_i, \ro{\OO_{X}}{U_i} ) \iso (\spec A_i, \OO_{\spec A_i})
.\]
:::

:::{.remark}
This produces the right objects, but not the correct morphisms. 
This is a subtle issue!
With this definition, a morphism of schemes would be a morphism of ringed spaces $(f, f^\#)$ with $f\in \Top(X, Y)$ and $f^\# \in \Sh\slice Y(\OO_Y, f_* \OO_X)$, where $f^\#$ is supposed to capture "pullback of functions".
The issue: $f^\#$ may not notice that $p \mapsvia{f} f(p)$!
In particular, it may not preserve the fact that $f(p) = 0$.

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 Hartshorne exercises for how this issue can actually arise.
:::

:::{.remark}
Let $(f, f^\#)$ be a map of ringed spaces, then there is an induced map 
\[
f_p^\#: \OO_{Y, f(p)} &\to \OO_{X, p} \\
(U, s) &\mapsto (f\inv(U), f^\#(U)(s))
.\]
:::

:::{.definition title="Locally ringed space"}
A **locally ringed space** $(X, \OO_X)$ is a ringed space for which the stalks $\OO_{X, p} \in \Loc\Ring$ are local rings, i.e. there exists a unique maximal ideal.
:::

:::{.example title="of a locally ringed space"}
For $(X, \OO_X) \da(\spec A, \OO_{\spec A})$, we saw that $\OO_{X, p} = A\localize{p}$, which is local.
:::

:::{.definition title="Morphisms of locally ringed spaces"}
A **morphism of locally ringed spaces** is a morphism of ringed spaces 
\[
(f, F^\#): (X, \OO_X) \to (Y, \OO_Y)
\]
such that $f^\#_p: \OO_{Y, f(p)} \to \OO_{X, p}$ is a homomorphism of local rings, i.e. $f^\#(\mfm_{f(p)}) \subseteq \mfm_p$.

> Here we should also require that $f^\# \neq 0$.

:::

:::{.remark}
Morally: this extra condition enforces that pulling back functions vanishing at $f(p)$ yields functions which vanish at $p$.
:::

:::{.remark}
Alternatively one could require that $(f^\#)\inv(\mfm_p) = \mfm_{f(p)}$, and (claim) this is equivalent to the above definition.
Use that $(f^\#)\inv(\mfm_p)$ is a prime ideal containing $\mfm_p$.
:::

:::{.example title="of a locally ringed space"}
Take $(X, \OO_X) \da (\RR, C^0(\wait; \RR))$.
Why this is in $\Loc\Ringedspace$: write a stalk as 
\[
C_p^0 = \ts{(f, I) \st I\ni p \text{ an interval}, f\in \Top(I, \RR)}\slice\sim
.\]
Why is this local?
Take $\mfm_p \da \ts{(f, I) \st f(p) = 0}$, which is maximal since $C_p^0/\mfm \cong \RR$ is a field.
Then $\mfm_p^c = \ts{(f, I) \st f(p) \neq 0}$, and any continuous function that isn't zero at $p$ is nonzero in some neighborhood $J \contains I$, so $\ro{f}{J}\neq 0$ anywhere.
Then $(f, I) \sim (\ro{f}{J}, J)$, which is invertible in the ring, so any element in the complement is a unit.
:::

:::{.example title="?"}
Consider 
\[
(\RR, C^0(\wait; \RR)) \mapsvia{(f, f^\#)} (\RR, C^\infty(\wait; \RR))
.\]
Take $f = \id$ and the inclusion
\[
f^\# : C^\infty(\wait; \RR)\injects \id_* C^0(\wait; \RR) = C^0(\wait; \RR)
.\]
Then 
\[
f_p^\#: C_p^\infty(\wait; \RR) \to C_p^0(\wait; \RR)
.\]
satisfies $f_p^\#(\mfm^\infty_{\id(p)}) = \mfm^0_p$.
We also have $(f_p^\#)\inv (\mfm_p^0) = \mfm_p^\infty$, since the germs are just equal.
:::

:::{.definition title="Scheme"}
A **scheme** $(X, \OO_X)$ is a locally ringed space which is locally isomorphic to $(\spec A, \OO_{\spec A})$ in $\Loc\Ringedspace$.
A **morphism of schemes** is a morphism in $\Loc\Ringedspace$.
:::

:::{.remark}
Note that we can restrict to opens, since this doesn't change the stalks.
:::

:::{.remark}
As a proof of concept that this is a good notion, we'll see that there's a fully faithful contravariant functor $\spec: \CRing \to \Sch$, so 
\[
\spec(\Mor_\Ring(B, A)) = \Mor_\Sch( \spec A, \spec B)
.\]
:::