# Wednesday, October 13

:::{.remark}
Result from last time: there doesn't exist a surjection $k\polynomialring{f_1,\cdots, f_m} \surjects k\powerseries{x}$ for any finite collection $\ts{f_i}_{i=1}^m$ of polynomials.
This can be checked by just considering their dimension as a $k\dash$modules, where $\dim_k LHS = \# \NN$ and $\dim_k RHS = \# \RR$.

We said that a morphism $X \mapsvia{f} Y$ is **locally finite type** if it is locally modeled on $\spec A\to \spec B$ with $B\to A$ a finitely-generated ring morphism, and is **finite** if locally modeled on $B \to A$ module-finite.
Note that in the first case, we require $f\inv(U) \contains U\to V$, but in the latter $U = f\inv(V)$.
:::

:::{.example title="?"}
As an example, the map $D(x) \to \AA^1\slice k$ was not finite since $k[x] \to k[x, x\inv]$ is not module-finite, despite the fact that this geometrically corresponds to $\AA^1\smz \injects \AA^1$:

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:::{.example title="Not finite type: $\spec$ of a local ring of a variety"}
Let $p \in V$ be a $k\dash$variety for $k$ an infinite field, which we can assume to be affine (so $\kxn/I$ for $I$ reduced).
Then $\OO_{V, p}$ is not not finitely-generated as a $k\dash$algebra, and $\spec \OO_{V, p}\to \spec k$ is not of finite type.
Consider the local ring of $\AA^1\slice k$ at the prime ideal $p \da \gens{x}$, then $k[x]\plocalize{p} = \ts{f/g \st g\not\in p}$, so $g(0) \neq 0$ for such $g$.
Note that this is not $k[x] \localize{x}$!

Idea: there are only finitely many denominators: if 
\[
\phi: k\polynomialring{f_1/g_1, \cdots, f_r/g_r} \surjects k[x]\plocalize{p}
,\]
then $\im \phi$ contains contains those $f/g$ such that $V(g) \subseteq \union V(g_i)$, so such a $\phi$ can not exist.
Note that this is still true for $k$ a finite field, just not by this proof.
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## Open/Closed Subschemes

:::{.definition title="Open subschemes"}
Given $X\in \Sch$, an **open subscheme** of $X$ is an open subset $U \subseteq \realize{X}$ with structure sheaf $\OO_U = \ro{\OO_X}{U}$.
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:::{.remark}
Why does $(U, \OO_U) \in \LRS$ form a scheme?
One needs to check that it's locally isomorphic to the spectrum of a ring: let $\ts{X_i}\covers X$ be an open affine cover, then $U_i\da U \intersect X_i$ is open in $U$ and in $X_i$, so can be covered by distinguished opens $V_{ij}$.
But then $\ts{V_{ij}}\covers U$ is a cover by affine schemes.
The inclusion $(U, \OO_U) \injects (X, \OO_X)$ is clearly a morphism in $\LRS$.
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:::{.definition title="Open Immersion"}
The inclusion above is an **open immersion**.
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:::{.remark}
A small dictionary

| AG        | Rest of Math |
|-----------|--------------|
| Immersion | Embedding    |
| Nothing!  | Immersion    |

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:::{.remark}
Here I write $\realize{X} \da \mathrm{sp}(X)$ as an alternative for Hartshorne's notation.
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:::{.definition title="Closed immersion"}
A **closed immersion** is a morphism $X \mapsvia{f} Y\in \Sch$ such that 

1. The induced map $\realize{X}\to \realize{Y} \in \Top$ is a homeomorphism onto a closed subset of $\realize{Y}$, 
2. $f^\#: \OO_Y \surjects f_* \OO_X \in \Sh(Y)$ is surjective.

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:::{.remark}
Set $U = D(f) \subseteq \spec A$ defines an open immersion $\spec A\localize{f} \to \spec A$.
So this corresponds to the ring map $A\injects A\localize{f}$ since $\spec A\localize{f} \subseteq \spec A$ are those ideals not containing $f$.
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:::{.example title="?"}
Consider $U \da \AA^2\slice k\sm\ts{\tv{x, y}} \injects \AA^2\slice k$.
Then $\ts{\tv{x, y}} = V(x, y)$, and this is a subscheme of an affine scheme which is not itself affine.
One can use that $\dim D(f)^c \geq 1$

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:::{.exercise title="?"}
Prove that this is not affine.
Hint: check $\OO_U = k[x,y]$,[^closed_regular]
and use that for any $X \in \Aff\Sch$ we have $\OO_X(X) = R$.
However, $\spec k[x, y] = \AA^2\neq U$.

[^closed_regular]: 
This says that any regular function on $U$ actually extends to all of $\AA^2$

:::

> Check "affinization"? It fills in holes of at codimension at most 2, and satisfies a universal property.
> Consider $X = \AA^1\times \PP^1$.

:::{.remark}
All examples are locally of this form: $F:X\injects Y = \spec A$ where $\realize{X}\to U \subseteq \realize{Y}$ maps homeomorphically onto a closed subset.
Recall that the closed subsets are of the form $U = V(I)$, and here we need $f^\#: \OO_Y \to f_* \OO_X$ surjective.
Let $X = \spec A/I$, and recall that every surjective ring map is of the form $A\to A/I$.
Here $q: A\surjects A/I$ where $\mfp \mapsfrom q\inv(\mfp)$, so we get some map $\spec A/I\to \spec A$, and this is homeomorphism onto $V(I) \subseteq \spec A$:
\[
\spec A/I &\to V(I) \\
\mfp &\mapsto q\inv(\mfp) \contains I \\
q(\mfq) &\mapsfrom \mfq
.\]
We also get an induced map $A\localize{g} \to (A/I)\localize{g}$, which is precisely 
\[
f^\#(D(g)): \OO_Y(D(g)) \to \OO_X(f\inv(D(g)))
\]
and is thus surjective.
Since it's surjective on a basis, by gluing it'll be surjective on the entire space.
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