# Monday, October 11

:::{.remark}
Last time: Noetherian isn't a purely topological property.
Today: another guiding principle in AG is that one can put properties on schemes, or alternatively on morphisms, usually to $\spec k$.
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:::{.definition title="Finite type morphisms"}
A morphism $X \mapsvia{f} Y$ of schemes is **locally of finite type** if there exists an affine open cover $\mcv\covers Y$ with $V_i = \spec B_i$ such that there exists an affine open cover $\mcu \covers f\inv(\mcv)$ so $U_{ij}\contains f\inv(V_i)$ where $U_{ij} = \spec A_{ij}$ such that the restricted function $f: U_{ij} \to V_{i}$ is induced by a ring morphism $B_{i } \to A_{ij}$ finitely generated over $B_i$.
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:::{.remark}
This globalizes the notion of being a finitely generated ring, essentially by covering the scheme morphisms by ring morphisms with the desired property.
As a special case, let $X \in \Sch\slice{k}$, so there is a morphism $X\to \spec k$.
Let $X = \union_j U_j$ with $U_j = \spec A_j$, then we want that the map $\spec A_{j}\to \spec k$ is induced by a finitely-generated ring morphism $k\to A_j$, so $A_j$ is a finitely-generated $k\dash$algebra.
So this is like having a sheaf of $k\dash$algebras.
As an abuse of notation/terminology, we say that $X$ is **finite type** over $k$ (since the target doesn't need to be covered).
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:::{.warnings}
Note that a subring of a finitely generated ring need not be finitely generated!
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:::{.example title="?"}
Let $A\in \kalg^{\fg}$, then $\spec A$ is finite type over $k$.
One can change the definition from "there exists an open cover" to "for all open covers" -- this amounts to checking localizations of ring maps.
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:::{.example title="?"}
Consider $X \da \Proj k[x, y] = \PP^1\slice k$, then recall that $(D(f), \ro{\OO_X}{D(f)}) \cong \spec R\localize{f}$.
Then 
\[
\PP^1\slice k = D(x) \disjoint_f D(y) \cong \AA^1\slice k\disjoint_f \AA^1\slice k
,\]
glued along inversion.
Then $k[x,y]\localize{x} \cong k\polynomialring{y\over x}$ and $k[x, y]\localize{y} \cong k\polynomialring{x\over y}$.
One can check that $\PP^1\slice k\to \spec k$ is finite type, and this works for $\PP^n$ as well.

Note that if $S\da \kxn/I$ for $I$ a homogeneous ideal, then $\Proj S$ is also finite type over $k$.
We can write $\Proj S = \Union_{i=0}^n D(\bar x_i)$, since taking complements yields $\emptyset = \Intersect_{i=0}^n V(\bar x_i)$, which is the set of homogeneous prime ideals $p\in \spec S$ with $p\contains \bar x_i$ for all $i$ and $p\not\contains S^+$ the irrelevant ideal, which is empty.
So $S\localize{x_i}$ is a finitely generated ring, with generators of the form $x_j/x_i$.
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:::{.example title="a non-example"}
$\spec k\formalpowerseries{t} \to \spec k$ is not a finite type morphism, i.e. $k\fps{t}$ is not a finitely generated $k\dash$algebra.
Toward a contradiction suppose $k\fps{t} = \gens{f_1, \cdots, f_r}$, so there is a ring map $k\adjoin{f_1,\cdots, f_r} \surjects k\fps{t}$.
Take $k= \QQ$, or any countable field, then the LHS is countable but the right-hand side is not.
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:::{.definition title="Finite morphisms"}
Recall that a morphism $\phi:B\to A \in \CRing$ is a **finite morphism** if $A$ is finitely-generated as a $B\dash$module.
A morphism $X \mapsvia{f} Y\in \Sch$ is **finite** iff there exists an affine open cover $\mcv\covers Y$ with $V_i = \spec B_i$ and $f\inv (V_i) = \spec A_i$, and the induced ring maps $B_i\to A_i$ are finite.
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:::{.remark}
Here the module structure is $b\cdot a \da \phi(b)a$.
Note that finite type required finite generation as rings, so $B(g_1,\cdots, g_r)\surjects A$, but here we require that any $a\in A$ is of the form $a = \sum_{i=1}^r \phi(b_i) a_i$ for some $b_i$.
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:::{.example title="?"}
Consider $X \da \spec R \da \spec \CC[x, y]/ \gens{y^2-f(x)}$ where $f$ has no repeated roots, which yields a hyperelliptic curve.
This is a reduced ring, so $X$ is the scheme associated to a variety.
Letting $r_i$ be the roots of $f$, we have the following:

\begin{tikzpicture}
\fontsize{45pt}{1em} 
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\end{tikzpicture}

Consider the function $f:X\to Y$ induced by the following ring map:
\[
\CC[x] &\to R \\
x &\mapsto x
,\]
which is projection onto the axis.
Note that $R \cong \CC[x] \gens{1} \oplus \CC[x] \gens{y}$ as a $\CC[x]\dash$module.
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:::{.example title="?"}
Consider $\spec \CC[x,y,z]/\gens{xyz - 1} \to \spec \CC[x]$, whose real locus yields a hyperboloid:

\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-10-11_12-18.pdf_tex} };
\end{tikzpicture}

Note that finite type should approximately be spec of finite type $k\dash$algebras, i.e. essentially varieties, where for finiteness the fibers should be finite. 
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:::{.example title="?"}
Consider $\spec \CC[x, x\inv] \to \spec \CC[x]$, i.e. $\GG_m \injects \AA^1$.
However, $\CC[x, x\inv]$ is not finitely-generated as a $\CC[x]$ modules, even though it has finite fibers.
Given any finite set of generators, one can take $\CC[x]\gens{f_i\over x^{k_i}}$ which doesn't contain $1/x^{\max k_i + 1}$.
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:::{.remark}
We'll define subschemes soon.
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