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\begin{document}
\title{Linear Algebraic Groups: Lecture 9}
\date{October 20, 2016}
\author{John Simanyi}
\maketitle
\section{Expanding on Algebraic Geometry}
As we hinted at in the last lecture, the relationship between algebraic
sets and affine schemes can be broadened. Alexander Grothendieck really
refined the idea of affine schemes into a much more general notion
of schemes, and the inclusions can be seen as follows:
\begin{center}
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\node (E) at (0,3) {\begin{tabular}{c} Algebraic Sets \\$ S\subseteq k^n$ \end{tabular}};
\node (N2) at (0.8,1.9) {};
\node (N) at (0,2) {Affine (Algebraic) Varieties};
\node (H2) at (0.8,0.95) {};
\node (H) at (0,1) {Affine Schemes};
\node (EE) at (3,2.5) {\begin{tabular}{c} Projective (Algebraic) Varieties \\$ S\subseteq kP^n$ \end{tabular}};
\node (P) at (3,1.5) {Quasiprojective Varieties};
\node (S) at (3,0.5) {Schemes};
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\par\end{center}
Previously, we looked at the duality between geometry and commutative
algebra, and determined we could study algebraic sets by looking at
the algebras of regular functions on those sets, which we called ``affine
algebras''. We also encountered some difficulties, including the
need to restrict to only algebraically closed fields, as well as fields
without nilpotents. Grothendieck's scheme-based approach includes
everything we discussed, but removes these restrictions. We will extend
our understanding of the above diagram as the course progresses. For
now, we have:
\medskip{}
\begin{myFramedBox}
\begin{thm*}
If $k$ is an algebraically closed field (such as $\mathbb{C}$ or
$\overline{\mathbb{F}}_{q}$), then the following conditions on a
commutative algebra $A$ over $k$ are equivalent:
\begin{enumerate}
\item $A\cong k[S]$ for some algebraic set $S\subseteq X,$ where $X$
is a finite dimensional vector space.
\item $A\cong k[X]/J,$ where $J$ is a radical ideal (so if $V^{n}\in J$
for some $n$, then $V\in J).$
\item $A$ is finitely generated, and has no nonzero nilpotents.
\end{enumerate}
If any of these hold, we call $A$ an \uline{affine algebra}.
\end{thm*}
\end{myFramedBox}
\vspace{-0.3cm}
\begin{proof}
\emph{(Sketch) }The equivalence of (1) and (2) follows from our ``nicer''
version of Hilbert's Nullstellensatz. Given some algebraic set $S\subseteq X$,
set $J=I(S),$ the set of polynomial functions which vanish on $S$.
Conversely, given a radical ideal $J,$ set
\[
S\ =\ \left\{ x\in X:P(x)=0\text{ for all }P\in J\right\} .
\]
By definition, this is an algebraic set.
(2) and (3) are also equivalent. If $A$ is a finitely generated commutative
algebra without nilpotents, choose generators $x_{1},\ldots,x_{n}$
of $A$, and form the polynomial algebra $k\left[x_{1},\ldots,x_{n}\right]$.
Then
\[
A\ \cong\ k\left[x_{1},\ldots,x_{n}\right]/J,
\]
where $J$ is the kernel of the algebra homomorphism from $A$ to
$k\left[x_{1},\ldots,x_{n}\right].$ Note that the homomorphism is
onto, as $\{x_{i}\}$ are generators, and $J$ is a radical ideal,
as $A$ has no nonzero nilpotents.
\end{proof}
So we can think of an algebraic set as corresponding to an affine
algebra, and use affine algebras as stand-ins for algebraic sets -
but they are defined intrinsically via (3).
Now, algebras are abstract, and we usually think of algebraic sets
as a ``shape''. Therefore, we call an affine algebra an \uline{affine
variety} when we want to think of it as a stand-in for an algebraic
set. If we're thinking of these objects as equivalent, what do we
mean by a ``map'' between algebraic sets?
\medskip{}
\begin{myFramedBox}
\begin{defn*}
Given two algebraic sets, $S\subseteq X$ and $S'\subseteq X'$, a
function $\varphi:S\rightarrow S'$ is \uline{regular} if there's
a regular function $\Phi:X\rightarrow X'$ (a function that is defined
via coordinate functions that are polynomial functions) such that
$\Phi\bigr|_{S}=\varphi.$
\end{defn*}
\end{myFramedBox}
\vspace{-0.3cm}
\begin{defn*}
How do we think about a regular map between algebraic sets intrinically,
as a map between commutative algebras? Here's where duality comes
in.
\end{defn*}
\medskip{}
\begin{myFramedBox}
\begin{thm*}
Given a regular map $\varphi:S\rightarrow S'$ , we get an algebra
homomorphism
\[
\varphi^{*}:k[S']\rightarrow k[S],
\]
the pullback if $\varphi.$ This is given in the usual manner: for
any $f\in k[S']$ and any $x\in S,$
\[
\varphi^{*}(f)(x)\ =\ f\left(\varphi(x)\right).
\]
Conversely, any homomorphism $\alpha:k[S']\rightarrow k[S]$ comes
from a unique regular map $\varphi:S\rightarrow S'$ in this way (so
$\alpha=\varphi^{*}$).
\end{thm*}
\end{myFramedBox}
\vspace{-0.3cm}
\begin{proof}
\emph{(Sketch) }If $f\in k[S'],$ then $\varphi^{*}(f)=f\circ\varphi$
is a composite of regular (and therefore polynomial) functions, so
it is a regular function. Thus,
\[
\varphi^{*}(f)\ \in\ k[S].
\]
Moreover, it's an algebra homomorphism. For $a,b\in k$ and $f,g,h\in k[S'],$
\begin{eqnarray*}
\varphi^{*}(af+bg\cdot h) & = & (af+bg\cdot h\}\circ\varphi\\
& = & \left(af\right)\circ\varphi+\left(bg\right)\circ\varphi\cdot h\circ\varphi\\
& = & a\left(f\circ\varphi\right)+b\left(g\circ\varphi\right)\cdot\left(h\circ\varphi\right)\\
& = & a\left(\varphi^{*}f\right)+b\left(\varphi^{*}g\right)\cdot\left(\varphi^{*}h\right).
\end{eqnarray*}
Conversely, suppose we have an algebra homomorphism $\alpha:k[S']\rightarrow k[S].$
The points of $S$ are in one-to-one correspondence with algebra homomorphisms
$\psi:A\rightarrow k.$ In fact, any $x\in S$ gives such a homomorphism
through the evaluation map,
\[
\psi_{x}(f)\ :=\ f(x).
\]
However, the hard part is showing that it's actually a one-to-one
correspondence. Using our $\alpha,$ let $\psi_{x}:k[S]\rightarrow k$
be the evaluation map . We then form
\[
\psi\circ\alpha:k[S']\rightarrow k,
\]
noting that this comes from a unique point in $S',$ which we call
$\varphi(x).$ This gives a map
\[
\varphi:S\rightarrow S',
\]
then we can check that it is indeed regular.
\end{proof}
The slick way to think of all of this is that
\[
\text{AlgSet}_{k}\ \cong\ \text{AffAlg}_{k}^{\textrm{op}},
\]
where $\text{AlgSet}_{k}$ is ``the land of geometry'' - the category
with algebraic sets as objects, and regular maps are morphisms - while
$\text{AffAlg}_{k}$ is ``the land of commutative algebras'' - the
category with affine algebras as objects, and algebra homomorphisms
as morphisms.
Recall that $\bullet^{\textrm{op}}$ just flips morphism arrows. We
then have a contravariant functor
\begin{eqnarray*}
\text{AlgSet}_{k} & \ \rightarrow\ & \text{AffAlg}_{k}\\
S & \ \mapsto\ & k[S],\\
\varphi:S\rightarrow S' & \ \mapsto\ & \varphi^{*}:k[S']\rightarrow k[S].
\end{eqnarray*}
This is an equivalence of categories (fully faithful, essentially
surjective functor, if you'd like to look that up). This is the duality
between geometry and commutative algebra.
\section{Schemes Appear}
There are three problems here.
\begin{itemize}
\item This construct only works if $k$ is algebraically complete (the reason
why algebraic geometers love $\mathbb{C}$ or $\overline{\mathbb{F}}_{q}$,
instead of $\mathbb{R}$ or $\mathbb{F}_{q}$).
\item Affine algebras are finitely generated, so we are restricted to looking
at finite dimensional algebraic sets (as subsets of some finite dimensional
vector space).
\item Affine Algebras can't have nilpotents, which particularly bothered
Grothendieck.
\end{itemize}
The last point excludes things like
\begin{align*}
k[x]/\left\langle x^{n}\right\rangle \ & =\ \{\text{Taylor polynomials of degree }n-1\}.\\
& =\ \{P\in k[x]:P(x)=\sum_{i=0}^{n-1}a_{i}x^{i},\text{with all }a_{i}\in k\}.
\end{align*}
According to Grothendieck, these algebras were meant to be associated
to some sort of a space. For example, $k[x]/\left\langle x^{2}\right\rangle $
is the collection of first order Taylor approximations, and is called
the \uline{first order infinitesimal line}.
\pagebreak{}
Whereas polynomial functions are smooth, looking at an equivalence
class $[f]\in k[x]/\left\langle x^{2}\right\rangle $ allows us to
just look at a point with a slope.
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\par\end{center}
In this manner, nilpotents describe the infinitesimal: ``If $x^{n}=0,$
$x$ may not be zero, bit it is so small that its $n^{\textrm{th}}$
power is zero''. You can actually redo calculus and differential
geometry using infinitesimals, and this approach was taken by \href{http://home.math.au.dk/kock/}{Andres Kock to develop synthetic differential geometry}.
Grothendieck resolved all three problem in one swipe, by
\medskip{}
\begin{myFramedBox}
\begin{defn*}
Let $\text{CommAlg}_{k}$ be the category with \uline{all} commutative
algebras over a field $k$ as objects, and \uline{all} algebraic
homomorphisms as morphisms. Define the category of affine schemes,
by $\text{CommAlg}_{k}^{\textrm{op}},$ with objects called\uline{
affine schemes over $k$}.
\end{defn*}
\end{myFramedBox}
\vspace{-0.3cm}
So Grothendieck invented a new space called an affine scheme, but
we don't really know what they look like. We do, however, know what
the algebra of regular functions (into $k$) looks like - it's a commutative
algebra.
Grothendieck went further, but we won't likely approach schemes in
this course. He was prolific during the 1940's and 50's in France,
a hotspot which produced \emph{Elements of Algebraic Geometry}, a
giant tome that developed much of our modern approach.
\end{document}