%! TEX root = AG2.tex
% vim: tw=50
% 11/10/2024 10AM

% Introductory reading
%   B. Hassett
%   M. Reid
% Commutative Algebra texts
%   Atiyah + MacDonald
%   Matsumura     Commutative Algebra, Commutative
%                                     Ring Theory

% AG texts:
%   Hartshorne, Algebraic Geometry, 1977
%   Gortz-Wedhorn Algebraic Geometry I, 2010
%   Ravi Vakil, The rising sea: Foundations of
%   algebraic geometry

\newpage
\setcounter{section}{-1}
\section{Motivation}

In Part II \courseref{AG} we defined an affine
algebraic variety by:

\begin{definition*}[Affine algebraic variety]
  We fix an algebraically closed field $k$
  and defined affine $n$-space $\Abb^n_k \defeq
  k^n$, and for an ideal $I \le k[x_1, \ldots,
  x_n]$ we defined
  \[ Z(I) \defeq \{(a_1, \ldots, a_n) \in
  \Abb^n \st f(a_1, \ldots, a_n) = 0 ~\forall f
  \in I\} \subseteq \Abb^n_k .\]
  We define a topology on $\Abb^n$ by taking the
  closed sets to be the sets of the form $Z(I)
  \subseteq \Abb^n$.
\end{definition*}

We will introduce \emph{schemes}.

Why schemes?

Why not varieties?

\begin{enumerate}[(1)]
  \item With varieties, we always work with
    algebraically closed fields. For example,
    take $k = \Rbb$, $I = (x^2 + y^2 + 1)
    \subseteq \Rbb[x, y]$. Then $Z(I) =
    \emptyset$.
  \item Number theory? We study Diophantine
    equations, i.e. $I \subseteq \Zbb[x_1, \ldots,
    x_n]$. So $Z(I) \subseteq \Zbb^n$.
  \item Even if $k$ is algebraically closed, we
    lose information when we pass from $I$ to
    $Z(I)$.
    For example, $I = (x^2) \subseteq \Cbb[x]$.
    Then $Z(I) = \{0\} = Z(x)$.

    Recall Hilbert's Nullstellensatz, which states
    that the set of polynomials in $k[x_1, \ldots,
    x_n]$ vanishing on $Z(I)$ for $I \subseteq
    k[x_1, \ldots, x_n]$ is the \emph{radical} of
    $I$.

    But it is natural ideals like $(x^2)$.

    $(y - x^2, y - \alpha) \subseteq \Cbb[x, y]$
    ($\alpha \in \Cbb$).
    \begin{center}
      \includegraphics[width=0.6\linewidth]{images/9deb3be75ff44e0e.png}
    \end{center}
    If $x = 0$, $(x^2 - y, y - \alpha) = (x^2, y)$.
\end{enumerate}

\subsection{Categorical philosophy}

(Read definition on Wikipedia or see
\courseref{CT}).

Let $\mathbf{Sets}$ be the category of sets.

$\mathbf{Sets}$ is the category with objects being
all sets, with morphisms between objects maps of
sets. If $X$ and $Y$ are sets, we write $\Hom(X,
Y)$ for the set of maps between $X$ and $Y$. Note
that there is a bijection
\begin{align*}
  \Hom(\{*\}, X)
  &\to X \\
  (f : \{*\} \to X)
  &\mapsto f(*)
\end{align*}
Let's use this philosophy to understand points on
affine algebraic varieties.

$\Abb_k^0$ is a point. If $X$ is an affine
variety, then the points of $X$ should be in $1-1$
correspondence with $\Hom(\Abb_k^0, X)$. Giving a
morphism between affine varieties is easy. Denote
by $A(X)$ (sometimes $k[X]$) the coordinate ring
($A(X) = k[x_1, \ldots, x_n] / I(X)$, where $I(X)$ is
the ideal of functions that vanish on $X$). This
is a $k$-algebra. We showed that if $X$ and $Y$
are affine varieties, then $\Hom(X, Y) =
\Hom_{\text{$k$-alg}}(A(Y), A(X))$ (see Part II
\courseref{AG} or handout). So
\[ \Hom(\Abb_k^0, X) =
\Hom_{\text{$k$-alg}}(k[x_1, \ldots, x_n] / I(X),
k) .\]
Note that giving a $k$-algebra homomorphism 
\[ k[x_1, \ldots, x_n] \to k \]
can be done by specifying the images of $x_i$ can
be done by specifying the images of $x_i$, say
$x_i \mapsto a_i$ such that for any $f \in I(X)$,
$f(a_1, \ldots, a_n) = 0$. So there is a $1-$
correspondence between such $k$-algebra
homomorphisms and points of $X$.

If $k$ is algebraically closed, the maximal ideals
of $k[x_1, \ldots, x_n]$ are precisely the ideals
of the form $(x_1 - a_1, \ldots, x_n - a_n)$ for
$(a_1, \ldots, a_n) \in \Abb^n$ (a form of Hilbert
Nullstellensatz) and the maximal ideals of $A(X)$
are of the form $(x_1 - a_1, \ldots, x_n - a_n)
\bmod I(X)$ with $(a_1, \ldots, a_n) \in X$. Thus
we have a $1-1$ correspondence between points of
$X$ and maximal ideals of $A(X)$.
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/2b4760156e344021.png}
\end{center}
Now suppose $k$ is \emph{not} algebraically
closed. Let's consider $k$-algebra homomorphisms
\[ k[x_1, \ldots, x_n] / I(X) = A(X) \to L \]
where $L$ is an extension of $k$. These are given
by $x_i \mapsto a_i$ with $f(a_1, \ldots, a_n) =
0$ for all $f \in I(X)$.
Thus
\[
  \Hom_k(A(X), L) = \{(a_1, \ldots, a_n) \in
  \Abb_L^n \st f(a_1, \ldots, a_n) = 0 ~\forall f
  \in I(X)\}
.\]