\documentclass{amsart}
\newenvironment{emquote}{\begin{quote}\em}{\end{quote}}
\newcommand{\superset}{\supset}
\newcommand{\bbZ}{{\mathbb Z}}
\newcommand{\bbA}{{\mathbb A}}
\newcommand{\bone}{{\mathbf 1}}
\newcommand{\tensor}{\otimes}
\DeclareMathOperator{\Spec}{Spec}
\DeclareMathOperator{\Hom}{Hom}
\DeclareMathOperator{\Fin}{{\bf Fin}}
\usepackage{hyperref}
\begin{document}
\title[Part II]{Affine Schemes}
\author{Kapil Hari Paranjape}
\maketitle
\section{Two Descriptions of the Category of Affine Schemes}
Affine algebraic geometry could be described as the study of solution of a
finite system of polynomial equations in a (finite) number of variables.
The objects of the category are then systems of equations of the form
\begin{eqnarray*}
f_1(X_1,\dots,X_p) &=& 0 \\
\dots \\
f_q(X_1,\dots,X_p) &=& 0
\end{eqnarray*}
A morphism from the system to another system should be a way of
transforming the solutions of one set of equations into solutions of the
other set of equations. Given two systems
$(f_1(X_1,\dots,X_p),\dots,f_q(X_1,\dots,X_p))$ and
$(g_1(Y_1,\dots,Y_r),\dots,g_s(Y_1,\dots,Y_r))$, a morphism from the first
system to the second system is a collection of polynomials
$(h_1(X_1,\dots,X_p),\dots,h_r(X_1,\dots,X_p))$ such that ``whenever''
$(a_1,\dots,a_p)$ satisfies the first system
\[
(b_1,\dots,b_r)= (h_1(a_1,\dots,a_p),\dots,h_r(a_1,\dots,a_p))
\]
gives a solution of the second system. Moreover, two such morphisms
$(h_1(X_1,\dots,X_p),\dots,h_r(X_1,\dots,X_p))$ and
$({h'}_1(X_1,\dots,X_p),\dots,{h'}_r(X_1,\dots,X_p))$ are equal if the
values are equal, i.~e.\
\[
(h_1(a_1,\dots,a_p),\dots,h_r(a_1,\dots,a_p))
= ({h'}_1(a_1,\dots,a_p),\dots,{h'}_r(a_1,\dots,a_p))
\]
``whenever'' $(a_1,\dots,a_p)$ satisfy the first system of equations
$(f_1,\dots, f_q)$.
The crucial step is to understand the meaning of ``whenever''. After
all, if we only look at solution in integers or rational numbers or some
finite field, then there may not be ``enough'' solutions to characterise
the system. Thus there are three possible interpretations of this
pseudo-definition.
\subsection{Ring theoretic definition}
In this case we assume that the interpretation of ``whenever'' is in the
formal algebraic sense by ``substitution''.
Let $R$ be the ring
\[
R = \bbZ[X_1,\dots,X_p]/(f_1(X_1,\dots,X_p),\dots,f_q(X_1,\dots,X_p))
\]
where we are examining equations with coefficients in integers or {\em
Diophantine equations}; when we are looking at equations with
coefficients in a field $k$ we can replace $\bbZ$ by $k$. Clearly, a
soltution {\em in any ring (resp. $k$-algebra) $S$} of the system of
equations can be interpreted as a ring homomorphism (resp. $k$-algebra
homomorphism) $R\to S$. Similarly,
\[
R' = \bbZ[Y_1,\dots,Y_r]/(g_1(Y_1,\dots,Y_r),\dots,g_s(Y_1,\dots,Y_r))
\]
A ``substitution'' $(h_1(X_1,\dots,X_p),\dots,h_r(X_1,\dots,X_p))$ is
a ring homomorphism
\[
\bbZ[Y_1,\dots, Y_r] \to \bbZ[X_1,\dots, X_p]
\]
The condition that it gives a morphism from the first system of
solutions to the other says that $g_i(h_1,\dots,h_r)$ should become zero
in the quotient $R$ of $\bbZ[X_1,\dots, X_p]$. In other words, this
substition corresponds to a homomorphism of rings $R'\to R$. Moreover,
equality of $(h_1,\dots, h_r)$ with $({h'}_1,\dots, {h'}_r)$ corresponds
to equality of the resulting homomorphism $R'\to R$.
The algebraic definition of the category of Affine Schemes (of finite
type over $\bbZ$ or a field $k$) is thus the opposite category of the
category of rings of the above type; note that the homomorphism of rings
goes in the opposite direction! Just to avoid confusion the term
$\Spec(R)$ is used when a ring is considered as an affine scheme.
\subsection{An arithmetic-geometric approach}
The above approach is mildly dis-satisfying to the number theorist or
the geometer since it does not seem to involve the ``solutions'' of the
equations in any way! The classical approach of looking at solutions in
algebraic numbers or complex numbers is also unsatisfactory since it
does not distinguish between the equation $X^2=0$ and $X=0$ which {\em
must} be distinguished for a number of reasons. Another problem with the
above approach is that it does not easily generalise to more general
(i.~e.\ non-affine) algebraic schemes.
One solution is to look at solutions in finite rings (for equations with
coefficients in $\bbZ$) or finite dimensional $k$-algebras (for
equations with coefficients in a field $k$). The latter has an
interesting interpretation as rings of commuting matrices. Let $\Fin$
denote this category of rings or $k$-algebras as the case may be.
Any system $(f_1(X_1,\dots,X_p),\dots,f_q(X_1,\dots,X_p))$ of
equations as above gives a functor from $\Fin$ to the category of sets
which takes a finite ring $A$ to the set $V(f_1,\dots,f_q)(A)$ of
solutions of the equations with $(a_1,\dots,a_p)$ with the $a_i$ in $A$.
Let $V(f_1,\dots,f_q)$ denote this functor.
Let $\bbA^k$ denote the functor from $\Fin$ to sets that assigns to
each finite ring $A$ the set of {\em all} $k$-tuples $(a_1,\dots,a_k)$
with entries in $A$; thus $\bbA^k$ corresponds to $0$ equations in $k$
variables! Clearly $V(f_1,\dots,f_q)$ is a sub-functor of $\bbA^p$ in a
natural way.
A system of polynomials $(h_1,\dots,h_r)$ as above is said to give a
morphism $V(f_1,\dots,f_q)\to V(g_1,\dots,g_s)$ if ``whenever'' in the
above pseudo-definition is interpreted with solutions in finite rings in
$A$. The polynomials $h_i$ define a natural transformation
$\bbA^p\to\bbA^r$ and we have a morphism $V(f_1,\dots,f_q)(A) \to
V(g_1,\dots,g_s)$ if this natural transformation takes
$V(f_1,\dots,f_q)(A)$ to $V(g_1,\dots,g_s)(A)$ for every ring $A$ in
$\Fin$. Two morphisms $(h_1,\dots,h_r)$ and $({h'}_1,\dots,{h'}_r)$ are
equal if the induced maps on $V(f_1,\dots,f_q)(A)$ are equal; note that
this {\em need not} imply that the maps are equal on all tuples
$(a_1,\dots,a_p)$.
\subsection{The equivalence}
If $R$ is the ring associated with the system $(f_1,\dots,f_q)$ then it
is clear that $V(f_1,\dots,f_q)(A)$ can be identified with the ring
homomorphisms $R\to A$. In particular it follows that a morphism
$\Spec(R)\to\Spec(R')$, which can be identified with a homomorphism
$R'\to R$, gives rise to a set map $\Hom(R,A)\to\Hom(R',A)$. There is
thus a natural functor from the category of $\Spec(R)$'s to the
category of $V(f_1,\dots,f_q)$'s. We need to prove that this is an
isomorphism of categories.
Since the objects of the two categories correspond to each other in a
natural way, this proof reduces to checking the identification of
the morphisms. On the one hand we need to see that given two distinct
homomorphisms $R'\to R$ the induced maps $\Hom(R,A)\to\Hom(R',A)$ are
different for at least one ring $A$ in $\Fin$. On the other hand we need
to show that if $(h_1,\dots,h_r)$ gives a morphism from
$V(f_1,\dots,f_q)$ to $V(g_1,\dots,g_s)$ then it gives a homomorphism
$R'\to R$.
Now let us assume the following result which will prove later. For any
ring $R$ of the type considered above the kernels, of all homomorphisms
from $R$ to a ring $A$ in $\Fin$, have zero as their common
interesection; i.~e.\
\[ \bigcap_{f:R\to A} \ker(f) = 0 \]
Assuming this we will be able to demonstrate what we want.
Let $h=(h_1,\dots, h_r)$ be a collection of polynomials. This induces a
homomorphism $\bbZ[Y_1,\dots,Y_r]\to\bbZ[X_1,\dots,X_p]$ (or with $\bbZ$
replaced by $k$ in case we are in the situation of $k$-algebras). To get
a homomorphism $R'\to R$ we need the condition that the ideal
$g_1,\dots, g_s)$ generated by the $g_j$'s is sent {\em into} the ideal
$(f_1,\dots, f_q)$ generated by the $f_i$'s; let $I_h$ denote the image
ideal. The condition that we have a morphism $V(f_1,\dots,f_q)\to
V(g_1,\dots,g_s)$ is the condition that every homomorphism $R\to A$
vanishes on $I_h$. By the result we have assumed this means that the
image of $I_h$ in $R$ is zero; in other words $I_h$ is contained in the
ideal $(f_1,\dots, f_q)$ as required to obtain a homomorphism $R'\to R$.
If $h$ and $h'$ induce distinct homomorphisms $R'\to R$, then there is at
least one $i$ such that the image of $h_i$ in $R$ is different from the
image of ${h'}_i$ in $R$. Consider the non-zero element ${h'}_i-h_i$ of
the ring $R$. By the result assumed there is a homomorphism ring $A$ in
$\Fin$ and a homomorphism $R\to A$ such that the image of this element
in $A$ is non-zero. In other words, the images of $h_i$ and ${h'}_i$ in
$A$ are distinct. Thus the composite induced homomorphisms from $R'$ to
$A$ are distinct; hence the induced maps $\Hom(R,A)\to\Hom(R',A)$ are
distinct as required.
\subsection{Closed morphisms}
A special class of morphisms of affine schemes is
$\Spec(R/I)\to\Spec(R)$ where $R\to R/I$ is the quotient by an ideal.
This is called a closed subscheme; the ``sub'' part is because
$\Spec(R/I)(A)\subset\Spec(R)(A)$ consists of those homomorphisms that
vanish on $I$.
If $I$ and $J$ are ideals in the ring $R$, then $I+J$ is also an
ideal. Clearly, an element of $\Spec(R)(A)$ that lies in $\Spec(R/I+J)(A)$
lies in $\Spec(R/I)(A)$; similarly for $J$. Thus the intersection of two
closed subschemes is also a closed subscheme.
More generally we can take the sum of any collection of ideals of $R$ to
show that the intersection of any collection of closed subschemes is a
closed subscheme.
The union of closed subschemes is a more tricky affair. The
union of $\Spec(R/I)(A)$ and $\Spec(R/J)(A)$ (as subsets of
$\Spec(R)(A)$) is not associated with an ideal in general. The natural
ideal to consider is $I\cap J$ but there may be more elements in
$\Spec(R/I\cap J)(A)$ then those coming from $\Spec(R/I)(A)$ and
$\Spec(R/J)(A)$; the whole is greater than the sum of the parts!
However, if we restrict $A$ to be a field then the kernel of $R\to A$ is
a maximal ideal (hence prime) and thus if it contains $I\cap J$ then it
must contain either $I$ or $J$. Alternatively we could look for the
smallest closed subscheme (equivalently, the largest ideal $K$) such
that it contains the union for all $A$; in other words, what is the
largest ideal $K$ which lies in the kernel of all homomorphisms $R\to A$
that vanish on either $I$ or $J$; we will later show that this is the
ideal $I\cap J$.
With these two definitions we see that closed subschemes satisfy the
usual properties of closed-ness.
\subsection{Images and inverse images}
Let $f:R\to R'$ be a homomorphism of rings and $I$ an ideal in $R$.
Given a ring homomorphism $g:R'\to A$ such that the
composite homomorphism $R\to R'\to A$ vanishes on $I$, we see that
$g(f(I)S)=0$. Conversely, any homomorphism $g$ with the latter property
gives a homomorphism $g\circ f:R\to A$ that vanishes on $I$.
Thus the inverse image of $\Spec(R/I)(A)$ under
$\Spec(R')(A)\to\Spec(R)(A)$ is $\Spec(R'/(f(I)S))(A)$.
Given any homomorphism $f:R\to R'$ and a closed subscheme $\Spec(R'/J)$
of $\Spec(R')$ we could ask for its ``image'' under
$\Spec(R')\to\Spec(R)$. As above this is a bit tricky.
One possible definition would be to take $I=f^{-1}(J)$ and call
$\Spec(R/I)$ its image. However, this need not always be the image in
the ``functorial'' sense. In other words, for a finite ring $A$ the
image of the composite $\Spec(R'/J)(A)\to\Spec(R')(A)\to\Spec(R)(A)$
need not be $\Spec(R/I)(A)$. In fact, $\Spec(R')(A)\to\Spec(R)(A)$ need
not be surjective. We thus have the following weaker notion which is
adequate.
Given a non-zero morphism $R/I\to A$ with $A$ finite, we ask for a
non-zero ring homomorphism $A\to B$ such that the composite $R/I\to A\to
B$ also factors as $R/I\to R'/J\to B$. In this case we can call
$\Spec(R/I)$ the ``true'' image of $\Spec(R)$. In case the true image of
$\Spec(R'/J)$ is $\Spec(R/f^{-1}(J))$ for all ideals $J$ in $\Spec(R')$
we say that the morphism $\Spec(R')\to\Spec(R)$ is {\em closed}.
\subsection{Finite morphisms}
A special class of morphisms of affine schemes where ``image'' is better
behaved is given by $f:R\to S$ where $S$ becomes a finitely generated
$R$-module under $f$; such a morphism is called ``finite''.
Let $x_1$, \dots, $x_r$ be system of generators of $S$ as an $R$ module.
By using this as a ``basis'', to every alement $a$ of $S$ we can an
associate an $r\times r$ matrix $A$ with entries $a_{ij}$ in $R$ by the
formula \[ a\cdot x_i = \sum_j a_{ij} x_j \] Let $B$ be the adjoint of
the matrix $A-a\bone_r$ so that $B(A-a\bone_r)=\det(A-a\bone_r)\cdot
\bone_r$. We see that $\det(A-a\bone_r)x_i=0$ for all $i$. Thus it is
the 0 element of $S$. The polynomial $\det(A-T\bone_r)$ is a polynomial
of degree $r$ over $R$ with leading coefficient a unit. In other words
it is a monic polynomial. Thus we have seen that each element of $S$
satisfies a monic polynomial over $R$; such an element is called an {\em
integral} element.
Conversely, if $R\to S$ is a ring homomorphism and $x$ in $S$ is
integral over $R$, we see that $R[x]\subset S$ is a quotient of the
finitely generated $R$-module $R[T]/(f(T))$ for some monic polynomial
$f$ satisfied by $x$. It is clear that finite-ness is transitive so we
see that any finitely generated $R$-algebra $S$ (all algebras in our
current examples) such that every element is integral over $R$ (such a
ring is called an integral extension of $R$) is also finite over $R$.
Now, if $J$ is an ideal in $S$ then $R/f^{-1}(J)\to S/J$ is also finite;
in addition it is injective. We can now show that it is a ``true'' image
in the sense given above. We can replace $R$ by $R/f^{-1}(J)$ and $S$ by
$S/J$ so we have an injective finite morphism $R\to S$. Given a non-zero
homomorphism $R\to A$ with $A$ finite we want to show that there is a
non-zero ``lift'' as above. Let $Q$ be the kernel of $R\to A$; it is
enough to show that $S/f(Q)S$ is non-zero and finite for we can then
take $B$ to be $A\tensor_{R/Q} S/f(Q)S$. The lemma that does this is
another guise of Nakayama's lemma. Note that $S/f(Q)S$ is finite over
the finite ring $R/Q$ and hence it is itself finite. Thus we only need
to show that it is non-zero.
We will prove the slightly more general result that follows. Let $f:R\to
S$ be injective and finite as above. Let $I$ be any ideal in $R$, then
we claim that $IS\cap R$ is contained in $\sqrt{I}$, the radical of $I$.
Let $x_1$,\dots, $x_r$ be a set of generators of $S$ as an $R$ module.
If $b$ lies in $f(I)\cap R$ then we have expressions $bx_i=\sum_j f(a_{ij}) x_j$ with
$a_{ij}$ in $I$. But then multiplying on the left with the adjoint of
the matrix $b\bone_r - (a_{ij})$ we see that $\det(b\bone_r - (a_{ij}))$
acts as 0 on all the $x_i$. In other words $b$ satisfies the
characteristic polynomial $P(T)$ of the matrix $(a_{ij})$, which is a
monic polynomial whose coefficients, as polynomials in the $a_{ij}$, lie
in $I$. This means that $b^n$ lies in $I$ or that $b$ lies in $\sqrt{I}$
as required.
\subsection{Hilbert's basis theorem}
We started with a finite system of equations but then ``generalised'' to
an arbitrary ideal. Is this truly a generalisation? Hilbert proved
otherwise. In other words we claim that any ideal in the polynomial ring
is finitely generated; in other words that the polynomial ring is
Noetherian. The proof proceeds by induction. It is obvious
that a field $k$ (with only two ideals) is Noetherian; the Noetherian
property for $\bbZ$ follows from Euclid's algorithm. Let $R$ be a ring
which is Noetherian, we claim that $R[X]$ is also Noetherian.
Before starting, let us note that if $R\to S$ is finite then $S$ is also
Noetherian. This is because it is already Noetherian as an $R$ module.
The general case is modelled on this.
Let $I$ be an ideal in $R[X]$. We want to show that this ideal is
finitely generated.
If $I$ contains a polynomial $f(X)$ whose leading coefficient is a unit
(a monic polynomial), then we note that $R[X]/(f(X))$ is isomorphic to
$R^{\deg(f)}$ as an $R$ module; thus the problem of finite generation of
$I/(f(X))$ is reduced to the Noetherian-ness of finitely generated
modules over $R$.
More generally, we proceed as follows. Let $I[X^{-1}]$ denote the ideal
generated by $I$ in the ring $R[X,X^{-1}]$; let $I_{\infty}$ be the
intersection of $I[X^{-1}]$ with the subring $R[X^{-1}]$ of
$R[X,X^{-1}]$; then the {\em content} $c(I)$ of the ideal $I$ is the
image of $I_{\infty}$ in $R=R[X^{-1}]/(X^{-1})$; it is also the ideal
consisting of the leading coefficients of elements of $I$. Since $c(I)$
is an ideal in $R$ it is finitely generated. Let $f_1$, \dots, $f_r$ be
elements of $I$ whose leading coefficients generate $c(I)$; we can
assume that all the degrees $\deg(f_i)$ are equal to some fixed positive
integer $k$ by replacing $f_i$ by a multiple of it with some power of
$X$ if necessary. If $g$ is an element of $I$ of degree at least $k$ and
$a$ is its leading coefficient then, by writing $a$ as an $R$-linear
combination of the generators of $c(I)$, we can find a linear
combination of the form $\sum a_i f_i(X) X^l$ which has the same
leading coefficient as $g$; by subtracting this element from $g$ we
obtain an element of $I$ of small degree. Hence any element of $I$ can
be written as a linear combination of the $f_i$ and an element of $I$
whose degree is less than $k$. As before the elements of $I$ whose
degree is less than $k$ form a Noetherian module over $R$; we can
therefore find a finite $R$-basis for this collection of elements. This
$R$-basis combined with the polynomials $f_1$, \dots, $f_r$ then
generate the ideal $I$.
\end{document}