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\newcommand{\bbZ}{{\mathbb Z}}
\newcommand{\bbA}{{\mathbb A}}
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\DeclareMathOperator{\Ann}{Ann}
\DeclareMathOperator{\Fin}{{\bf Fin}}
\DeclareMathOperator{\Irr}{{\bf Irr}}
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\begin{document}

\title[Part V]{Dimension Theory}

\author{Kapil Hari Paranjape}
\maketitle
%\appendix
\section{Dimension Theory}
An affine variety associated with an Artinian ring should be said
to have dimension 0 since there is nothing smaller that we consider
in algebraic geometry. As we go to higher dimensions we shall see
below that there appear to be a number of different ways of approaching
the notion of dimension; luckily for us these all turn out to be the
same!

\subsection{Chains of Prime ideals}
If $\Spec(R)$ has dimension bigger than 0 then it should have a proper
subvariety; moreover we should ignore ``nilpotent-thickenings'' when
looking for subvarieties; for example, we should not consider
$\Spec(k[X,Y]/(Y))$ to be a proper subvariety of $\Spec(k[X,Y]/(Y^2)$.
From the point of view of components in the sense introduced earlier,
this easily translates into the notion of a proper subvariety $\Spec(R/I)$
contained in $\Spec(R)$ as that associated with an ideal $I$ such
that none of the minimal prime ideals containing $I$ are minimal
prime ideals of $R$. (We could also be a bit more restrictive and insist
that no associated prime of $I$ should be an associated prime for $(0)$;
but this works only for finitely generated rings $R$ as we shall see
below.)

A sequence of varieties such that each is properly contained in the
above sense corresponds to a sequence $I_{0}\subset I_{1}\cdots I_{k}$
such that the minimal primes containing $I_{r}$ are distinct from
those that contain $I_{r+1}$. In particular we could take all the
$I_{j}$'s to be prime. The first definition of dimension $p(R)$
of a ring is the maximum of the lengths of strictly monotonic chains
of prime ideals.

\subsection{Complete Intersections}
Any element of an Artinian ring is either a zero divisor or is a unit.
From this one would assume that a ring $R$ has dimension bigger than
0 if and only if we can find a function that is not a zero divisor or a
unit. However, the localisation of $k[X,Y]/(X^{2},XY)$ at $(X,Y)$ has
the property that every element of its maximal ideal is a zero
divisor---but the ring is not Artinian. Fortunately, this is not a ring
of the type we are considering---our rings are quotients of the
polynomial ring in finitely many variables (either over $\bbZ$ or over a
field $k$).  For such a ring $R$ we can define a complete intersection
subvariety as one defined by a sequence $(a_{1},\dots,a_{r})$ where
$a_{k}$ is not a zero divisor or unit in the ring
$R/(a_{1},\dots,a_{k-1})$. The dimension $s(R)$ can then be defined by
considering the maximum over the length of such sequences. 

A relation of this to the previous definition can be understood by
considering the slightly larger class of all sequences $(a_1,\dots,a_r)$
such that $a_k$ is not a zero divisor or unit in the ring
$R/\sqrt{(a_1,\dots,a_{k-1})}$. Thus there is at least one prime ideal
that contains $(a_1,\dots,a_k)$ that is not a minimal prime for
$(a_1,\dots,a_{k-1})$. Let $s_0(R)$ denote the maximal length of a
sequence of this kind. Then $s(R)\leq s_0(R)\leq p(R)$.

\subsection{Finite morphisms}
By definition the varieties $\Spec(R)$ that we are considering are
closed subvarietes of $\bbA^n$ for some $n$; such an inclusion can be
identified with a surjective homomorphism $k[X_1,\dots, X_n]\to R$ Now,
the dimension of $\bbA^n$ {\em ought} to be $n$ so that dimension of $R$
in this case ought to be less than $n$. How does one formalise this
approach to dimension.

First of all we need to expand from surjective homomorphisms to finite
homomorphisms. If $R\to S$ is a finite ring homomorphism then each
``point'' of $\Spec(R)$ has ``finitely many'' points of $\Spec(S)$ above
it the following sense. Let $I$ be any ideal of $R$ such that $R/I$ is
finite, then $S/IS$ is also finite. Thus we can see the dimension of $S$
should be bounded above by the dimension of $R$.

Let $R\to S$ be a finite homomorphism of rings and $Q_1\subset Q_2$ be a
pair of distinct primes in $S$. Replacing $R$ by $R/P_1$ and $S$ by
$S/Q_1$, we have an injective finite homomorphism of domains. The
norm of a non-zero element $a$ of $S$ is a non-zero element of $aS\cap
R$. Thus $f^{-1}(Q_2)$ is distinct from $f^{-1}(Q_1)$.

Now, if $R\to S$ is finite and injective and $I$ is any ideal in $R$ we
have seen that $IS\cap R$ is contained in the radical of $I$. In
particular, if $P$ is prime then $PS\cap R=P$. It follows easily that
there is a (minimal) prime ideal $A$ containing $PS$ such that $Q\cap
S=P$. 

What we have shown is that for any finite morphism $R\to S$ we have
$p(R)\geq p(S)$ and if in addition the morphism is injective then
$p(R)=p(S)$.

We can define the dimension $f(R)$ of a ring $R$ to be the smallest
$n$ such that there is a finite ring homomorphism $k[X_1,\dots,X_n]\to
R$. We then have $p(R)\leq p(k[X_1,\dots,X_{f(R)}]$.

Thus, in order to prove that all the notions of dimensions coincide we
need to show that $f(R)\leq s(R)$ and that the $p$ dimension of the
polynomial ring in $n$ variables is in fact $n$. We will first prove
this by reformulating these notions for homogeneous rings.

\subsection{Homogeneous rings}
Let us restrict our attention to homogeneous $k$-algebras $R$ where
$R_0=k$. In this case the irrelevant ideal $\Irr(R)$ is a maximal ideal
which contains all homogeneous ideals. In the definitions above we
should restrict our attention to homogeneous ideals and homogeneous
elements. We should also ignore the irrelevant ideal(s).

We define $P(R)$ to be the maximum length of a strictily monotic
sequence of homogeneous prime ideals in $R$ none of which is the
irrelevant ideal.

We define $S_0(R)$ to be the maximum length of a sequence $a_0$, \dots,
$a_n$ of homogeneous elements of positive degree such that $a_k$ is not
a zero-divisor in the ring $R/\sqrt{(a_0,\dots,a_{k-1})}$ for any
$k$.

We define $S(R)$ to be the maximum length of a sequence $a_1$, \dots,
$a_n$ of homogeneous elements of positive degree such that for each $k$,
$a_k$ is not a zero-divisor in the ring $R/((a_0,\dots,a_{k-1}):\Irr(R)^s)$
for any $s$. Equivalently, if $R_k$ denotes the ring
$R/(a_0,\dots,a_{k})$, then $a_k$ is not a zero-divisor in
$R_{k-1}/N_{R_{k-1}}$, where $N_R$ denotes the ``small'' irrelevant
ideal of a graded ring $R$.

Finally, we define $F(R)$ to be the smallest $n$ for which there is a
graded finite homomorphism $k[X_0,\dots,X_n]\to R$.

The above collection of inequalities remains valid for these
definitions. We will now establish a relation between $F(R)$ and $S(R)$.

\subsection{Noether Normalisation}
Let $R$ be a finitely generated graded ring.
Let us assume that $N_R\neq R$ or equivalently that there are associated
primes of $R$ that do not contain the irrelevant ideal. Let $P$ be a
prime associated with the irrelevant ideal (this is just the lift to $R$
of an associated prime of $(0)$ in $R_0$). Then any $P$-primary ideal
contains a power of the irrelevant ideal and so $P$ is not contained in
any prime associated with $N_R$. Let $a_P$ be a homogeneous element of
$P$ that does not lie in any associated prime of $N_R$. The product $a$
of the $a_P$'s as $P$ varies over the primes associated with the
irrelevant ideal is not in any associated prime of $N_R$. On the other
hand some power of $a$ lies in the irrelevant ideal. To conclude, there
is a homogeneous element of the irrelevant ideal that is not a
zero-divisor modulo $N_R$.

We can go modulo such an element and iterate the process. This way we
obtain a sequence of elements $z_0$, \dots, $z_k$ such that if
$R_k=R/(z_0,\dots,z_{k})$, then $z_{k+1}$ is not a zero-divisor in
$R_k/N_{R_k}$. Since $R$ is Noetherian we see that this process must
stop; at that stage $R_k=N_{R_k}$, or equivalently, the irrelevant ideal
of $R_k$ is nilpotent.

In particular, at that stage $R_k$ is finite over $R_0$. As we have seen
earlier, this is equivalent to the assertion that $R$ is finite over the
polynomial ring $R_0[Z_0,\dots,Z_{k}]$. As a consequence, we see that
$F(R)\leq S(R)$.

For future reference let us note that we could have started with {\em
any} element $z_0$ of the irrelevant ideal which is not a zero-divisor
in $R/N_R$.

Finally, we need to show that the resulting homomorphism
$k[Z_0,\dots,Z_n]\to R$ is injective. We will prove this now.

\subsection{All definitions are equal}
For a graded $k$-algebra $R$ with $R_0=k$ we have proved
\[
	F(R) \leq S(R) \leq S_0(R) \leq P(R) \leq
			P(k[X_0,\dots,X_{F(R)}]
\]
Thus, all that remains is to prove that the $p$-dimension of a
graded polynomial ring in $n+1$ variables is $n$. We will in addition
prove that if $z_0$,\dots, $z_r$ is a maximal sequence in $R$ as above then
$k[Z_0,\dots,Z_n]\to R$ is injective as well.

This is clear for $n=0$. Let us assume that we have proved the result
for all smaller dimensions. Let $f$ be a non-zero homogeneous element of
positive degree in $k[X_0,\dots,X_n]$ and $R=k[X_0,\dots,X_n]/f$. Since
any minimal prime that contains $(f)$ is non-zero, we see that
$r=P(R)<P(k[X_0,\dots,X_n]$. By the induction hypothesis, we have a
graded inclusion $k[Y_0,\dots,Y_r]\to R$ which is finite. We lift the
images of $Y_i$ to the polynomial ring and add $f$ to this to obtain a
homomorphism $k[Y_0,\dots,Y_{r+1}]\to k[X_0,\dots,X_n]$ which is finite
as well. Any element in the kernel of this must go to 0 in $R$ as well,
but then it must be a multiple of $Y_r$. Since $f$ is not a zero-divisor
in the polynomial ring (being non-zero!) this is impossible. Thus, we
have a finite inclusion of one polynomial ring into another; by the
theory of transcendence we must have $r+1=n$. On the other hand we can
choose any homogeneous $f$ of positive degree. Let $(0\subset P_1 \cdots
P_s$ be a maximal chain of homogeneous prime ideals in
$k[X_0,\dots,X_n]$ and $f$ a non-zero element of $P_1$. We see that
$r=P(R)\geq s-1$ in this case. It follows that $s\leq n$ and so
$P(k[X_0,\dots,X_n])\leq n$ as required.

In the course of the proof we have proved two important facts that we
also note. First of all a single non-zero divisor in $R/N_R$ reduces the
dimension by exactly 1. Secondly, given a maximal sequence as in the
definition of $S(R)$ we obtain a graded inclusion $k[Z_0,\dots,Z_n]\to
R$ which is also finite; in particular, all such such sequences have
length equal to the dimension.

\subsection{Noether Normalisation for general rings}
Let $R$ be a finitely generated ring. Expression $R$ as a quotient of
some polynomial ring we can define $R_n$ as the elements of degree at
most $n$. Let $R^h$ denote the corresponding homogenisation and $T$ be
the element of $R^h$ that represents 1 in $R_1$. As proved earlier $T$
is not a zero divisor in $R_h$ so $N_{R_h}=0$. Thus we can find a
maximal sequence $z_0=T$, $z_1$, \dots $z_n$ for $R^h$ which gives a
finite inclusion $k[Z_0,\dots,Z_n]\to R^h$. The ideal $(Z_0-1)$ is a
radical ideal in the polynomial ring and so $(T-1)\cap
k[Z_0,\dots,Z_n]=(Z_0-1)$.  The map
\[
	k[Z_0,\dots,Z_n]=k[Z_0,\dots,Z_n]/(Z_0-1) \to R= R^h/(T-1)
\]
is thus a finite inclusion. Thus we have produced a finite inclusion of a
polynomial ring into $R$ as required.

\subsection{Comparison of dimension for general rings}
Now let $R$ be any finitely generated $k$-algebra. We have already shown
that 
\[ s(R) \leq s_0(R) \leq p(R) \leq p(k[X_1,\dots,X_{f(R)}] \]
Further, it is clear that if $R^h$ is a homogenisation of $R$ then
$p(R)\leq P(R^h)$ since a sequence of prime ideals in $P$ gives a
sequence of homogeneous prime ideals in $R^h$ which are all distinct
from the irrelevant ideal. Thus, we see that $p(k[X_1,\dots,X_n]\leq n$.
On the other hand we have a natural sequence 
\[ (0)\subset (X_1)\subset \dots\subset (X_1,\dots,X_n) \] 
of prime ideals in the polynomial ring so $p(k[X_1,\dots,X_n]=n$ as
required. We now only need to show that $f(R)\leq s(R)$.

Let $a_1,$, \dots, $a_n$ be a maximal sequence in $R$ as in the
definition of $s(R)$; then $R/(a_1,\dots,a_n)$ consists of units and
zero divisors. Let $R^n$ be the homogenisation of $R$ and $T$ be the
element that corresponds to the element 1 in $R_1$. We put
$z_i=a_iT^{\deg(a_i)}$ and note that $(z_1,\dots,z_i)$ is the
homogenisation of the ideal $(a_1,\dots,a_i)$. It follows that if
$R_k=R^h/(z_1,\dots,z_k)$, then $R_k$ is the homogenisation of
$R/(a_1,\dots,a_k)$. So $N_{R_k}=0$, $z_{k+1}$ is not a zero divisor in
this ring and $T$ is not a zero-divisor in $R_n$.  In order to prove
that $s(R)=S(R^h)$ it then suffices to show that $R_n/T$ is finite.
This is what we prove next.

\subsection{Hilbert's Nullstellensatz}
Let $R$ be a finitely generated $k$-algebra which consists only of units
and zero-divisors. We claim that $R$ is finite.

Any non-unit lies in an associated prime of $(0)$ in $R$; let $N$ denote
the collection of maximal elements among these. If $M$ is a maximal
ideal of $R$ which is not in this collection, then we can find an
element of $M$ which is not in any of the elements of $N$. Such an
element can not be a zero-divisor or a unit which contradicts the
hypothesis. It follows that $R$ has finitely many maximal ideals.

We have seen above that there is a finite inclusion $k[Z_1,\dots,Z_n]\to
R$ for some $n$; it is enough to should that $n=0$. Given finitely many
non-constant polynomials $f_1$,\dots $f_k$, the expression
$1+f_1\cdots f_k$ gives a non-constant polynomial which does not lie in
any of the ideals generated by a subset of the $f_i$'s unless that
sub-ideal is the full ring. Thus, there are infinitely many maximal
ideals in $k[Z_1]$; it follows that there are infinitely many maximal
ideals in $k[Z_1,\dots,Z_n]$ if $n>0$. This proves the result.

\subsection{The Hilbert function}
For a finitely generated graded ring $R$ with $R_0$ a finite ring, the
Hilbert function $h_R$ is defined by setting $h_R(n)$ to be the length
of the $R_n$ as an $R_0$ module. Let us concern ourselves with the case
when $R_0$ is a field $k$; the other case is simlar. We claim that there
is a polynomial $P_R(T)$ such that $P_R(n)=h_R(n)$ for all $n$
sufficiently large; moreover, the degree of this polynomial is the
graded dimenion of $R$.

Let $z_0$, \dots, $z_r$ be a minimal system of homogeneous generators 
for a graded $k$-algebra $R$ as considered above. From
what we have seen above, the resulting homomorphism $k[Z_0,\dots,Z_r]\to
R$ is finite and the dimension of $R$ is also $r$. It follows that the
homomorphism is also injective.

We prove our assertion by induction on the dimension of $R$. When $R$ is
finite the assertion is clear since $h_R(n)=0$ for $n$ sufficiently
large. Now suppose that $Z_r$ has degree $d$ and let $S=R/(Z_r)$. We see that
\[	\dim(R_n) = \dim(Z_rR_{n-d})+ \dim(S_n) \]
By induction we have a polynomial $P_S(T)$ of degree $r-1$ such that $\dim(S_n)=P_S(n)$
for all sufficently large $n$. We also have
\[	\dim(Z_rR_{n-d}) = \dim(R_{n-d}) - \dim (\Ann(Z_r)_{n-d}) \]
The ideal $\Ann(Z_r)$  is a finitely generated module module over
$k[Z_0,\dots,Z_{r-1}]$ and so by induction $\dim(\Ann(Z_r)_{n-d})$ is
represented by a polynomial of degree at most $r-1$. It follows
that we obtain (for all $n$ sufficiently large)  an identity
$h_R(n)-h_R(n-d) = Q_r(n)$ where $Q_r(T)$ is a polynomial of degree $r$.
By ``discrete integration'' we observe that this implies the result.

\end{document}