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!

1.1. 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} I_{1}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.

1.2. 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 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},...,a_{r})
where a_{k} is not a zero divisor or unit in the ring R/(a_{1},...,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},...,a_{r}) such that a_{k} is not a zero divisor or unit
in the ring R/. Thus there is at least one prime ideal that contains
(a_{1},...,a_{k}) that is not a minimal prime for (a_{1},...,a_{k-1}). Let s_{0}(R) denote the maximal
length of a sequence of this kind. Then s(R) __<__ s_{0}(R) __<__ p(R).

1.3. Finite morphisms. By definition the varieties Spec(R) that we are considering are
closed subvarietes of ^{n} for some n; such an inclusion can be identified with a surjective
homomorphism k[X_{1},...,X_{n}] R Now, the dimension of ^{n} 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 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 S be a finite homomorphism of rings and Q_{1} 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 R. Thus f^{-1}(Q_{2}) is distinct from f^{-1}(Q_{1}).

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

What we have shown is that for any finite morphism R S we have p(R) __>__ 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},...,X_{n}] R. We then have
p(R) __<__ p(k[X_{1},...,X_{f(R)}].

Thus, in order to prove that all the notions of dimensions coincide we need to show
that f(R) __<__ 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.

1.4. 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}, ..., a_{n} of
homogeneous elements of positive degree such that a_{k} is not a zero-divisor in the ring
R/ for any k.

We define S(R) to be the maximum length of a sequence a_{1}, ..., 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},...,a_{k-1}) : Irr(R)^{s}) for any s. Equivalently, if R_{k} denotes the ring R/(a_{0},...,a_{k}),
then a_{k} is not a zero-divisor in R_{k-1}/N_{Rk-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},...,X_{n}] R.

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

1.5. Noether Normalisation. Let R be a finitely generated graded ring. Let us assume
that N_{R}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}, ..., z_{k} such that if R_{k} = R/(z_{0},...,z_{k}), then z_{k+1} is not a
zero-divisor in R_{k}/N_{Rk}. Since R is Noetherian we see that this process must
stop; at that stage R_{k} = N_{Rk}, 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},...,Z_{k}]. As a
consequence, we see that F(R) __<__ S(R).

For future reference let us note that we could have started with 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},...,Z_{n}] R is
injective. We will prove this now.

1.6. All definitions are equal. For a graded k-algebra R with R_{0} = k we have
proved

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},..., z_{r} is a maximal sequence in
R as above then k[Z_{0},...,Z_{n}] 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},...,X_{n}] and R = k[X_{0},...,X_{n}]/f. Since any minimal prime that contains (f) is
non-zero, we see that r = P(R) < P(k[X_{0},...,X_{n}]. By the induction hypothesis, we
have a graded inclusion k[Y _{0},...,Y _{r}] 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},...,Y _{r+1}] k[X_{0},...,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 P_{1}P_{s} be a maximal chain of
homogeneous prime ideals in k[X_{0},...,X_{n}] and f a non-zero element of P_{1}. We see that
r = P(R) __>__ s - 1 in this case. It follows that s __<__ n and so P(k[X_{0},...,X_{n}]) __<__ 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},...,Z_{n}] R which is also finite; in particular, all such such sequences have length
equal to the dimension.

1.7. 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_{Rh} = 0. Thus we can find a maximal sequence z_{0} = T, z_{1}, ...z_{n} for R^{h}
which gives a finite inclusion k[Z_{0},...,Z_{n}] R^{h}. The ideal (Z_{0} - 1) is a radical
ideal in the polynomial ring and so (T - 1) k[Z_{0},...,Z_{n}] = (Z_{0} - 1). The
map

is thus a finite inclusion. Thus we have produced a finite inclusion of a polynomial ring into R as required.

1.8. Comparison of dimension for general rings. Now let R be any finitely generated k-algebra. We have already shown that

Further, it is clear that if R^{h} is a homogenisation of R then p(R) __<__ 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},...,X_{n}] __<__ n. On the
other hand we have a natural sequence

of prime ideals in the polynomial ring so p(k[X_{1},...,X_{n}] = n as required. We now only
need to show that f(R) __<__ s(R).

Let a_{1},, ..., a_{n} be a maximal sequence in R as in the definition of s(R); then
R/(a_{1},...,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_{i}T^{deg(ai)} and
note that (z_{1},...,z_{i}) is the homogenisation of the ideal (a_{1},...,a_{i}). It follows that if
R_{k} = R^{h}/(z_{1},...,z_{k}), then R_{k} is the homogenisation of R/(a_{1},...,a_{k}). So N_{Rk} = 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.

1.9. 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},...,Z_{n}] R for some n; it is
enough to should that n = 0. Given finitely many non-constant polynomials f_{1},...f_{k},
the expression 1 + f_{1}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},...,Z_{n}] if n > 0. This proves the
result.

1.10. 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}, ..., 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},...,Z_{r}] 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

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

The ideal Ann(Z_{r}) is a finitely generated module module over k[Z_{0},...,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.