Next: Bibliography
Up: Some Lectures on Number
Previous: A..1 Counterexample
In order to compare the given definition of schemes with the
``classical'' one, we will prove the following theorem:
Theorem 18
Let
f :
RS be a homomorphism between finitely generated rings so
that for every finite ring
A, the induced map
Hom(
S,
A)
Hom(
R,
A) is a bijection. Then
f is an isomorphism.
In the paragraphs below R and S will always denote rings
satisfying the conditions of the theorem. We first prove a special
case:
Lemma 19
Let
f :
RS be a homomorphism of finite rings so that for every
finite ring
A the induced map
Hom(
S,
A)
Hom(
R,
A) is a
bijection. Then
f is an isomorphism.
Proof.
Taking
A =
R we see that there is a homomorphism
g :
SR such
that the composite
gof :
RSR is identity. For any
finite ring
A, consider the chain of maps
Hom(
R,
A)
Hom(
S,
A)
Hom(
R,
A)
The second map is a bijection by assumption. The composite is the
identity and in particular, a bijection. It follows that
Hom(
R,
A)
Hom(
S,
A) is a bijection as well. Now, taking
A =
S we
see that we also have a homomorphism
h :
RS so that the
composite homomorphism
SgRhS is the identity. We then
have
f = id_{S}of = hogof = hoid_{R} = h
Thus
fog = id
_{S} and
gof = id
_{R}, hence
f and
g are
isomorphisms.
Next, we show that the above condition is ``inherited'' by quotients.
Lemma 20
Let
f :
RS be as above. Let
I be an ideal in
R, then we
obtain a homomorphism
R/
IS/
f (
I)
S. For any finite ring
A, the
induced map
Hom(
S/
f (
I)
S,
A)
Hom(
R/
I,
A) is a bijection.
Proof.
Consider the diagram
Hom(S, A) 

Hom(R, A) 



Hom(S/f (I)S, A) 

Hom(R/I, A) 
The top row is a bijection. Let
g :
S/
f (
I)
SA be any element in
the bottom left corner then the corresponding element
h :
SA in
the top left corner satisfies
h(
f (
I)
S) = 0. Thus
hof :
RA
satisfies
hof (
I) = 0. Thus it factors through a homomorphism
e :
R/
IA. Thus we see that the elements in the bottom left
corner are mapped to elements in the bottom right corner.
Conversely, let
g :
R/
IA be an element in the bottom right
corner and
h :
RA be its image in the top right corner; then
h(
I) = 0. By assumption there is a homomorphism
e :
SA such that
h =
eof. It follows
e(
f (
I)) = 0 and thus
e(
f (
I)
S) = 0. Thus
e
factors through an element
d :
S/
f (
I)
SA in the bottom left
corner. In other words we have a bijection
Hom(
S/
f (
I)
S,
A)
Hom(
R/
I,
A).
Combining the above two lemmas we see that if I is any ideal in
R such that R/I and S/f (I)S are finite, then the map
R/IS/f (I)S is an isomorphism. We will now show that if R/I is finite
then S/f (I)S is ``automatically'' finite as well.
Lemma 21
Let
f :
RS be as in the theorem. For any maximal ideal
m in
R, the ideal
f (
m)
S in
S also a maximal ideal.
Proof.
Since
R is finitely generated
R/
m is a finite field by Hilbert's
Nullstellensatz. Thus
Hom(
S,
R/
m)
Hom(
R,
R/
m) is a bijection and
so the homomorphism
RR/
m must factor through
S; moreover,
this factorisation is unique. Let
n be the kernel of this
factorisation. Then
n is a maximal ideal containing
f (
m)
S such
that
R/
mS/
n is an isomorphism. Now, let
n' be any maximal
ideal in
S containing
f (
m)
S. Then, the composite
RSS/
n' factors through
R/
m. Thus,
S/
n' is a finite field
extension of
R/
m. If this extension has degree > 1 then if
q is
the cardinality of
R/
m, the map
x x^{q} is a nontrivial
automorphism of
S/
n' which is identity on
R/
m. Thus we obtain
two maps
SS/
n' which restrict to the same map
RS/
n'
contradicting the hypothesis. Thus
R/
mS/
n' is an isomorphism.
But then this isomorphism gives a map
SR/
m which restricts to
the natural map
RR/
m; there is a unique such map by
hypothesis. Since that map has kernel
n, we see that
n' =
n.
In other words, we see that f (m)S is contained in a unique maximal
ideal n in S. Thus S/f (m)S is an Artinian ring. By the earlier
discussion we see that
R/mS/f (m)S is an isomorphism. In other
words f (m)S = n is a maximal ideal for every maximal ideal m in
R. Conversely, if n is any maximal ideal in S, then
f^{1}(n) = m is the kernel of the composite
RSS/n which
is a map to a finite field; hence m is a maximal ideal. It follows
that every maximal ideal in S is of the form f (m)S for a
maximal ideal m in R.
Now, if I is any ideal such that R/I is finite then there are
finitely many maximal ideals m_{1}, ..., m_{k} and positive integers
r_{1}, ..., r_{k} such that
I m_{1}^{r1 . }m_{2}^{r2 ... }m_{k}^{rk}. As seen above
n_{i} = f (m_{i})S is a
maximal ideal. The relations
f (
I)
S f (
m_{1}^{r1 ... }m_{k}^{rk})
S =
n_{1}^{r1 ... }n_{k}^{rk}
shows that the ring S/f (I)S is finite as well. It follows that for
any ideal I such that R/I is finite, the map
R/IS/f (I)S is
an isomorphism.
On the other hand suppose J is any ideal in S such that S/J is
finite and let
I = f^{1}(J); then R/I is a subring of S/J and
thus also finite. We have seen above that this implies that
R/IS/f (I)S is an isomorphism. But the inverse image of J/f (I)S under
this is the zero ideal in R/I. Thus we have J = f (I)S. To
summarise,
Lemma 22
Let
f :
RS be as in the conditions of the theorem. The map
I f (
I)
S is a oneone correspondence between ideals of
finite index in
R and ideals of finite index in
S. The map
J f^{1}(
J) is the inverse correspondence from ideals
J
in
S to ideals in
R. Moreover, the natural homomorphism
R/
IS/
f (
I)
S is an isomorphism for such ideals.
Thus the original condition has been restated intrinsically in terms
of ideals. Next we wish to prove that the given homomorphism is
``closed''. That is to say given a prime ideal Q in S, let
m be a maximal ideal in R that contains the prime ideal
P = f^{1}(Q). We wish to prove that there is a maximal
ideal n in S which contains Q and satisfies
f^{1}(n) = m.
To do this we can restrict our attention to
R/PS/f (Q)S. Since
f^{1}(f (P)S) f^{1}(Q) = P, the latter homomorphism is also injective.
Lemma 23
Let
f :
RS be an injective homomorphism of finitely generated
rings with
R a domain. We have a factoring of
f as follows
RR[
X_{1},...,
X_{a}] =
R_{1}R_{1}[
t_{1},...,
t_{b}] =
R_{2}S
where
 R_{1} is a polynomial ring over R.
 There is a nonzero element r of R_{1} such that for each
i, the element
rt_{i} R_{2} satisfies a monic polynomial over
R_{1}. Other than this relation there are no further relations
among the t_{i} in R_{2}.
 R_{2}S is the quotient by an ideal that intersects R_{1}
in the zero ideal.
Proof.
Since
S is finitely generated we can choose a maximal collection
of elements
X_{1}, ...,
X_{a} of
S that are algebraically
independent over (the quotient field of)
R. Then
R_{1} =
R[
X_{1},...,
X_{a}] is the polynomial ring over
R and is a
subring of
S. The remaining generators of
S are algebraically
dependent on the
X_{i}'s. Thus each of them satisfies an equation of
the form
r_{0}T^{d} +
r_{1}T^{d  1} + ... +
r_{d} for some elements
r_{j} in
R_{1}. Moreover, we can assume that
r_{0} is nonzero in such an
equation. Let
r be the product in
R_{1} of polynomials
r_{0}
corresponding to different generators of
S. Since
R is a
domain, so is
R_{1} and the polynomial
r is nonzero. For each
generator
S choose a polynomial of the above form with leading
coefficient
r (one such such clearly exists) and let
R_{2} be the
ring obtained from
R_{1} by adjoining the roots of these equations.
We have a natural map
R_{2}S; let
be the kernel. Since
R_{1}S factors through
R_{2} and is injective, it follows that
intersects
R_{1} in the zero ideal.
Let Q_{1}, ..., Q_{r} be the minimal primes in S or equivalently a
minimal primes in R_{2} that contains the kernel of R_{2}S. Since
R_{1} meets this kernel in the zero ideal, the intersection of the
prime ideals
Q_{i} R_{1} in R_{1} is a nilpotent ideal. Since R_{1}
is a domain there is an index i such that
Q_{i} R_{1} = (0). Let Q
denote the prime ideal Q_{i} for any such index i.
Let m be a maximal ideal in R such that r is not contained in
the prime ideal
m[X_{1},..., X_{a}] of R_{1}. Since R_{1} is a domain
we see that
Q_{r} (R_{1})_{r} is the zero ideal. Now, (R_{2})_{r} is a
finite free module over (R_{1})_{r} and so (by the going up theorem)
there is a prime ideal Q' in R_{2} which contains Q and restricts
to
m[X_{1},..., X_{a}] in R_{1}. Similarly, for any maximal ideal n'
in R_{1} that contains
m[X_{1},..., X_{a}] and does not contain r,
there is a maximal ideal n in R_{2} that contains Q' (and hence
Q) that lies over n'.
Now, if a > 0 (i. e. R R_{1}) then there are at least two (in
fact infinitely many) such maximal ideals n'. But then we see that
we have at least two maximal ideals in S that lie over a given
maximal ideal m in R contradicting lemma 22. Thus we must
have R = R_{1}.
Again, if is another minimal prime in R_{2} that contains
the kernel of R_{2}S and such that
R_{1} = (0), then
as above we can find a prime ideal
which contains
and lies over m and is distinct from Q'. Now there are
distinct maximal ideals n' and
in R_{2}, that contain
Q' and
respectively. This again contradicts
lemma 22. It follows that there is a unique minimal
prime Q containing the kernel of R_{2}S such that
Q R = (0).
Now suppose that Q_{0} is another miminal prime in S, or
equivalently a minimal prime in R_{2} that contains the kernel of
R_{2}S. We must have
Q_{0} R (0). However, we have the
lemma
Lemma 24
Let
f :
RS be a homomorphism of finitely generated rings with
R a domain. Let
Q be a minimal prime in
S such that
f^{1}(
Q) is nonzero. Then there is a maximal ideal
n in
S
and an integer
k such that if
m =
f^{1}(
n), then
R/
m^{k}S/
n^{k}
is not an isomorphism.
Proof.
Let
x be an element of all the minimal primes of
S other than
Q. Replacing
S by its localisation
S_{x} at
x, we can
assume that
Q is the unique minimal prime in
S. Then
Q consists of nilpotent elements. Since
f^{1}(
Q) is
nonzero and
R is a domain it follows that
RS has a nonzero
kernel. Now let
n be
any maximal ideal in
S and
m =
f^{1}(
m). The homomorphism of local rings
R_{m}S_{n} has a
nonzero kernel. The result follows by the ArtinRees lemma.
On the other hand, for our given homomorphism RS we know that
R/m^{k}S/n^{k} must be an isomorphism for all k. It follows that
there is no such prime ideal Q_{0} in S.
We have thus proved that there is a unique prime ideal Q in S that
lies over a given prime ideal P in R and
f^{1}(Q) = P. The
``closed''ness condition is an immediate corollary.
Let us note that if R[X] is the polynomial ring over a ring R,
then
Hom(R[X], A) is naturally identified with
Hom(R, A)×A. Thus, if
g : R[X]S[X] denotes the natural extension of the
above homomorphism to the corresponding polynomial rings then, for any
finite ring the induced map
Hom(S[X], A)Hom(R[X], A) is a
bijection whenever
Hom(S, A)Hom(R, A) is a bijection. In
particular, we can apply the above lemmas to the homomorphism g as
well.
Lemma 25
Let
f :
RS be as in the theorem and
g :
R[
X]
S[
X] be the
induced homomorphism on polynomial rings in one variable. Let
be any element of
S and
be the ideal
(
X 
)
S[
X] in
S[
X]. Let
be the ideal
g^{1}((
X 
a)
S[
X]). Then
contains a
monic
polynomial.
Proof.
Let
A be any ring and
be an ideal in the polynomial ring
A[
X]. Let
denote the ideal
^{ . }A[
X,
X^{1}]
in the ring
A[
X,
X^{1}]. We have
= {
P(
X)
^{ . }X^{n}
P(
X)
and
n 0 an integer }
Let
be the restriction
A[
X^{1}] of this
ideal to
A[
X^{1}]. We have
= {
P(
X)
^{ . }X^{d}
P(
X)
and
d = deg(
P(
X))}
The
content
c(
) of the ideal
is defined as
the image of
in
A[
X^{1}]/(
X^{1}) =
A. Clearly,
c(
) = {
a A
P(
X)
such that
P(
X) =
aX^{d} + lower degree terms }
Returning to the rings
R and
S let us use the subscripts 1 and 2
to denote the above constructions applied to ideals in
R[
X] and
S[
X]; specifically to the ideals
and
.
We want to show that the content
c() of the ideal
in R[X] is the unit ideal. Suppose that
c() m for
some maximal ideal m in R. The ideal
= m[X^{1}] + X^{1}R[X^{}1] is then a maximal ideal in
R[X^{1}] which contains
. Moreover, by the above
description of
it is clear that
= g_{2}^{1}(), where
g_{2} : R[X^{1}]S[X^{1}] is
the natural homomorphism. Applying the ``goingup'' which has been
proved above, it follows that there should exist a prime ideal
containing
such that
g_{2}^{1}() = . But
is the ideal
generated by
1  X^{1} and X^{1} lies in
. Thus would have to be the unit ideal which
contradicts its primality. It follows that
c() is the unit
ideal.
From this lemma we see that S is integral over R. Now the
result that
R/mS/f (m)S is an isomorphism for all maximal ideals
m implies theorem 18 by Nakayama's lemma.
Next: Bibliography
Up: Some Lectures on Number
Previous: A..1 Counterexample
Kapil Hari Paranjape
20021020