Proof: We will actually prove the contrapositive. So, assume that
R[x] is not noetherian. Let
I be
an ideal in
R that is not finitely
generated. Choose an element
f1∈ I of minimal
positive degree. Inductively, we can choose elements
fk+1
∈ I\ (f1,…, fk) that have minimal
positive degree. Let
nk denote the degree of
fk, and let
ak denote its
leading coefficient. Observe first that
n1 ≤ n2 ≤ n3 ≤ ….
Now consider the chain of ideals
(a1)⊂ (a1, a2)⊂(a1,a2,a3)
⊂…
contained in
R. This chain of ideals cannot be
stationary. For, suppose that
(a1, …, ak) = (a1,
…, ak, ak+1). Then we can write
ak+1 = ∑i=1k bi ai
for some
bi∈ R. Now consider the polynomial
g = fk+1 - ∑i=1k bi xnk+1-ni
fi.
By the construction of the sequence of
fi's, we
know that
g ∈ I \ (f1, …,
fk). However, we have constructed
g in
a way that kills off the leading coefficient, leaving a polynomial of
degree smaller than the degree of
fk+1. This is
a contradiction. So, we have shown that if
R[x] is
not noetherian, then
R is not noetherian. The
theorem follows.
Proof: By the previous lemma, it suffices to prove this when
I is a maximal ideal. Let
L=k[x1,
…, xn]/I be the quotient field. If
L=k, then we're done. To see this, write
ai∈ k for the image
ai ≡ xi
mod I. Thus, each polynomial
xi-ai∈
I and
(x1-a1, …, xn-an) ⊂
I.
Since the former ideal is maximal, this containment
is an equality, and it is easy to see that
Z(I)={(a1,
…, an)}.
It now suffices to prove the following proposition.
Proposition Let k be a field, and let
L be an extension field of k
that is finitely generated as a k-algebra. Then
L is an algebraic extension of
k.
Proof: We can write
L=k[a1,…,an]. If
n=1, then either
a1 is an
algebraic element or
L≈ k[a1] is not a
field. So, by induction, we may assume that
L=k(a1)[a2,
…, an] is an algebraic extension of the field
k(a1). We may also assume that
a1 is transcendental over
k
(otherwise, we are done).
For all
2 ≤ i ≤ n, we have an equation
aini + bi,1 aini-1 +…+bi, ni =
0,
with coefficients
bi,j∈ k(a1). We can
choose an element
b∈ k(a1) large enough to
clear all the denominators of the
bi,j, so that
(bai)ni +bbi,1 (bai)ni-1
+…+bnibi,ni = 0.
In particular,
L is an integral extension of
k(a1). (In other words, it is generated by the
integral elements
ba2,…,ban.) In
particular, given any element
z∈ L, there
exists a non-negative integer
N such that
bNz is integral over
k[a1]. To
see this, one needs a series of lemmas.
Lemma If s is an integral element over a ring
R, then R[s] is finitely
generated as an R-module.
Proof: Obvious.
Lemma Let R⊂ S be an extension of
commutative rings with 1. If S is finitely
generated as an R-module, then every element
s∈ S is integral over R.
Proof: Let
t1,…,tm be module
generators. Write each product
sti = ∑j=1m rijtj,
with coefficients
rij∈ R.
Rewriting this with Kronecker deltas yields
∑j=1m (δijs-rij)wj = 0
for all
i=1, …,m. Looking at this equation
in an appropriate vector space over the field of fractions of
S, one gets a nontrivial solution to a certain
matrix equation. Therefore, one has
0 = det(δijs-rij) = sm + (terms of lower
degree),
a monic equation satisfied by
s with coefficients
in
R.
Lemma The R⊂ S be an inclusion of
commutative rings with 1. Then the set of elements of
S that are integral over R forms
a subring of S.
Proof: Let a, b∈ S be
integral elements. Since b is certainly integral
over R[a], we have that R[a]/R
and R[a,b]/R[a] are finitely generated
modules. Thus, R[a,b] is a finitely generated
R-module. Thus, the elements a+b and ab∈ R[a,b] are integral
over R, and the result follows.
Returning to the proof of the proposition, we see that any
z∈ L
can be written as a polynomial in
a2,…,an; multiplying by a
large power of
b writes it as a combination of things that are known
to be integral over the polynomial ring
k[ai], and it follows that
bNz is also integral.
Now suppose
f∈ k[a1] is any polynomial that is relatively prime
to
b. Then
1/f∈ L, but none of the elements
bN/f can possibly
be integral over
k[a1]. This is a contradiction; therefore,
a1
could not have been transcendental, and the proposition (and the Weak
Nullstellensatz) follows.
The previous theorem shows that every proper ideal cuts out a nonempty
algebraic set. The next theorem will characterize the ideals
associated to algebraic sets.
Definition Let
I be an ideal in a ring
R. The radical of
I is defined
as
Rad(I) = {f∈ R : ∃ N>0, fN∈ I}.
Theorem (Hilbert's Nullstellensatz) Let
k be an
algebraically closed field. Let
J⊂ k[x1, …,
xn] be any ideal. Then
I(Z(J)) = Rad(J).
Proof: Since
I(X) is always radical and
J⊂ I(Z(J)), the inclusion
Rad(J)⊂ I(Z(J)) is trivial. So, let
g∈ I(Z(J)) Now use Hilbert's Basis Theorem to
choose generators
f1,…,fs of
J. Consider the auxiliary ideal
J' = (f1,…,fs,gxn+1-1) ⊂ k[x1,…,xn+1].
Since
g=0 whenever all
fi=0, we see that
Z(J')=∅. By
the Weak Nullstellensatz,
J'=k[x1,…,xn+1]. Thus, there is a
relation
1 = (∑1s aifi) + b(gxn+1-1)
for some
b, ai ∈ k[x1,…,xn+1]. Substitute
y=1/xn+1
in this equation and clear denominators to get
ym = ∑ a'i fi + b'(g-y)
with coefficients now living in
k[x1,…,xn,y]. Now specialize
by setting
y=g to obtain
gm = ∑ a"i fi ∈ J ⊂ k[x1,…,xn].
Thus,
g∈ Rad(J) and the theorem is proved.
Corollary Two algebraic sets X1, X2 ∈ An are equal if and only if
I(X1)=I(X2).
Corollary There is a natural one-to-one inclusion-reversing correspondence
between algebraic sets in An and radical ideals in
k[x1,…,xn]. Under this correspondence, points are associated
with maximal ideals.
Definition Let
X be an algebraic set in
An. Define the ring of
regular functions on
X to be
A(X) = k[x1, …, xn]/I(X).
Notice that the assignment
X|→ A(X) shows
that
X determines
A(X)
uniquely. However, unlike the ideal
I(X), the ring
A(X) does not directly determine
X as a closed subset of
An. The problem, of course, is that there is
no canonical way to recover the generators of the polynomial ring from
the quotient ring. If you choose two different sets of generators for
A(X), you get two distinct algebraic sets, possibly
embedded in different affine spaces, with the same coordinate ring.
We can already see, however, that
A(X) has the
potential to recover
X in a more abstract form. For
example, the points of
X are in one-to-one
correspondence with the maximal ideals of
k[x1, …,
xn] that contain
I(X); by the usual
yoga of the isomorphism theorems, these are, in turn, in one-to-one
correspondence with the maximal ideals of
A(X).
Comments on this web site should be addressed to the
author:
Kevin R. Coombes
Department of Biomedical Informatics
The Ohio State University
Columbus, Ohio 43210