**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

`f`_{1}∈ I of minimal
positive degree. Inductively, we can choose elements

`f`_{k+1}
∈ I\ (f_{1},…, f_{k}) that have minimal
positive degree. Let

`n`_{k} denote the degree of

`f`_{k}, and let

`a`_{k} denote its
leading coefficient. Observe first that

`n`_{1} ≤ n_{2} ≤ n_{3} ≤ ….

Now consider the chain of ideals

`(a`_{1})⊂ (a_{1}, a_{2})⊂(a_{1},a_{2},a_{3})
⊂…

contained in

`R`. This chain of ideals cannot be
stationary. For, suppose that

`(a`_{1}, …, a_{k}) = (a_{1},
…, a_{k}, a_{k+1}). Then we can write

`a`_{k+1} = ∑_{i=1}^{k} b_{i} a_{i}

for some

`b`_{i}∈ R. Now consider the polynomial

` g = f`_{k+1} - ∑_{i=1}^{k} b_{i} x^{nk+1}-n_{i}
f_{i}.

By the construction of the sequence of

`f`_{i}'s, we
know that

`g ∈ I \ (f`_{1}, …,
f_{k}). However, we have constructed

`g` in
a way that kills off the leading coefficient, leaving a polynomial of
degree smaller than the degree of

`f`_{k+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[x`_{1},
…, x_{n}]/I be the quotient field. If

`L=k`, then we're done. To see this, write

`a`_{i}∈ k for the image

`a`_{i} ≡ x_{i}
mod I. Thus, each polynomial

`x`_{i}-a_{i}∈
I and

`(x`_{1}-a_{1}, …, x_{n}-a_{n}) ⊂
I.

Since the former ideal is maximal, this containment
is an equality, and it is easy to see that

`Z(I)={(a`_{1},
…, a_{n})}.
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[a`_{1},…,a_{n}]. If

`n=1`, then either

`a`_{1} is an
algebraic element or

`L≈ k[a`_{1}] is not a
field. So, by induction, we may assume that

`L=k(a`_{1})[a_{2},
…, a_{n}] is an algebraic extension of the field

`k(a`_{1}). We may also assume that

`a`_{1} is transcendental over

`k`
(otherwise, we are done).
For all

`2 ≤ i ≤ n`, we have an equation

` a`_{i}^{ni} + b_{i,1} a_{i}^{ni-1} +…+b_{i, ni} =
0,

with coefficients

`b`_{i,j}∈ k(a_{1}). We can
choose an element

`b∈ k(a`_{1}) large enough to
clear all the denominators of the

`b`_{i,j}, so that

` (ba`_{i})^{ni} +bb_{i,1} (ba_{i})^{ni-1}
+…+b^{ni}b_{i,ni} = 0.

In particular,

`L` is an integral extension of

`k(a`_{1}). (In other words, it is generated by the

integral elements
`ba`_{2},…,ba_{n}.) In
particular, given any element

`z∈ L`, there
exists a non-negative integer

`N` such that

`b`^{N}z is integral over

`k[a`_{1}]. 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

`t`_{1},…,t_{m} be module
generators. Write each product

`st`_{i} = ∑_{j=1}^{m} r_{ij}t_{j},

with coefficients

`r`_{ij}∈ R.
Rewriting this with Kronecker deltas yields

`∑`_{j=1}^{m} (δ_{ij}s-r_{ij})w_{j} = 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(δ`_{ij}s-r_{ij}) = s^{m} + (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

`a`_{2},…,a_{n}; 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[a`_{i}], and it follows that

`b`^{N}z is also integral.
Now suppose

`f∈ k[a`_{1}] is any polynomial that is relatively prime
to

`b`. Then

`1/f∈ L`, but none of the elements

`b`^{N}/f can possibly
be integral over

`k[a`_{1}]. This is a contradiction; therefore,

`a`_{1}
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, f`^{N}∈ I}.

**Theorem** (Hilbert's Nullstellensatz) Let

`k` be an
algebraically closed field. Let

`J⊂ k[x`_{1}, …,
x_{n}] 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

`f`_{1},…,f_{s} of

`J`. Consider the auxiliary ideal

`
J' = (f`_{1},…,f_{s},gx_{n+1}-1) ⊂ k[x_{1},…,x_{n+1}].

Since

`g=0` whenever all

`f`_{i}=0, we see that

`Z(J')=∅`. By
the Weak Nullstellensatz,

`J'=k[x`_{1},…,x_{n+1}]. Thus, there is a
relation

`
1 = (∑`_{1}^{s} a_{i}f_{i}) + b(gx_{n+1}-1)

for some

`b, a`_{i} ∈ k[x_{1},…,x_{n+1}]. Substitute

`y=1/x`_{n+1}
in this equation and clear denominators to get

`
y`^{m} = ∑ a'_{i} f_{i} + b'(g-y)

with coefficients now living in

`k[x`_{1},…,x_{n},y]. Now specialize
by setting

`y=g` to obtain

`
g`^{m} = ∑ a"_{i} f_{i} ∈ J ⊂ k[x_{1},…,x_{n}].

Thus,

`g∈ Rad(J)` and the theorem is proved.

**Corollary** Two algebraic sets `X`_{1}, X_{2} ∈ A^{n} are equal if and only if
`I(X`_{1})=I(X_{2}).

**Corollary** There is a natural one-to-one inclusion-reversing correspondence
between algebraic sets in `A`^{n} and radical ideals in
`k[x`_{1},…,x_{n}]. Under this correspondence, points are associated
with maximal ideals.

**Definition** Let

`X` be an algebraic set in

`A`^{n}. Define the ring of

regular functions on

`X` to be

` A(X) = k[x`_{1}, …, x_{n}]/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

`A`^{n}. 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[x`_{1}, …,
x_{n}] 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