2Rings
IB Groups, Rings and Modules
2.8 Noetherian rings
We now revisit the idea of Noetherian rings, something we have briefly mentioned
when proving that PIDs are UFDs.
Definition (Noetherian ring). A ring is Noetherian if for any chain of ideals
I
1
⊆ I
2
⊆ I
3
⊆ ··· ,
there is some N such that I
N
= I
N+1
= I
N+2
= ···.
This condition is known as the ascending chain condition.
Example. Every finite ring is Noetherian. This is since there are only finitely
many possible ideals.
Example. Every field is Noetherian. This is since there are only two possible
ideals.
Example. Every principal ideal domain (e.g.
Z
) is Noetherian. This is easy to
check directly, but the next proposition will make this utterly trivial.
Most rings we love and know are indeed Noetherian. However, we can
explicitly construct some non-Noetherian ideals.
Example. The ring
Z
[
X
1
, X
2
, X
3
, ···
] is not Noetherian. This has the chain of
strictly increasing ideals
(X
1
) ⊆ (X
1
, X
2
) ⊆ (X
1
, X
2
, X
3
) ⊆ ··· .
We have the following proposition that makes Noetherian rings much more
concrete, and makes it obvious why PIDs are Noetherian.
Definition (Finitely generated ideal). An ideal
I
is finitely generated if it can
be written as I = (r
1
, ··· , r
n
) for some r
1
, ··· , r
n
∈ R.
Proposition. A ring is Noetherian if and only if every ideal is finitely generated.
Every PID trivially satisfies this condition. So we know every PID is Noethe-
rian.
Proof. We start with the easier direction — from concrete to abstract.
Suppose every ideal of
R
is finitely generated. Given the chain
I
1
⊆ I
2
⊆ ···
,
consider the ideal
I = I
1
∪ I
2
∪ I
3
∪ ··· .
This is obviously an ideal, and you will check this manually in example sheet 2.
We know I is finitely generated, say I = (r
1
, ··· , r
n
), with r
i
∈ I
k
i
. Let
K = max
i=1,··· ,n
{k
i
}.
Then r
1
, ··· , r
n
∈ I
K
. So I
K
= I. So I
K
= I
K+1
= I
K+2
= ···.
To prove the other direction, suppose there is an ideal
I C R
that is not
finitely generated. We pick
r
1
∈ I
. Since
I
is not finitely generated, we know
(r
1
) 6= I. So we can find some r
2
∈ I \ (r
1
).
Again (
r
1
, r
2
)
6
=
I
. So we can find
r
3
∈ I \
(
r
1
, r
2
). We continue on, and then
can find an infinite strictly ascending chain
(r
1
) ⊆ (r
1
, r
2
) ⊆ (r
1
, r
2
, r
3
) ⊆ ··· .
So R is not Noetherian.
When we have developed some properties or notions, a natural thing to ask
is whether it passes on to subrings and quotients.
If
R
is Noetherian, does every subring of
R
have to be Noetherian? The
answer is no. For example, since
Z
[
X
1
, X
2
, ···
] is an integral domain, we can
take its field of fractions, which is a field, hence Noetherian, but Z[X
1
, X
2
, ···]
is a subring of its field of fractions.
How about quotients?
Proposition. Let
R
be a Noetherian ring and
I
be an ideal of
R
. Then
R/I
is
Noetherian.
Proof.
Whenever we see quotients, we should think of them as the image of a
homomorphism. Consider the quotient map
π : R → R/I
x 7→ x + I.
We can prove this result by finitely generated or ascending chain condition. We
go for the former. Let
J C R/I
be an ideal. We want to show that
J
is finitely
generated. Consider the inverse image
π
−1
(
J
). This is an ideal of
R
, and is
hence finitely generated, since
R
is Noetherian. So
π
−1
(
J
) = (
r
1
, ··· , r
n
) for
some r
1
, ··· , r
n
∈ R. Then J is generated by π(r
1
), ··· , π(r
n
). So done.
This gives us many examples of Noetherian rings. But there is one important
case we have not tackled yet — polynomial rings. We know
Z
[
X
] is not a PID,
since (2
, X
) is not principal. However, this is finitely generated. So we are not
dead. We might try to construct some non-finitely generated ideal, but we are
bound to fail. This is since
Z
[
X
] is a Noetherian ring. This is a special case of
the following powerful theorem:
Theorem (Hilbert basis theorem). Let
R
be a Noetherian ring. Then so is
R[X].
Since Z is Noetherian, we know Z[X] also is. Hence so is Z[X, Y ] etc.
The Hilbert basis theorem was, surprisingly, proven by Hilbert himself. Before
that, there were many mathematicians studying something known as invariant
theory. The idea is that we have some interesting objects, and we want to look
at their symmetries. Often, there are infinitely many possible such symmetries,
and one interesting question to ask is whether there is a finite set of symmetries
that generate all possible symmetries.
This sounds like an interesting problem, so people devoted much time, writing
down funny proofs, showing that the symmetries are finitely generated. However,
the collection of such symmetries are often just ideals of some funny ring. So
Hilbert came along and proved the Hilbert basis theorem, and showed once and
for all that those rings are Noetherian, and hence the symmetries are finitely
generated.
Proof.
The proof is not too hard, but we will need to use both the ascending
chain condition and the fact that all ideals are finitely-generated.
Let
I C R
[
X
] be an ideal. We want to show it is finitely generated. Since we
know R is Noetherian, we want to generate some ideals of R from I.
How can we do this? We can do the silly thing of taking all constants of
I
,
i.e.
I ∩ R
. But we can do better. We can consider all linear polynomials, and
take their leading coefficients. Thinking for a while, this is indeed an ideal.
In general, for n = 0, 1, 2, ···, we let
I
n
= {r ∈ R : there is some f ∈ I such that f = rX
n
+ ···} ∪ {0}.
Then it is easy to see, using the strong closure property, that each ideal
I
n
is an
ideal of
R
. Moreover, they form a chain, since if
f ∈ I
, then
Xf ∈ I
, by strong
closure. So I
n
⊆ I
n+1
for all n.
By the ascending chain condition of
R
, we know there is some
N
such that
I
N
=
I
N+1
=
···
. Now for each 0
≤ n ≤ N
, since
R
is Noetherian, we can write
I
n
= (r
(n)
1
, r
(n)
2
, ··· , r
(n)
k(n)
).
Now for each r
(n)
i
, we choose some f
(n)
i
∈ I with f
(n)
i
= r
(n)
i
X
n
+ ···.
We now claim the polynomials
f
(n)
i
for 0
≤ n ≤ N
and 1
≤ i ≤ k
(
n
) generate
I.
Suppose not. We pick g ∈ I of minimal degree not generated by the f
(n)
i
.
There are two possible cases. If deg g = n ≤ N, suppose
g = rX
n
+ ··· .
We know r ∈ I
n
. So we can write
r =
X
i
λ
i
r
(n)
i
for some λ
i
∈ R, since that’s what generating an ideal means. Then we know
X
i
λ
i
f
(n)
i
= rX
n
+ ··· ∈ I.
But if
g
is not in the span of the
f
(j)
i
, then so isn’t
g −
P
i
λ
i
f
(n)
i
. But this has
a lower degree than g. This is a contradiction.
Now suppose
deg g
=
n > N
. This might look scary, but it is not, since
I
n
= I
N
. So we write the same proof. We write
g = rX
n
+ ··· .
But we know r ∈ I
n
= I
N
. So we know
r =
X
I
λ
i
r
(N)
i
.
Then we know
X
n−N
X
i
λ
i
f
(n)
i
= rX
N
+ ··· ∈ I.
Hence
g − X
n−N
P
λ
i
f
(N)
i
has smaller degree than
g
, but is not in the span of
f
(j)
i
.
As an aside, let
E ⊆ F
[
X
1
, X
2
, ··· , X
n
] be any set of polynomials. We view
this as a set of equations
f
= 0 for each
f ∈ E
. The claim is that to solve the
potentially infinite set of equations
E
, we actually only have to solve finitely
many equations.
Consider the ideal (
E
)
C F
[
X
1
, ··· , X
n
]. By the Hilbert basis theorem, there
is a finite list f
1
, ··· , f
k
such that
(f
1
, ··· , f
k
) = (E).
We want to show that we only have to solve
f
i
(
x
) = 0 for these
f
i
. Given
(α
1
, ··· , α
n
) ∈ F
n
, consider the homomorphism
φ
α
: F [X
1
, ··· , X
n
] → F
X
i
7→ α
i
.
Then we know (
α
1
, ··· , α
n
)
∈ F
n
is a solution to the equations
E
if and only
if (
E
)
⊆ ker
(
ϕ
α
). By our choice of
f
i
, this is true if and only if (
f
1
, ··· , f
k
)
⊆
ker
(
ϕ
α
). By inspection, this is true if and only if (
α
1
, ··· , α
n
) is a solution to
all of
f
1
, ··· , f
k
. So solving
E
is the same as solving
f
1
, ··· , f
k
. This is useful
in, say, algebraic geometry.