2Noetherian algebras
III Algebras
2.3 Injective modules and Goldie’s theorem
The goal of this section is to prove Goldie’s theorem.
Theorem
(Goldie’s theorem)
.
Let
A
be a right Noetherian algebra with no
non-zero ideals all of whose elements are nilpotent. Then
A
embeds in a finite
direct sum of matrix algebras over division algebras.
The outline of the proof is as follows — given any
A
, we embed
A
in an
“injective hull”
E
(
A
). We will then find that similar to what we did in Artin–
Wedderburn, we can decompose
End
(
E
(
A
)) into a direct sum of matrix algebras
over division algebras. But we actually cannot. We will have to first quotient
End(E(A)) by some ideal I.
On the other hand, we do not actually have an embedding of
A
∼
=
End
A
(
A
)
into
End
(
E
(
A
)). Instead, what we have is only a homomorphism
End
A
(
A
)
→
End
(
E
(
A
))
/I
, where we quotient out by the same ideal
I
. So actually the two
of our problems happen to cancel each other out.
We will then prove that the kernel of this map contains only nilpotent
elements, and then our hypothesis implies this homomorphism is indeed an
embedding.
We begin by first constructing the injective hull. This is going to involve
talking about injective modules, which are dual to the notion of projective
modules.
Definition
(Injective module)
.
An
A
-module
E
is injective if for every diagram
of A-module maps
0 M N
E
θ
φ
ψ
,
such that
θ
is injective, there exists a map
ψ
that makes the diagram commute.
Equivalently, Hom( · , E) is an exact functor.
Example. Take A = k. Then all k-vector spaces are injective k-modules.
Example. Take A = k[X]. Then k(X) is an injective k[X]-module.
Lemma. Every direct summand of an injective module is injective, and direct
products of injectives is injective.
Proof. Same as proof for projective modules.
Lemma. Every A-module may be embedded in an injective module.
We say the category of
A
-modules has enough injectives. The dual result for
projectives was immediate, as free modules are projective.
Proof.
Let
M
be a right
A
-module. Then
Hom
k
(
A, M
) is a right
A
-module via
(fa)(x) = f(ax).
We claim that Hom
k
(A, M ) is an injective module. Suppose we have
0 M
1
N
1
Hom
k
(A, M )
θ
φ
We consider the k-module diagram
0 M
1
N
1
M
θ
α
β
where
α
(
m
1
) =
φ
(
m
1
)(1). Since
M
is injective as a
k
-module, we can find the
β
such that α = βθ. We define ψ : N
1
→ Hom
k
(A, M ) by
ψ(n
1
)(x) = β(n
1
x).
It is straightforward to check that this does the trick. Also, we have an embedding
M → Hom
k
(A, M ) by m 7→ (φ
n
: x 7→ mx).
The category theorist will write the proof in a line as
Hom
A
( · , Hom
k
(A, M ))
∼
=
Hom
k
( · ⊗
A
A, M )
∼
=
Hom
k
( · , M ),
which is exact since M is injective as a k-module.
Note that neither the construction of
Hom
k
(
A, M
), nor the proof that it is
injective requires the right
A
-modules structure of
M
. All we need is that
M
is
an injective k-module.
Lemma.
An
A
-module is injective iff it is a direct summand of every extension
of itself.
Proof.
Suppose
E
is injective and
E
0
is an extension of
E
. Then we can form
the diagram
0 E E
0
E
id
ψ
,
and then by injectivity, we can find ψ. So
E
0
= E ⊕ ker ψ.
Conversely, suppose
E
is a direct summand of every extension. But by the
previous lemma, we can embed
E
in an injective
E
0
. This implies that
E
is a
direct summand of E
0
, and hence injective.
There is some sort of “smallest” injective a module embeds into, and this
is called the injective hull, or injective envelope. This is why our injectives are
called
E
. The “smallness” will be captured by the fact that it is an essential
extension.
Definition
(Essential submodule)
.
An essential submodule
M
of an
A
-module
N
is one where
M ∩ V 6
=
{
0
}
for every non-zero submodule
V
of
N
. We say
N
is an essential extension of M .
Lemma. An essential extension of an essential extension is essential.
Proof.
Suppose
M < E < F
are essential extensions. Then given
N ≤ F
, we
know
N ∩ E 6
=
{
0
}
, and this is a submodule of
E
. So (
N ∩ E
)
∩ M
=
N ∩ M 6
= 0.
So F is an essential extension of M .
Lemma. A maximal essential extension is an injective module.
Such maximal things exist by Zorn’s lemma.
Proof.
Let
E
be a maximal essential extension of
M
, and consider any embedding
E → F
. We shall show that
E
is a direct summand of
F
. Let
S
be the set of
all non-zero submodules
V
of
F
with
V ∩ E
=
{
0
}
. We apply Zorn’s lemma to
get a maximal such module, say V
1
.
Then
E
embeds into
F/V
1
as an essential submodule. By transitivity of
essential extensions,
F/V
1
is an essential extension of
M
, but
E
is maximal. So
E
∼
=
F/V
1
. In other words,
F = E ⊕ V
1
.
We can now make the following definition:
Definition
(Injective hull)
.
A maximal essential extension of
M
is the injective
hull (or injective envelope) of M, written E(M).
Proposition.
Let
M
be an
A
-module, with an inclusion
M → I
into an injective
module. Then this extends to an inclusion E(M) → I.
Proof. By injectivity, we can fill in the diagram
0 M E(M )
I
ψ
.
We know
ψ
restricts to the identity on
M
. So
ker ψ ∩ M
=
{
0
}
. By Since
E
(
M
)
is essential, we must have ker ψ = 0. So E(M) embeds into I.
Proposition.
Suppose
E
is an injective essential extension of
M
. Then
E
∼
=
E(M ). In particular, any two injective hulls are isomorphic.
Proof.
By the previous lemma,
E
(
M
) embeds into
E
. But
E
(
M
) is a maximal
essential extension. So this forces E = E(M).
Using what we have got, it is not hard to see that
Proposition.
E(M
1
⊕ M
2
) = E(M
1
) ⊕ E(M
2
).
Proof.
We know that
E
(
M
1
)
⊕ E
(
M
2
) is also injective (since finite direct sums
are the same as direct products), and also
M
1
⊕ M
2
embeds in
E
(
M
1
)
⊕ E
(
M
2
).
So it suffices to prove this extension is essential.
Let V ≤ E(M
1
) ⊕ E(M
2
). Then either V /E(M
1
) 6= 0 or V/E(M
2
) 6= 0.
We wlog it is the latter. Note that we can naturally view
V
E(M
2
)
≤
E(M
1
) ⊕ E(M
2
)
E(M
2
)
∼
=
E(M
1
).
Since M
1
⊆ E(M
1
) is essential, we know
M
1
∩ (V/E(M
2
)) 6= 0.
So there is some
m
1
+
m
2
∈ V
such that
m
2
∈ E
(
M
2
) and
m
1
∈ M
1
. Now
consider
{m ∈ E(M
2
) : am
1
+ m ∈ V for some a ∈ A}.
This is a non-empty submodule of
E
(
M
2
), and so contains an element of
M
2
,
say n. Then we know am
1
+ n ∈ V ∩ (M
1
⊕ M
2
), and we are done.
The next two examples of injective hulls will be stated without proof:
Example. Take A = k[X], and M = k[X]. Then E(M) = k(X).
Example.
Let
A
=
k
[
X
] and
V
=
k
be the trivial module, where
X
acts by 0.
Then
E(V ) =
k[X, X
−1
]
Xk[X]
,
which is a quotient of A-modules. We note V embeds in this as
V
∼
=
k[X]
Xk[X]
→
k[X, X
−1
]
Xk[X]
.
Definition
(Uniform module)
.
A non-zero module
V
is uniform if given non-zero
submodules V
1
, V
2
, then V
1
∩ V
2
6= {0}.
Lemma. V is uniform iff E(V ) is indecomposable.
Proof.
Suppose
E
(
V
) =
A ⊕ B
, with
A, B
non-zero. Then
V ∩ A 6
=
{
0
}
and
V ∩B 6
=
{
0
}
since the extension is essential. So we have two non-zero submodules
of V that intersect trivially.
Conversely, suppose
V
is not uniform, and let
V
1
, V
2
be non-zero submod-
ules that intersect trivially. By Zorn’s lemma, we suppose these are maximal
submodules that intersect trivially. We claim
E(V
1
) ⊕ E(V
2
) = E(V
1
⊕ V
2
) = E(V )
To prove this, it suffices to show that
V
is an essential extension of
V
1
⊕ V
2
, so
that E(V ) is an injective hull of V
1
⊕ V
2
.
Let
W ≤ V
be non-zero. If
W ∩
(
V
1
⊕ V
2
) = 0, then
V
1
⊕
(
V
2
⊕ W
) is a larger
pair of submodules with trivial intersection, which is not possible. So we are
done.
Definition
(Domain)
.
An algebra is a domain if
xy
= 0 implies
x
= 0 or
y
= 0.
This is just the definition of an integral domain, but when we have non-
commutative algebras, we usually leave out the word “integral”.
To show that the algebras we know and love are indeed domains, we again
do some deformation.
Lemma.
Let
A
be a filtered algebra, which is exhaustive and separated. Then
if gr A is a domain, then so is A.
Proof.
Let
x ∈ A
i
\ A
i−1
, and
y ∈ A
j
\ A
j−1
. We can find such
i, j
for any
elements
x, y ∈ A
because the filtration is exhaustive and separated. Then we
have
¯x = x + A
i−1
6= 0 ∈ A
i
/A
i−1
¯y = y + A
j−1
6= 0 ∈ A
j
/A
j−1
.
If
gr A
is a domain, then we deduce
¯x¯y 6
= 0. So we deduce that
xy 6∈ A
i+j−1
. In
particular, xy 6= 0.
Corollary. A
n
(k) and U(g) are domains.
Lemma.
Let
A
be a right Noetherian domain. Then
A
A
is uniform, i.e.
E
(
A
A
)
is indecomposable.
Proof.
Suppose not, and so there are
xA
and
yA
non-zero such that
xA ∩ yA
=
{0}. So xA ⊕ yA is a direct sum.
But
A
is a domain and so
yA
∼
=
A
as a right
A
-module. Thus
yxA ⊕ yyA
is
a direct sum living inside yA. Further decomposing yyA, we find that
xA ⊕ yxA ⊕ y
2
xA ⊕ · · · ⊕ y
n
xA
is a direct sum of non-zero submodules. But this is an infinite strictly ascending
chain as n → ∞, which is a contradiction.
Recall that when we proved Artin–Wedderburn, we needed to use Krull–
Schmidt, which told us the decomposition is unique up to re-ordering. That
relied on the endomorphism algebra being local. We need something similar
here.
Lemma.
Let
E
be an indecomposable injective right module. Then
End
A
(
E
)
is a local algebra, with the unique maximal ideal given by
I = {f ∈ End(E) : ker f is essential}.
Note that since
E
is indecomposable injective, given any non-zero
V ≤ E
, we
know
E
(
V
) embeds into, and hence is a direct summand of
E
. Hence
E
(
V
) =
E
.
So
ker f
being essential is the same as saying
ker f
being non-zero. However,
this description of the ideal will be useful later on.
Proof.
Let
f
:
E → E
and
ker f
=
{
0
}
. Then
f
(
E
) is an injective module, and
so is a direct summand of
E
. But
E
is indecomposable. So
f
is surjective. So it
is an isomorphism, and hence invertible. So it remains to show that
I = {f ∈ End(E) : ker f is essential}
is an ideal.
If
ker f
and
ker g
are essential, then
ker
(
f
+
g
)
≥ ker f ∩ ker g
, and the
intersection of essential submodules is essential. So ker(f + g) is also essential.
Also, if
ker g
is essential, and
f
is arbitrary, then
ker
(
f ◦ g
)
≥ ker g
, and is
hence also essential. So I is a maximal left ideal.
The point of this lemma is to allow us to use Krull–Schmidt.
Lemma.
Let
M
be a non-zero Noetherian module. Then
M
is an essential
extension of a direct sum of uniform submodules N
1
, · · · , N
r
. Thus
E(M )
∼
=
E(N
1
) ⊕ · · · E(N
r
)
is a direct sum of finitely many indecomposables.
This decomposition is unique up to re-ordering (and isomorphism).
Proof.
We first show any non-zero Noetherian module contains a uniform one.
Suppose not, and
M
is in particular not uniform. So it contains non-zero
V
1
, V
0
2
with
V
1
∩ V
0
2
= 0. But
V
0
2
is not uniform by assumption. So it contains non-zero
V
2
and V
0
3
with zero intersection. We keep on repeating. Then we get
V
1
⊕ V
2
⊕ · · · ⊕ V
n
is a strictly ascending chain of submodules of M, which is a contradiction.
Now for non-zero Noetherian
M
, pick
N
1
uniform in
M
. Either
N
1
is essential
in
M
, and we’re done, or there is some
N
0
2
non-zero with
N
1
∩ N
0
2
= 0. We pick
N
2
uniform in N
0
2
. Then either N
1
⊕ N
2
is essential, or. . .
And we are done since M is Noetherian. Taking injective hulls, we get
E(M ) = E(N
1
) ⊕ · · · ⊕ E(N
r
),
and we are done by Krull–Schmidt and the previous lemma.
This is the crucial lemma, which isn’t really hard. This allows us to define
yet another dimension for Noetherian algebras.
Definition
(Uniform dimension)
.
The uniform dimension, or Goldie rank of
M is the number of indecomposable direct summands of E(M).
This is analogous to vector space dimensions in some ways.
Example.
The Goldie rank of domains is 1, as we showed
A
A
is uniform. This
is true for A
n
(k) and U(g).
Lemma.
Let
E
1
, · · · , E
r
be indecomposable injectives. Put
E
=
E
1
⊕ · · · ⊕ E
r
.
Let I = {f ∈ End
A
(E) : ker f is essential}. This is an ideal, and then
End
A
(E)/I
∼
=
M
n
1
(D
1
) ⊕ · · · ⊕ M
n
s
(D
s
)
for some division algebras D
i
.
Proof. We write the decomposition instead as
E = E
n
1
1
⊕ · · · ⊕ E
n
r
r
.
Then as in basic linear algebra, we know elements of
End
(
E
) can be written as
an r × r matrix whose (i, j)th entry is an element of Hom(E
n
i
i
, E
n
j
j
).
Now note that if
E
i
6
∼
=
E
j
, then the kernel of a map
E
i
→ E
j
is essential in
E
i
. So quotienting out by I kills all of these “off-diagonal” entries.
Also
Hom
(
E
n
i
i
, E
n
i
i
) =
M
n
i
(
End
(
E
i
)), and so quotienting out by
I
gives
M
n
i
(End(E
i
)/{essential kernel})
∼
=
M
n
i
(D
i
), where
D
i
∼
=
End(E
i
)
essential kernel
,
which we know is a division algebra since I is a maximal ideal.
The final piece to proving Goldie’s theorem is the following piece
Lemma.
If
A
is a right Noetherian algebra, then any
f
:
A
A
→ A
A
with
ker f
essential in A
A
is nilpotent.
Proof. Consider
0 < ker f ≤ ker f
2
≤ · · · .
Suppose
f
is not nilpotent. We claim that this is a strictly increasing chain.
Indeed, for all n, we have f
n
(A
A
) 6= 0. Since ker f is essential, we know
f
n
(A
A
) ∩ ker f 6= {0}.
This forces ker f
n+1
> ker f
n
, which is a contradiction.
We can now prove Goldie’s theorem.
Theorem
(Goldie’s theorem)
.
Let
A
be a right Noetherian algebra with no
non-zero ideals all of whose elements are nilpotent. Then
A
embeds in a finite
direct sum of matrix algebras over division algebras.
Proof. As usual, we have a map
A End
A
(A
A
)
x left multiplication by x
For a map A
A
→ A
A
, it lifts to a map E(A
A
) → E(A
A
) by injectivity:
0 A
A
E(A
A
)
A
A
E(A
A
)
f
θ
f
0
θ
We can complete the diagram to give a map
f
0
:
E
(
A
A
)
→ E
(
A
A
), which
restricts to
f
on
A
A
. This is not necessarily unique. However, if we have two
lifts
f
0
and
f
00
, then the difference
f
0
− f
00
has
A
A
in the kernel, and hence has
an essential kernel. So it lies in I. Thus, if we compose maps
A
A
End
A
(A
A
) End(E(A
A
))/I .
The kernel of this consists of
A
which when multiplying on the left has essential
kernel. This is an ideal all of whose elements is nilpotent. By assumption, any
such ideal vanishes. So we have an embedding of
A
in
End
(
E
(
A
A
))
/I
, which we
know to be a direct sum of matrix algebras over division rings.
Goldie didn’t present it like this. This work in injective modules is due to
Matlis.
We saw that (right Noetherian) domains had Goldie rank 1. So we get that
End
(
E
(
A
))
/I
∼
=
D
for some division algebra
D
. So by Goldie’s theorem, a right
Noetherian algebra embeds in a division algebra. In particular, this is true for
A
n
(k) and U(g).