1Artinian algebras
III Algebras
1.1 Artinian algebras
We continue writing down some definitions. We already defined left and right
Artinian algebras in the introduction. Most examples we’ll meet are in fact
finite-dimensional vector spaces over
k
. However, there exists some more perverse
examples:
Example. Let
A =
r s
0 t
: r ∈ Q, s, t ∈ R
Then this is right Artinian but not left Artinian over
Q
. To see it is not left
Artinian, note that there is an ideal
I =
0 s
0 0
: s ∈ R
∼
=
R
of
A
, and a matrix
r s
0 t
acts on this on the left by sending
0 s
0
0 0
to
rs
0
. Since
R
is an infinite-dimensional
Q
-vector space, one sees that there is an
infinite strictly descending chain of ideals contained in I.
The fact that it is right Artinian is a direct verification. Indeed, it is not
difficult to enumerate all the right ideals, which is left as an exercise for the
reader.
As in the case of commutative algebra, we can study the modules of an
algebra.
Definition
(Module)
.
Let
A
be an algebra. A left
A
-module is a
k
-vector space
M and a bilinear map
A ⊗ M M
a ⊗ m xm
such that (
ab
)
m
=
a
(
bm
) for all
a, b ∈ A
and
m ∈ M
. Right
A
-modules are
defined similarly.
An
A
-
A
-bimodule is a vector space
M
that is both a left
A
-module and a
right
A
-module, such that the two actions commute — for
a, b ∈ A
and
x ∈ M
,
we have
a(xb) = (ax)b.
Example.
The algebra
A
itself is a left
A
-module. We write this as
A
A
, and
call this the left regular representation. Similarly, the right action is denoted
A
A
.
These two actions are compatible by associativity, so A is an A-A-bimodule.
If we write
End
k
(
A
) for the
k
-linear maps
A → A
, then
End
k
is naturally a
k
-
algebra by composition, and we have a
k
-algebra homomorphism
A → End
k
(
A
)
that sends
a ∈ A
to multiplication by
a
on the left. However, if we want
to multiply on the right instead, it is no longer a
k
-algebra homomorphism
A → End
k
(A). Instead, it is a map A → End
k
(A)
op
, where
Definition
(Opposite algebra)
.
Let
A
be a
k
-algebra. We define the opposite
algebra
A
op
to be the algebra with the same underlying vector space, but with
multiplication given by
x · y = yx.
Here on the left we have the multiplication in
A
op
and on the right we have the
multiplication in A.
In general, a left A-module is a right A
op
-module.
As in the case of ring theory, we can talk about prime ideals. However, we
will adopt a slightly different definition:
Definition
(Prime ideal)
.
An ideal
P
is prime if it is a proper ideal, and if
I
and J are ideals with IJ ⊆ P , then either I ⊆ P or J ⊆ P .
It is an exercise to check that this coincides in the commutative case with
the definition using elements.
Definition
(Annihilator)
.
Let
M
be a left
A
-module and
m ∈ M
. We define
the annihilators to be
Ann(m) = {a ∈ A : am = 0}
Ann(M) = {a ∈ A : am = 0 for all m ∈ M } =
\
m∈M
Ann(m).
Note that
Ann
(
m
) is a left ideal of
A
, and is in fact the kernel of the
A
-module
homomorphism
A → M
given by
x 7→ xm
. We’ll denote the image of this map
by Am, a left submodule of M, and we have
A
Ann(m)
∼
=
Am.
On the other hand, it is easy to see that
Ann
(
M
) is an fact a (two-sided) ideal.
Definition
(Simple module)
.
A non-zero module
M
is simple or irreducible if
the only submodules of M are 0 and M.
It is easy to see that
Proposition.
Let
A
be an algebra and
I
a left ideal. Then
I
is a maximal left
ideal iff A/I is simple.
Example. Ann(m) is a maximal left ideal iff Am is irreducible.
Proposition.
Let
A
be an algebra and
M
a simple module. Then
M
∼
=
A/I
for some (maximal) left ideal I of A.
Proof.
Pick an arbitrary element
m ∈ M
, and define the
A
-module homomor-
phism
ϕ
:
A → M
by
ϕ
(
a
) =
am
. Then the image is a non-trivial submodule,
and hence must be
M
. Then by the first isomorphism theorem, we have
M
∼
=
A/ ker ϕ.
Before we start doing anything, we note the following convenient lemma:
Lemma.
Let
M
be a finitely-generated
A
module. Then
M
has a maximal
proper submodule M
0
.
Proof.
Let
m
1
, · · · , m
k
∈ M
be a minimal generating set. Then in particular
N
=
hm
1
, · · · , m
k−1
i
is a proper submodule of
M
. Moreover, a submodule
of
M
containing
N
is proper iff it does not contain
m
k
, and this property is
preserved under increasing unions. So by Zorn’s lemma, there is a maximal
proper submodule.
Definition
(Jacobson radical)
.
The
J
(
A
) of
A
is the intersection of all maximal
left ideals.
This is in fact an ideal, and not just a left one, because
J(A) =
\
{maximal left ideals} =
\
m∈M,M simple
Ann(m) =
\
M simple
Ann(M),
which we have established is an ideal. Yet, it is still not clear that this is
independent of us saying “left” instead of “right”. It, in fact, does not, and this
follows from the Nakayama lemma:
Lemma
(Nakayama lemma)
.
The following are equivalent for a left ideal
I
of
A.
(i) I ≤ J(A).
(ii)
For any finitely-generated left
A
-module
M
, if
IM
=
M
, then
M
= 0,
where
IM
is the module generated by elements of the form
am
, with
a ∈ I
and m ∈ M.
(iii) G = {1 + a : a ∈ I} = 1 + I is a subgroup of the unit group of A.
In particular, this shows that the Jacobson radical is the largest ideal satisfying
(iii), which is something that does not depend on handedness.
Proof.
–
(i)
⇒
(ii): Suppose
I ≤ J
(
A
) and
M 6
= 0 is a finitely-generated
A
-module,
and we’ll see that IM M .
Let
N
be a maximal submodule of
M
. Then
M/N
is a simple module,
so for any
¯m ∈ M/N
, we know
Ann
(
¯m
) is a maximal left ideal. So
J(A) ≤ Ann(M/N). So IM ≤ J(A)M ≤ N M.
–
(ii)
⇒
(iii): Assume (ii). We let
x ∈ I
and set
y
= 1 +
x
. Hence
1 =
y − x ∈ Ay
+
I
. Since
Ay
+
I
is a left ideal, we know
Ay
+
I
=
A
. In
other words, we know
I
A
Ay
=
A
Ay
.
Now using (ii) on the finitely-generated module
A/Ay
(it is in fact generated
by 1), we know that
A/Ay
= 0. So
A
=
Ay
. So there exists
z ∈ A
such
that 1 =
zy
=
z
(1 +
x
). So (1 +
x
) has a left inverse, and this left inverse
z
lies in
G
, since we can write
z
= 1
− zx
. So
G
is a subgroup of the unit
group of A.
–
(iii)
⇒
(i): Suppose
I
1
is a maximal left ideal of
A
. Let
x ∈ I
. If
x 6∈ I
1
,
then
I
1
+
Ax
=
A
by maximality of
I
. So 1 =
y
+
zx
for some
y ∈ I
1
and
z ∈ A
. So
y
= 1
− zx ∈ G
. So
y
is invertible. But
y ∈ I
1
. So
I
1
=
A
.
This is a contradiction. So we found that
I < I
1
, and this is true for all
maximal left ideals I
1
. Hence I ≤ J(A).
We now come to the important definition:
Definition (Semisimple algebra). An algebra is semisimple if J(A) = 0.
We will very soon see that for Artinian algebras, being semi-simple is equiva-
lent to a few other very nice properties, such as being completely reducible. For
now, we shall content ourselves with some examples.
Example. For any A, we know A/J(A) is always semisimple.
We can also define
Definition (Simple algebra). An algebra is simple if the only ideals are 0 and
A.
It is trivially true that any simple algebra is semi-simple — the Jacobson
radical is an ideal, and it is not
A
. A particularly important example is the
following:
Example.
Consider
M
n
(
k
). We let
e
i
be the matrix with 1 in the (
i, i
)th
entry and zero everywhere else. This is idempotent, i.e.
e
2
i
=
e
i
. It is also
straightforward to check that
Ae
i
=
0 · · · 0 a
1
0 · · · 0
0 · · · 0 a
2
0 · · · 0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 · · · 0 a
n
0 · · · 0
The non-zero column is, of course, the
i
th column. Similarly,
e
i
A
is the matrices
that are zero apart from in the
i
th row. These are in fact all left and all right
ideals respectively. So the only ideals are 0 and A.
As a left A-module, we can decompose A as
A
A =
n
M
i=1
Ae
i
,
which is a decomposition into simple modules.
Definition
(Completely reducible)
.
A module
M
of
A
is completely reducible
iff it is a sum of simple modules.
Here in the definition, we said “sum” instead of “direct sum”, but the
following proposition shows it doesn’t matter:
Proposition. Let M be an A-module. Then the following are equivalent:
(i) M is completely reducible.
(ii) M is the direct sum of simple modules.
(iii)
Every submodule of
M
has a complement, i.e. for any submodule
N
of
M
,
there is a complement N
0
such that M = N ⊕ N
0
.
Often, the last condition is the most useful in practice.
Proof.
– (i) ⇒ (ii): Let M be completely reducible, and consider the set
n
{S
α
≤ M} : S
α
are simple,
X
S
α
is a direct sum
o
.
Notice this set is closed under increasing unions, since the property of
being a direct sum is only checked on finitely many elements. So by Zorn’s
lemma, it has a maximal element, and let N be the sum of the elements.
Suppose this were not all of
M
. Then there is some
S ≤ M
such that
S 6⊆ N
. Then
S ∩ N ( S
. By simplicity, they intersect trivially. So
S
+
N
is a direct sum, which is a contradiction. So we must have
N
=
M
, and
M is the direct sum of simple modules.
– (ii) ⇒ (i) is trivial.
– (i) ⇒ (iii): Let N ≤ M be a submodule, and consider
n
{S
α
≤ M} : S
α
are simple, N +
X
S
α
is a direct sum
o
.
Again this set has a maximal element, and let
P
be the direct sum of those
S
α
. Again if
P ⊕ N
is not all of
M
, then pick an
S ≤ M
simple such that
S
is not contained in
P ⊕ N
. Then again
S ⊕ P ⊕ N
is a direct sum, which
is a contradiction.
–
(iii)
⇒
(i): It suffices to show that if
N < M
is a proper submodule,
then there exists a simple module that intersects
N
trivially. Indeed, we
can take
N
to be the sum of all simple submodules of
M
, and this forces
N = M .
To do so, pick an
x 6∈ N
, and let
P
be submodule of
M
maximal among
those satisfying
P ∩ N
= 0 and
x 6∈ N ⊕ P
. Then
N ⊕ P
is a proper
submodule of M . Let S be a complement. We claim S is simple.
If not, we can find a proper submodule
S
0
of
S
. Let
Q
be a complement of
N ⊕ P ⊕ S
0
. Then we can write
M = N ⊕ P ⊕ S
0
⊕ Q
x = n + p + s + q
.
By assumption,
s
and
q
are not both zero. We wlog assume
s
is non-
zero. Then
P ⊕ Q
is a larger submodule satisfying (
P ⊕ Q
)
∩ N
= 0 and
x 6∈ N ⊕
(
P ⊕ Q
). This is a contradiction. So
S
is simple, and we are
done.
Using these different characterizations, we can prove that completely reducible
modules are closed under the familiar ex operations.
Proposition.
Sums, submodules and quotients of completely reducible modules
are completely reducible.
Proof.
It is clear by definition that sums of completely reducible modules are
completely reducible.
To see that submodules of completely reducible modules are completely
reducible, let
M
be completely reducible, and
N ≤ M
. Then for each
x ∈ N
,
there is some simple submodule
S ≤ M
containing
x
. Since
S ∩ N ≤ S
and
contains x, it must be S, i.e. S ⊆ N . So N is the sum of simple modules.
Finally, to see quotients are completely reducible, if
M
is completely reducible
and N is a submodule, then we can write
M = N ⊕ P
for some P . Then M/N
∼
=
P , and P is completely reducible.
We will show that every left Artinian algebra is completely reducible over
itself iff it is semi-simple. We can in fact prove a more general fact for
A
-modules.
To do so, we need a generalization of the Jacobson radical.
Definition
(Radical)
.
For a module
M
, we write
Rad
(
M
) for the intersection
of maximal submodules of M, and call it the radical of M.
Thus, we have Rad(
A
A) = J(A) = Rad(A
A
).
Proposition.
Let
M
be an
A
-module satisfying the descending chain condition
on submodules. Then M is completely reducible iff Rad(M) = 0.
Proof.
It is clear that if
M
is completely reducible, then
Rad
(
M
) = 0. Indeed,
we can write
M =
M
α∈A
S
α
,
where each S
α
is simple. Then
J(A) ≤
\
α∈A
M
β∈A\{α}
S
β
= {0}.
Conversely, if
Rad
(
M
) = 0, we note that since
M
satisfies the descending
chain condition on submodules, there must be a finite collection
M
1
, · · · , M
n
of
maximal submodules whose intersection vanish. Then consider the map
M
n
M
i=1
M
M
i
x (x + M
1
, x + M
2
, · · · , x + M
n
)
The kernel of this map is the intersection of the
M
i
, which is trivial. So this
embeds
M
as a submodule of
L
M
M
i
. But each
M
M
i
is simple, so
M
is a submodule
of a completely reducible module, hence completely reducible.
Corollary.
If
A
is a semi-simple left Artinian algebra, then
A
A
is completely
reducible.
Corollary.
If
A
is a semi-simple left Artinian algebra, then every left
A
-module
is completely reducible.
Proof.
Every
A
-module
M
is a quotient of sums of
A
A
. Explicitly, we have a
map
M
m∈M
A
A M
(a
m
)
P
a
m
m
Then this map is clearly surjective, and thus M is a quotient of
L
M
A
A.
If
A
is not semi-simple, then it turns out it is rather easy to figure out radical
of M, at least if M is finitely-generated.
Lemma.
Let
A
be left Artinian, and
M
a finitely generated left
A
-module, then
J(A)M = Rad(M).
Proof.
Let
M
0
be a maximal submodule of
M
. Then
M/M
0
is simple, and is in
fact A/I for some maximal left ideal I. Then we have
J(A)
M
M
0
= 0,
since J(A) < I. Therefore J(A)M ≤ M
0
. So J(A)M ≤ Rad(M).
Conversely, we know
M
J(A)M
is an
A/J
(
A
)-module, and is hence completely
reducible as
A/J
(
A
) is semi-simple (and left Artinian). Since an
A
-submodule
of
M
J(A)M
is the same as an
A/J
(
A
)-submodule, it follows that it is completely
reducible as an A-module as well. So
Rad
M
J(A)M
= 0,
and hence Rad(M) ≤ J(A)M.
Proposition. Let A be left Artinian. Then
(i) J(A) is nilpotent, i.e. there exists some r such that J(A)
r
= 0.
(ii)
If
M
is a finitely-generated left
A
-module, then it is both left Artinian and
left Noetherian.
(iii) A is left Noetherian.
Proof.
(i)
Since
A
is left-Artinian, and
{J
(
A
)
r
:
r ∈ N}
is a descending chain of
ideals, it must eventually be constant. So
J
(
A
)
r
=
J
(
A
)
r+1
for some
r
. If
this is non-zero, then again using the descending chain condition, we see
there is a left ideal
I
with
J
(
A
)
r
I 6
= 0 that is minimal with this property
(one such ideal exists, say J(A) itself).
Now pick
x ∈ I
with
J
(
A
)
r
x 6
= 0. Since
J
(
A
)
2r
=
J
(
A
)
r
, it follows
that
J
(
A
)
r
(
J
(
A
)
r
x
)
6
= 0. So by minimality,
J
(
A
)
r
x ≥ I
. But the other
inclusion clearly holds. So they are equal. So there exists some
a ∈ J
(
A
)
r
with x = ax. So
(1 − a)x = 0.
But 1 − a is a unit. So x = 0. This is a contradiction. So J(A)
r
= 0.
(ii)
Let
M
i
=
J
(
A
)
i
M
. Then
M
i
/M
i+1
is annihilated by
J
(
A
), and hence
completely reducible (it is a module over semi-simple
A/J
(
A
)). Since
M
is a finitely generated left
A
-module for a left Artinian algebra, it satisfies
the descending chain condition for submodules (exercise), and hence so
does M
i
/M
i+1
.
So we know
M
i
/M
i+1
is a finite sum of simple modules, and therefore
satisfies the ascending chain condition. So
M
i
/M
i+1
is left Noetherian,
and hence M is (exercise).
(iii) Follows from (ii) since A is a finitely-generated left A-module.