1Artinian algebras
III Algebras
1.4 Projectives and blocks
In general, if
A
is not semi-simple, then it is not possible to decompose
A
as a
direct sum of simple modules. However, what we can do is to decompose it as a
direct sum of indecomposable projectives.
We begin with the definition of a projective module.
Definition
(Projective module)
.
An
A
-module is projective
P
if given modules
M and M
0
and maps
P
M
0
M 0
α
θ
,
then there exists a map
β
:
P → M
0
such that the following diagram commutes:
P
M
0
M 0
α
β
θ
.
Equivalently, if we have a short exact sequence
0 N M
0
M 0,
then the sequence
0 Hom(P, N) Hom(P, M
0
) Hom(P, M) 0
is exact.
Note that we get exactness at
Hom
(
P, N
) and
Hom
(
P, M
0
) for any
P
at all.
Projective means it is also exact at Hom(P, M).
Example. Free modules are always projective.
In general, projective modules are “like” free modules. We all know that free
modules are nice, and most of the time, when we want to prove things about
free modules, we are just using the property that they are projective. It is also
possible to understand projective modules in an algebro-geometric way — they
are “locally free” modules.
It is convenient to characterize projective modules as follows:
Lemma. The following are equivalent:
(i) P is projective.
(ii) Every surjective map φ : M → P splits, i.e.
M
∼
=
ker φ ⊕ N
where N
∼
=
P .
(iii) P is a direct summand of a free module.
Proof.
– (i) ⇒ (ii): Consider the following lifting problem:
P
M P 0
φ
,
The lifting gives an embedding of
P
into
M
that complements
ker φ
(by
the splitting lemma, or by checking it directly).
–
(ii)
⇒
(iii): Every module admits a surjection from a free module (e.g. the
free module generated by the elements of P )
–
(iii)
⇒
(i): It suffices to show that direct summands of projectives are
projective. Suppose P is is projective, and
P
∼
=
A ⊕ B.
Then any diagram
A
M
0
M 0
α
θ
,
can be extended to a diagram
A ⊕ B
M
0
M 0
˜α
θ
,
by sending
B
to 0. Then since
A ⊕ B
∼
=
P
is projective, we obtain a lifting
A ⊕ B → M
0
, and restricting to A gives the desired lifting.
Our objective is to understand the direct summands of a general Artinian
k
-algebra
A
, not necessarily semi-simple. Since
A
is itself a free
A
-module, we
know these direct summands are always projective.
Since
A
is not necessarily semi-simple, it is in general impossible to decompose
it as a direct sum of simples. What we can do, though, is to decompose it as a
direct sum of indecomposable modules.
Definition
(Indecomposable)
.
A non-zero module
M
is indecomposable if
M
cannot be expressed as the direct sum of two non-zero submodules.
Note that since
A
is (left) Artinian, it can always be decomposed as a finite
sum of indecomposable (left) submodules. Sometimes, we are also interested in
decomposing A as a sum of (two-sided) ideals. These are called blocks.
Definition
(Block)
.
The blocks are the direct summands of
A
that are inde-
composable as ideals.
Example. If A is semi-simple Artinian, then Artin-Wedderburn tells us
A =
M
M
n
i
(D
i
),
and the M
n
i
(D
i
) are the blocks.
We already know that every Artinian module can be decomposed as a direct
sum of indecomposables. The first question to ask is whether this is unique. We
note the following definitions:
Definition
(Local algebra)
.
An algebra is local if it has a unique maximal left
ideal, which is J(A), which is the unique maximal right ideal.
If so, then
A/J
(
A
) is a division algebra. This name, of course, comes from
algebraic geometry (cf. local rings).
Definition
(Unique decomposition property)
.
A module
M
has the unique
decomposition property if
M
is a finite direct sum of indecomposable modules,
and if
M =
m
M
i=1
M
i
=
n
M
i=1
M
0
i
,
then n = m, and, after reordering, M
i
= M
0
i
.
We want to prove that
A
as an
A
-module always has the unique decomposi-
tion property. The first step is the following criterion for determining unique
decomposition property.
Theorem
(Krull–Schmidt theorem)
.
Suppose
M
is a finite sum of indecom-
posable
A
-modules
M
i
, with each
End
(
M
i
) local. Then
M
has the unique
decomposition property.
Proof. Let
M =
m
M
i=1
M
i
=
n
M
i=1
M
0
i
.
We prove by induction on
m
. If
m
= 1, then
M
is indecomposable. Then we
must have n = 1 as well, and the result follows.
For m > 1, we consider the maps
α
i
: M
0
i
M M
1
β
i
: M
1
M M
0
i
We observe that
id
M
1
=
n
X
i=1
α
i
◦ β
i
: M
1
→ M
1
.
Since
End
A
(
M
1
) is local, we know some
α
i
◦ β
i
must be invertible, i.e. a unit, as
they cannot all lie in the Jacobson radical. We may wlog assume
α
1
◦ β
1
is a
unit. If this is the case, then both
α
1
and
β
1
have to be invertible. So
M
1
∼
=
M
0
1
.
Consider the map id −θ = φ, where
θ : M M
1
M
0
1
M
L
m
i=2
M
i
M.
α
−1
1
Then φ(M
0
1
) = M
1
. So φ|
M
0
1
looks like α
1
. Also
φ
m
M
i=2
M
i
!
=
m
M
i=2
M
i
,
So
φ|
L
m
i=2
M
i
looks like the identity map. So in particular, we see that
φ
is
surjective. However, if φ(x) = 0, this says x = θ(x), So
x ∈
m
M
i=2
M
i
.
But then
θ
(
x
) = 0. Thus
x
= 0. Thus
φ
is an automorphism of
m
with
φ(M
0
1
) = φ(M
1
). So this gives an isomorphism between
m
M
i=2
M
i
=
M
M
1
∼
=
M
M
1
∼
=
n
M
i=2
M
0
i
,
and so we are done by induction.
Now it remains to prove that the endomorphism rings are local. Recall the
following result from linear algebra.
Lemma
(Fitting)
.
Suppose
M
is a module with both the ACC and DCC on
submodules, and let f ∈ End
A
(M). Then for large enough n, we have
M = im f
n
⊕ ker f
n
.
Proof. By ACC and DCC, we may choose n large enough so that
f
n
: f
n
(M) → f
2n
(M)
is an isomorphism, as if we keep iterating
f
, the image is a descending chain and
the kernel is an ascending chain, and these have to terminate.
If m ∈ M, then we can write
f
n
(m) = f
2n
(m
1
)
for some m
1
. Then
m = f
n
(m
1
) + (m − f
n
(m
1
)) ∈ im f
n
+ ker f
n
,
and also
im f
n
∩ ker f
n
= ker(f
n
: f
n
(M) → f
2n
(M)) = 0.
So done.
Lemma.
Suppose
M
is an indecomposable module satisfying ACC and DCC
on submodules. Then B = End
A
(M) is local.
Proof.
Choose a maximal left ideal of
B
, say
I
. It’s enough to show that if
x 6∈ I
,
then x is left invertible. By maximality of I, we know B = Bx + I. We write
1 = λx + y,
for some
λ ∈ B
and
y ∈ I
. Since
y ∈ I
, it has no left inverse. So it is not an
isomorphism. By Fitting’s lemma and the indecomposability of
M
, we see that
y
m
= 0 for some m. Thus
(1 + y + y
2
+ · · · + y
m−1
)λx = (1 + y + · · · + y
m−1
)(1 − y) = 1.
So x is left invertible.
Corollary.
Let
A
be a left Artinian algebra. Then
A
has the unique decompo-
sition property.
Proof.
We know
A
satisfies the ACC and DCC condition. So
A
A
is a finite
direct sum of indecomposables.
So if
A
is an Artinian algebra, we know
A
can be uniquely decomposed as a
direct sum of indecomposable projectives,
A =
M
P
j
.
For convenience, we will work with right Artinian algebras and right modules
instead of left ones. It turns out that instead of studying projectives in
A
, we
can study idempotent elements instead.
Recall that
End
(
A
A
)
∼
=
A
. The projection onto
P
j
is achieved by left
multiplication by an idempotent e
j
,
P
j
= e
j
A.
The fact that the
A
decomposes as a direct sum of the
P
j
translates to the
condition
X
e
j
= 1, e
i
e
j
= 0
for i 6= j.
Definition
(Orthogonal idempotent)
.
A collection of idempotents
{e
i
}
is or-
thogonal if e
i
e
j
= 0 for i 6= j.
The indecomposability of P
j
is equivalent to e
j
being primitive:
Definition
(Primitive idempotent)
.
An idempotent is primitive if it cannot be
expressed as a sum
e = e
0
+ e
00
,
where e
0
, e
00
are orthogonal idempotents, both non-zero.
We see that giving a direct sum decomposition of
A
is equivalent to finding
an orthogonal collection of primitive idempotents that sum to 1. This is rather
useful, because idempotents are easier to move around that projectives.
Our current plan is as follows — given an Artinian algebra
A
, we can quotient
out by
J
(
A
), and obtain a semi-simple algebra
A/J
(
A
). By Artin–Wedderburn,
we know how we can decompose
A/J
(
A
), and we hope to be able to lift this
decomposition to one of
A
. The point of talking about idempotents instead is
that we know what it means to lift elements.
Proposition.
Let
N
be a nilpotent ideal in
A
, and let
f
be an idempotent of
A/N ≡
¯
A. Then there is an idempotent e ∈ A with f = ¯e.
In particular, we know
J
(
A
) is nilpotent, and this proposition applies. The
proof involves a bit of magic.
Proof.
We consider the quotients
A/N
i
for
i ≥
1. We will lift the idempotents
successively as we increase
i
, and since
N
is nilpotent, repeating this process
will eventually land us in A.
Suppose we have found an idempotent
f
i−1
∈ A/N
i−1
with
¯
f
i−1
=
f
. We
want to find f
i
∈ A/N
i
such that
¯
f
i
= f.
For
i >
1, we let
x
be an element of
A/N
i
with image
f
i−1
in
A/N
i−1
.
Then since
x
2
− x
vansishes in
A/N
i−1
, we know
x
2
− x ∈ N
i−1
/N
i
. Then in
particular,
(x
2
− x)
2
= 0 ∈ A/N
i
. (†)
We let
f
i
= 3x
2
− 2x
3
.
Then by a direct computation using (
†
), we find
f
2
i
=
f
i
, and
f
i
has image
3
f
i−1
−
2
f
i−1
=
f
i−1
in
A/N
i−1
(alternatively, in characteristic
p
, we can use
f
i
= x
p
). Since N
k
= 0 for some k, this process gives us what we want.
Just being able to lift idempotents is not good enough. We want to lift
decompositions as projective indecomposables. So we need to do better.
Corollary. Let N be a nilpotent ideal of A. Let
¯
1 = f
1
+ · · · + f
r
with {f
i
} orthogonal primitive idempotents in A/N. Then we can write
1 = e
1
+ · · · + e
r
,
with {e
i
} orthogonal primitive idempotents in A, and ¯e
i
= f
i
.
Proof. We define a sequence e
0
i
∈ A inductively. We set
e
0
1
= 1.
Then for each
i >
1, we pick
e
0
i
a lift of
f
i
+
· · ·
+
f
t
∈ e
0
i−1
Ae
0
i−1
, since by
inductive hypothesis we know that f
i
+ · · · + f
t
∈ e
0
i−1
Ae
0
i−1
/N. Then
e
0
i
e
0
i+1
= e
0
i+1
= e
0
i+1
e
0
i
.
We let
e
i
= e
0
i
− e
0
i+1
.
Then
¯e
i
= f
i
.
Also, if j > i, then
e
j
= e
0
i+1
e
j
e
0
i+1
,
and so
e
i
e
j
= (e
0
i
− e
0
i+1
)e
0
i+1
e
j
e
0
i+1
= 0.
Similarly e
j
e
i
= 0.
We now apply this lifting of idempotents to
N
=
J
(
A
), which we know is
nilpotent. We know
A/N
is the direct sum of simple modules, and thus the
decomposition corresponds to
¯
1 = f
1
+ · · · + f
t
∈ A/J(A),
and these
f
i
are orthogonal primitive idempotents. Idempotent lifting then gives
1 = e
1
+ · · · + e
t
∈ A,
and these are orthogonal primitive idempotents. So we can write
A =
M
e
j
A =
M
P
i
,
where
P
i
=
e
i
A
are indecomposable projectives, and
P
i
/P
i
J
(
A
) =
S
i
is simple.
By Krull–Schmidt, any indecomposable projective isomorphic to one of these
P
j
.
The final piece of the picture is to figure out when two indecomposable
projectives lie in the same block. Recall that if
M
is a right
A
-module and
e
is
idempotent, then
Me
∼
=
Hom
A
(eA, M).
In particular, if M = fA for some idempotent f, then
Hom(eA, fA)
∼
=
fAe.
However, if e and f are in different blocks, say B
1
and B
2
, then
fAe ∈ B
1
∩ B
2
= 0,
since B
1
and B
2
are (two-sided!) ideals. So we know
Hom(eA, fA) = 0.
So if
Hom
(
eA, fA
)
6
= 0, then they are in the same block. The existence of a
homomorphism can alternatively be expressed in terms of composition factors.
We have seen that each indecomposable projective P has a simple “top”
P/P J(A)
∼
=
S.
Definition
(Composition factor)
.
A simple module
S
is a composition factor
of a module M if there are submodules M
1
≤ M
2
with
M
2
/M
1
∼
=
S.
Suppose
S
is a composition factor of a module
M
. Then we have a diagram
P
M
2
S 0
So by definition of projectivity, we obtain a non-zero diagonal map
P → M
2
≤ M
as shown.
Lemma.
Let
P
be an indecomposable projective, and
M
an
A
-module. Then
Hom(P, M) 6= 0 iff P/P J(A) is a composition factor of M.
Proof.
We have proven
⇒
. Conversely, suppose there is a non-zero map
f
:
P →
M. Then it factors as
S =
P
P J(A)
→
im f
(im f)J(A)
.
Now we cannot have
im f
= (
im f
)
J
(
A
), or else we have
im f
= (
im f
)
J
(
A
)
n
= 0
for sufficiently large
n
since
J
(
A
) is nilpotent. So this map must be injective,
hence an isomorphism. So this exhibits S as a composition factor of M .
We define a (directed) graph whose vertices are labelled by indecomposable
projectives, and there is an edge
P
1
→ P
2
if the top
S
1
of
P
1
is a composition
factor of P
2
.
Theorem.
Indecomposable projectives
P
1
and
P
2
are in the same block if and
only if they lie in the same connected component of the graph.
Proof.
It is clear that
P
1
and
P
2
are in the same connected component, then
they are in the same block.
Conversely, consider a connected component X, and consider
I =
M
P ∈X
P.
We show that this is in fact a left ideal, hence an ideal. Consider any
x ∈ A
. Then
for each P ∈ X, left-multiplication gives a map P → A, and if we decompose
A =
M
P
i
,
then this can be expressed as a sum of maps
f
i
:
P → P
i
. Now such a map can
be non-zero only if
P
is a composition factor of
P
i
. So if
f
i
6
= 0, then
P
i
∈ X
.
So left-multiplication by
x
maps
I
to itself, and it follows that
I
is an ideal.