3Modules
IB Groups, Rings and Modules
3.2 Direct sums and free modules
We’ve been secretly using the direct sum in many examples, but we shall define
it properly now.
Definition (Direct sum of modules). Let
M
1
, M
2
, ··· , M
k
be
R
-modules. The
direct sum is the R-module
M
1
⊕ M
2
⊕ ··· ⊕ M
k
,
which is the set M
1
× M
2
× ··· × M
k
, with addition given by
(m
1
, ··· , m
k
) + (m
0
1
, ··· , m
0
k
) = (m
1
+ m
0
1
, ··· , m
k
+ m
0
k
),
and the R-action given by
r · (m
1
, ··· , m
k
) = (rm
1
, ··· , rm
k
).
We’ve been using one example of the direct sum already, namely
R
n
= R ⊕ R ⊕ ··· ⊕ R
| {z }
n times
.
Recall we said modules are like vector spaces. So we can try to define things like
basis and linear independence. However, we will fail massively, since we really
can’t prove much about them. Still, we can define them.
Definition (Linear independence). Let
m
1
, ··· , m
k
∈ M
. Then
{m
1
, ··· , m
k
}
is linearly independent if
k
X
i=1
r
i
m
i
= 0
implies r
1
= r
2
= ··· = r
k
= 0.
Lots of modules will not have a basis in the sense we are used to. The next
best thing would be the following:
Definition (Freely generate). A subset S ⊆ M generates M freely if
(i) S generates M
(ii)
Any set function
ψ
:
S → N
to an
R
-module
N
extends to an
R
-module
map θ : M → N.
Note that if
θ
1
, θ
2
are two such extensions, we can consider
θ
1
−θ
2
:
M → N
.
Then
θ
1
− θ
2
sends everything in
S
to 0. So
S ⊆ ker
(
θ
1
− θ
2
)
≤ M
. So the
submodule generated by S lies in ker(θ
1
− θ
2
) too. But this is by definition M.
So
M ≤ ker
(
θ
1
− θ
2
)
≤ M
, i.e. equality holds. So
θ
1
− θ
2
= 0. So
θ
1
=
θ
2
. So
any such extension is unique.
Thus, what this definition tells us is that giving a map from
M
to
N
is
exactly the same thing as giving a function from S to N.
Definition (Free module and basis). An
R
-module is free if it is freely generated
by some subset S ⊆ M, and S is called a basis.
We will soon prove that if
R
is a field, then every module is free. However, if
R is not a field, then there are non-free modules.
Example. The
Z
-module
Z/
2
Z
is not freely generated. Suppose
Z/
2
Z
were
generated by some
S ⊆ Z/
2
Z
. Then this can only possibly be
S
=
{
1
}
. Then
this implies there is a homomorphism
θ
:
Z/
2
Z → Z
sending 1 to 1. But it does
not send 0 = 1 + 1 to 1 + 1, since homomorphisms send 0 to 0. So
Z/
2
Z
is not
freely generated.
We now want to formulate free modules in a way more similar to what we
do in linear algebra.
Proposition. For a subset
S
=
{m
1
, ··· , m
k
} ⊆ M
, the following are equivalent:
(i) S generates M freely.
(ii) S generates M and the set S is independent.
(iii) Every element of M is uniquely expressible as
r
1
m
1
+ r
2
m
2
+ ··· + r
k
m
k
for some r
i
∈ R.
Proof.
The fact that (ii) and (iii) are equivalent is something we would expect
from what we know from linear algebra, and in fact the proof is the same. So
we only show that (i) and (ii) are equivalent.
Let S generate M freely. If S is not independent, then we can write
r
1
m
1
+ ··· + r
k
m
k
= 0,
with
r
i
∈ R
and, say,
r
1
non-zero. We define the set function
ψ
:
S → R
by
sending
m
1
7→
1
R
and
m
i
7→
0 for all
i 6
= 1. As
S
generates
M
freely, this
extends to an R-module homomorphism θ : M → R.
By definition of a homomorphism, we can compute
0 = θ(0)
= θ(r
1
m
1
+ r
2
m
2
+ ··· + r
k
m
k
)
= r
1
θ(m
1
) + r
2
θ(m
2
) + ··· + r
k
θ(m
k
)
= r
1
.
This is a contradiction. So S must be independent.
To prove the other direction, suppose every element can be uniquely written
as
r
1
m
1
+
···
+
r
k
m
k
. Given any set function
ψ
:
S → N
, we define
θ
:
M → N
by
θ(r
1
m
1
+ ··· + r
k
m
k
) = r
1
ψ(m
1
) + ··· + r
k
ψ(m
k
).
This is well-defined by uniqueness, and is clearly a homomorphism. So it follows
that S generates M freely.
Example. The set
{
2
,
3
} ∈ Z
generates
Z
. However, they do not generate
Z
freely, since
3 · 2 + (−2) · 3 = 0.
Recall from linear algebra that if a set
S
spans a vector space
V
, and it is not
independent, then we can just pick some useless vectors and throw them away
in order to get a basis. However, this is no longer the case in modules. Neither
2 nor 3 generate Z.
Definition (Relations). If
M
is a finitely-generated
R
-module, we have shown
that there is a surjective
R
-module homomorphism
φ
:
R
k
→ M
. We call
ker
(
φ
)
the relation module for those generators.
Definition (Finitely presented module). A finitely-generated module is finitely
presented if we have a surjective homomorphism
φ
:
R
k
→ M
and
ker φ
is finitely
generated.
Being finitely presented means I can tell you everything about the module
with a finite amount of paper. More precisely, if
{m
1
, ··· , m
k
}
generate
M
and
{n
1
, n
2
, ··· , n
`
} generate ker(φ), then each
n
i
= (r
i1
, ···r
ik
)
corresponds to the relation
r
i1
m
1
+ r
i2
m
2
+ ··· + r
ik
m
k
= 0
in
M
. So
M
is the module generated by writing down
R
-linear combinations
of
m
1
, ··· , m
k
, and saying two elements are the same if they are related to one
another by these relations. Since there are only finitely many generators and
finitely many such relations, we can specify the module with a finite amount of
information.
A natural question we might ask is if
n 6
=
m
, then are
R
n
and
R
m
the same?
In vector spaces, they obviously must be different, since basis and dimension are
well-defined concepts.
Proposition (Invariance of dimension/rank). Let
R
be a non-zero ring. If
R
n
∼
=
R
m
as R-modules, then n = m.
We know this is true if
R
is a field. We now want to reduce this to the case
where R is a ring.
If
R
is an integral domain, then we can produce a field by taking the field of
fractions, and this might be a good starting point. However, we want to do this
for general rings. So we need some more magic.
We will need the following construction:
Let I C R be an ideal, and let M be an R-module. We define
IM = {am ∈ M : a ∈ I, m ∈ M } ≤ M.
So we can take the quotient module M/IM , which is an R-module again.
Now if b ∈ I, then its action on M/IM is
b(m + IM) = bm + IM = IM.
So everything in
I
kills everything in
M/IM
. So we can consider
M/IM
as an
R/I module by
(r + I) · (m + IM) = r · m + IM.
So we have proved that
Proposition. If
I C R
is an ideal and
M
is an
R
-module, then
M/IM
is an
R/I module in a natural way.
We next need to use the following general fact:
Proposition. Every non-zero ring has a maximal ideal.
This is a rather strong statement, since it talks about “all rings”, and we can
have weird rings. We need to use a more subtle argument, namely via Zorn’s
lemma. You probably haven’t seen it before, in which case you might want to
skip the proof and just take the lecturer’s word on it.
Proof.
We observe that an ideal
I C R
is proper if and only if 1
R
6∈ I
. So every
increasing union of proper ideals is proper. Then by Zorn’s lemma, there is a
maximal ideal (Zorn’s lemma says if an arbitrary union of increasing things is
still a thing, then there is a maximal such thing, roughly).
With these two notions, we get
Proposition (Invariance of dimension/rank). Let
R
be a non-zero ring. If
R
n
∼
=
R
m
as R-modules, then n = m.
Proof.
Let
I
be a maximal ideal of
R
. Suppose we have
R
n
∼
=
R
m
. Then we
must have
R
n
IR
n
∼
=
R
m
IR
m
,
as R/I modules.
But staring at it long enough, we figure that
R
n
IR
n
∼
=
R
I
n
,
and similarly for
m
. Since
R/I
is a field, the result follows by linear algebra.
The point of this proposition is not the result itself (which is not too inter-
esting), but the general constructions used behind the proof.