3Modules

IB Groups, Rings and Modules

3.1 Definitions and examples
Definition
(Module)
.
Let
R
be a commutative ring. We say a quadruple
(M, +, 0
M
, ·) is an R-module if
(i) (M, +, 0
M
) is an abelian group
(ii) The operation · : R × M M satisfies
(a) (r
1
+ r
2
) · m = (r
1
· m) + (r
2
· m);
(b) r · (m
1
+ m
2
) = (r · m
1
) + (r · m
2
);
(c) r
1
· (r
2
· m) = (r
1
· r
2
) · m; and
(d) 1
R
· M = m.
Note that there are two different additions going on addition in the ring
and addition in the module, and similarly two notions of multiplication. However,
it is easy to distinguish them since they operate on different things. If needed,
we can make them explicit by writing, say, +
R
and +
M
.
We can imagine modules as rings acting on abelian groups, just as groups
can act on sets. Hence we might say
R
acts on
M
to mean
M
is an
R
-module.
Example.
Let
F
be a field. An
F
-module is precisely the same as a vector space
over F (the axioms are the same).
Example. For any ring R, we have the R-module R
n
= R × R × ··· × R via
r · (r
1
, ··· , r
n
) = (rr
1
, ··· , rr
n
),
using the ring multiplication. This is the same as the definition of the vector
space F
n
for fields F.
Example. Let I C R be an ideal. Then it is a R-module via
r ·
M
a = r ·
R
a, r
1
+
M
r
2
= r
1
+
R
r
2
.
Also, R/I is an R-module via
r ·
M
(a + I) = (r ·
R
a) + I,
Example.
A
Z
-module is precisely the same as an abelian group. For
A
an
abelian group, we have
Z × A A
(n, a) 7→ a + ··· + a
| {z }
n times
,
a + ··· + a
| {z }
n times
= (a) + ··· + (a)
| {z }
n times
,
and adding something to itself 0 times is just 0.
This definition is essentially forced upon us, since by the axioms of a module,
we must have (1, a) 7→ a. Then we must send, say, (2, a) = (1 + 1, a) 7→ a + a.
Example.
Let
F
be a field and
V
a vector space over
F
, and
α
:
V V
be a
linear map. Then V is an F[X]-module via
F [X] × V V
(f, v) 7→ f(α)(v).
This is a module.
Note that we cannot just say that
V
is an
F
[
X
] module. We have to specify
the α as well. Picking a different α will give a different F[X]-module structure.
Example.
Let
φ
:
R S
be a homomorphism of rings. Then any
S
-module
M
may be considered as an R module via
R × M M
(r, m) 7→ φ(r) ·
M
m.
Definition
(Submodule)
.
Let
M
be an
R
-module. A subset
N M
is an
R
-submodule if it is a subgroup of (
M,
+
,
0
M
), and if
n N
and
r R
, then
rn N. We write N M .
Example.
We know
R
itself is an
R
-module. Then a subset of
R
is a submodule
if and only if it is an ideal.
Example.
A subset of an
F
-module
V
, where
F
is a field, is a
F
-submodule if
and only if it is a vector subspace of V .
Definition
(Quotient module)
.
Let
N M
be an
R
-submodule. The quotient
module M/N is the set of N-cosets in (M, +, 0
M
), with the R action given by
r · (m + N) = (r · m) + N.
It is easy to check this is well-defined and is indeed a module.
Note that modules are different from rings and groups. In groups, we had
subgroups, and we have some really nice ones called normal subgroups. We are
only allowed to quotient by normal subgroups. In rings, we have subrings and
ideals, which are unrelated objects, and we only quotient by ideals. In modules,
we only have submodules, and we can quotient by arbitrary submodules.
Definition
(
R
-module homomorphism and isomorphism)
.
A function
f
:
M
N
between
R
-modules is a
R
-module homomorphism if it is a homomorphism of
abelian groups, and satisfies
f(r · m) = r · f (m)
for all r R and m M.
An isomorphism is a bijective homomorphism, and two
R
-modules are
isomorphic if there is an isomorphism between them.
Note that on the left, the multiplication is the action in
M
, while on the
right, it is the action in N .
Example.
If
F
is a field and
V, W
are
F
-modules (i.e. vector spaces over
F
),
then an F-module homomorphism is precisely an F-linear map.
Theorem
(First isomorphism theorem)
.
Let
f
:
M N
be an
R
-module
homomorphism. Then
ker f = {m M : f(m) = 0} M
is an R-submodule of M. Similarly,
im f = {f(m) : m M} N
is an R-submodule of N. Then
M
ker f
=
im f.
We will not prove this again. The proof is exactly the same.
Theorem (Second isomorphism theorem). Let A, B M . Then
A + B = {m M : m = a + b for some a A, b B} M,
and
A B M.
We then have
A + B
A
=
B
A B
.
Theorem (Third isomorphism theorem). Let N L M. Then we have
M
L
=
M
N
L
N
.
Also, we have a correspondence
{submodules of M/N} {submodules of M which contain N }
It is an exercise to see what these mean in the cases where
R
is a field, and
modules are vector spaces.
We now have something new. We have a new concept that was not present
in rings and groups.
Definition
(Annihilator)
.
Let
M
be a
R
-module, and
m M
. The annihilator
of m is
Ann(m) = {r R : r · m = 0}.
For any set S M , we define
Ann(S) = {r R : r · m = 0 for all m S} =
\
mS
Ann(m).
In particular, for the module M itself, we have
Ann(M) = {r R : r · m = 0 for all m M} =
\
mM
Ann(m).
Note that the annihilator is a subset of
R
. Moreover it is an ideal if
r ·m
= 0 and
s ·m
= 0, then (
r
+
s
)
·m
=
r ·m
+
s ·m
= 0. So
r
+
s Ann
(
m
).
Moreover, if r · m = 0, then also (sr) · m = s · (r · m) = 0. So sr Ann(m).
What is this good for? We first note that any
m M
generates a submodule
Rm as follows:
Definition
(Submodule generated by element)
.
Let
M
be an
R
-module, and
m M. The submodule generated by m is
Rm = {r · m M : r R}.
We consider the R-module homomorphism
φ : R M
r 7→ rm.
This is clearly a homomorphism. Then we have
Rm = im(φ),
Ann(m) = ker(φ).
The conclusion is that
Rm
=
R/ Ann(m).
As we mentioned, rings acting on modules is like groups acting on sets. We can
think of this as the analogue of the orbit stabilizer theorem.
In general, we can generate a submodule with many elements.
Definition
(Finitely generated module)
.
An
R
-module
M
is finitely generated
if there is a finite list of elements m
1
, ··· , m
k
such that
M = Rm
1
+ Rm
2
+ ··· + Rm
k
= {r
1
m
1
+ r
2
m
2
+ ··· + r
k
m
k
: r
i
R}.
This is in some sense analogous to the idea of a vector space being finite-
dimensional. However, it behaves much more differently.
While this definition is rather concrete, it is often not the most helpful
characterization of finitely-generated modules. Instead, we use the following
lemma:
Lemma.
An
R
-module
M
is finitely-generated if and only if there is a surjective
R-module homomorphism f : R
k
M for some finite k.
Proof. If
M = Rm
1
+ Rm
2
+ ··· + Rm
k
,
we define f : R
k
M by
(r
1
, ··· , r
k
) 7→ r
1
m
1
+ ··· + r
k
m
k
.
It is clear that this is an
R
-module homomorphism. This is by definition
surjective. So done.
Conversely, given a surjection f : R
k
M, we let
m
i
= f(0, 0, ··· , 0, 1, 0, ··· , 0),
where the 1 appears in the ith position. We now claim that
M = Rm
1
+ Rm
2
+ ··· + Rm
k
.
So let m M. As f is surjective, we know
m = f(r
1
, r
2
, ··· , r
k
)
for some r
i
. We then have
f(r
1
, r
2
, ··· , r
k
) = f((r
1
, 0, ··· , 0) + (0, r
2
, 0, ··· , 0) + ··· + (0, 0, ··· , 0, r
k
))
= f(r
1
, 0, ··· , 0) + f(0, r
2
, 0, ··· , 0) + f(0, 0, ··· , 0, r
k
)
= r
1
f(1, 0, ··· , 0) + r
2
f(0, 1, 0, ··· , 0) + r
k
f(0, 0, ··· , 0, 1)
= r
1
m
1
+ r
2
m
2
+ ··· + r
k
m
k
.
So the m
i
generate M.
This view is a convenient way of thinking about finitely-generated modules.
For example, we can immediately prove the following corollary:
Corollary.
Let
N M
and
M
be finitely-generated. Then
M/N
is also finitely
generated.
Proof.
Since
m
is finitely generated, we have some surjection
f
:
R
k
M
.
Moreover, we have the surjective quotient map
q
:
M M/N
. Then we get the
following composition
R
k
M M/N,
f q
which is a surjection, since it is a composition of surjections. So
M/N
is finitely
generated.
It is very tempting to believe that if a module is finitely generated, then its
submodules are also finitely generated. It would be very wrong to think so.
Example.
A submodule of a finitely-generated module need not be finitely
generated.
We let
R
=
C
[
X
1
, X
2
, ···
]. We consider the
R
-module
M
=
R
, which is
finitely generated (by 1). A submodule of the ring is the same as an ideal.
Moreover, an ideal is finitely generated as an ideal if and only if it is finitely
generated as a module. We pick the submodule
I = (X
1
, X
2
, ···),
which we have already shown to be not finitely-generated. So done.
Example.
For a complex number
α
, the ring
Z
[
α
] (i.e. the smallest subring
of
C
containing
α
) is a finitely-generated as a
Z
-module if and only if
α
is an
algebraic integer.
Proof is left as an exercise for the reader on the last example sheet. This allows
us to prove that algebraic integers are closed under addition and multiplication,
since it is easier to argue about whether Z[α] is finitely generated.