Since the intersection of an arbitrary family of submodules of M is a submodule, we have the concept of a submodule generated by a subset.
Given any subset , let denote the set of all submodules of M containing S. Note that is non-empty since . Now, the submodule of M generated by S is:
This is the “smallest submodule of M containing S“; to be precise, (a) it is a submodule of M containing S, and (b) any submodule N of M containing S must also contain (S). With this, we see that given a family of submodules , the sum is simply the submodule of M generated by their union .
Now (S) has a very simple description.
For any subset S of M, the submodule consists of the set of all finite linear combinations
Thus we may sometimes call (S) the span of S (over the base ring A).
A module M is said to be finitely generated (or just finite) if for some finite subset S of M.
A finitely generated ideal is an ideal which is finitely generated as an A-module.
It is not true that every submodule of a finitely generated module must be finitely generated. In fact, there exist rings with ideals which are not finitely generated, so this gives where A = (1) is finitely generated but is not.
E.g. let ; more explicitly
Let be the set of all sums with only positive . This ideal is not finitely generated, since any finite subset is contained in for some n > 0 so lies outside (S).
Note that if is a finite subset, there is a unique A-linear
In fact, such a map exists for any finite sequence , i.e. some of the elements may repeat.
This map is surjective if and only if S generates M. Thus M is finitely generated if and only if there is a surjective A-linear map for some n > 0. When is f injective?
Let and f be as above; f is injective if and only if:
- for any , if , then all .
When that happens, we say that are linearly independent.
Proof. Easy exercise. ♦
We make two important definitions here, inherited from linear algebra.
- If S is linearly independent and generates M, we say that S is a basis.
- We say M is finite free if it has a finite subset which is a basis.
- The rank of a finite free M is the size of S for any basis .
Note: the zero module is a free module of rank zero since generates M.
Subtle question: is the rank well-defined? In other words, if finite subsets are both bases, must we have ? This question will be answered a few articles later.
Two More Isomorphism Theorems
Just like the case of group theory, we have the following.
Let M be any A-module.
- (Second Isomorphism Theorem) For any submodules we have
- (Third Isomorphism Theorem) If we have submodules , then treating as a submodule of we have
We have written as to avoid cluttering up the notation.
For the first claim, map by .
- The kernel of this map is .
- The map is surjective since any for is just .
Hence by the first isomorphism theorem, we get our result.
The second claim is similar: map via , which is a well-defined homomorphism. Clearly the map is surjective; the kernel is just N/P; now apply the first isomorphism theorem. ♦
Correspondence of Submodules
Finally, for any submodule P of M, there is a bijective correspondence between
- submodules of M containing P, and
- submodules of M/P.
Specifically, if N satisfies in the first case, it corresponds to in the second.
[Diagram: correspondence between submodules – arrows indicate inclusion.]
The correspondence respects:
- containment: if and only if ;
- intersection: ;
- sum: .
Tip: you can prove all these directly, or you can deduce the second and third properties from the first by noting: the intersection of submodules is the largest submodule contained in every . A similar property exists for the sum.
Finally the case for product is tricky: even if N contains P, may not contain P. However, does and we have:
Exercise. Prove this result.
For A-modules M and N, let be the set of all A-linear maps .
is an A-module, under the following operations. For A-linear maps and , we have
Proof. Easy exercise. ♦
This generalizes the case of vector spaces, where the set of all linear maps of vector spaces, forms a vector space.
The Hom construction thus creates a new A-module out of two existing ones. We have:
Let and be A-linear maps of A-modules. These induce A-linear maps
Proof. Easy exercise. ♦
Suppose for some , considered as an A-module (not a ring!). Then corresponds to the submodule of N. Indeed any A-linear map is determined by the image . This can be any element as long as .
[We usually say n is annihilated by . This will be covered in greater detail later.]
E.g. if and , then picks out the 2-torsion subgroup of N.
Now if is an A-linear map, the above map can be interpreted as follows:
where the composed map is simply g restricted to .
The upshot is: the process of taking the a-torsion elements of a module can be expressed by Hom. We write this process as , or simply . All these can be formalized in the language of category theory.