Commutative Algebra 8

Generated Submodule

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 S\subseteq M, let \Sigma denote the set of all submodules of M containing S. Note that \Sigma is non-empty since M\in \Sigma. Now, the submodule of M generated by S is:

(S) := \cap \{N : N\in \Sigma\}.

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 N_i \subseteq M, the sum \sum_i N_i is simply the submodule of M generated by their union \cup_i N_i.

Now (S) has a very simple description.

Lemma 1.

For any subset S of M, the submodule (S)\subseteq M consists of the set of all finite linear combinations

a_1 m_1 + \ldots + a_k m_k, \quad a_i \in A, m_i \in S.

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 M = (S) for some finite subset S of M.

finitely generated ideal is an ideal which is finitely generated as an A-module.

warningIt 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 \mathfrak a \subseteq A where A = (1) is finitely generated but \mathfrak a is not.

E.g. let A = \mathbb R[x^{1/n} : n = 1, 2, \ldots]; more explicitly

A = \{ c_0 x^{e_0} + c_1 x^{e_1} + \ldots + c_k x^{e_k} : c_0, \ldots, c_k \in \mathbb R, e_i \in \mathbb Q_{\ge 0} \}.

Let \mathfrak m\subset A be the set of all sums with only positive e_i. This ideal is not finitely generated, since any finite subset S \subset \mathfrak m is contained in x^{1/n} \mathbb R[x^{1/n}] for some n > 0 so x^{1/(n+1)} \in \mathfrak m lies outside (S).


Linear Dependence

Note that if S = \{m_1, \ldots, m_k\} \subseteq M is a finite subset, there is a unique A-linear

f : A^n \to M, \qquad (a_1, \ldots, a_n) \mapsto a_1 m_1 + \ldots + a_k m_k.

In fact, such a map exists for any finite sequence m_1, \ldots, m_k, 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 f : A^n \to M for some n > 0. When is f injective?

Proposition 1.

Let S = \{m_1, \ldots, m_k\} \subseteq M and f be as above; f is injective if and only if:

  • for any a_1, \ldots, a_k \in A, if a_1 m_1 + \ldots + a_k m_k = 0, then all a_i = 0.

When that happens, we say that m_1, \ldots, m_k 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 S\subseteq M.

Note: the zero module is a free module of rank zero since \emptyset generates M.

Subtle question: is the rank well-defined? In other words, if finite subsets S, S'\subseteq M are both bases, must we have |S| = |S'|? 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 N, N'\subseteq M we have

N / (N\cap N') \cong (N + N') / N', \quad n + (N\cap N') \mapsto n + N'.

  • (Third Isomorphism Theorem) If we have submodules P\subseteq N \subseteq M, then treating N/P as a submodule of M/P we have

(M/P)\ /\ (N/P) \cong\ M/N, \quad \overline m + (N/P) \mapsto m + N.


We have written m +P \in M/P as \overline m to avoid cluttering up the notation.


For the first claim, map N \to (N + N')/N' by n\mapsto n + N'.

  • The kernel of this map is \{n \in N : n + N' = 0 + N'\} = N\cap N'.
  • The map is surjective since any (n + n') + N' for n\in N, n'\in N' is just n+N'.

Hence by the first isomorphism theorem, we get our result.

The second claim is similar: map M/P \to M/N via m+P \mapsto m+N, 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 P \subseteq N \subseteq M in the first case, it corresponds to N/P \subseteq M/P in the second.


[Diagram: correspondence between submodules – arrows indicate inclusion.]

The correspondence respects:

  • containment: N/P \subseteq N'/P if and only if N\subseteq N';
  • intersection: \cap_i (N_i/P) = (\cap_i N_i)/P;
  • sum: \sum_i (N_i/P) = (\sum_i N_i)/P.

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 N_i is the largest submodule contained in every N_i. A similar property exists for the sum.

Finally the case for product is tricky: even if N contains P, \mathfrak a N may not contain P. However, \mathfrak a N + P does and we have:

\mathfrak a (N/P) = (\mathfrak a N + P)/P.

Exercise. Prove this result.


Hom Module

For A-modules M and N, let \mathrm{Hom}_A(M, N) be the set of all A-linear maps f:M\to N.

Proposition 2.

\mathrm{Hom}_A(M, N) is an A-module, under the following operations. For A-linear maps f, g : M\to N and a\in A, we have

  • f+g : M\to N, \ (f+g)(m) := f(m) + g(m),
  • a\cdot f : M\to N, \ (af)(m) := a\cdot f(m).

Proof. Easy exercise. ♦

This generalizes the case of vector spaces, where the set of all linear maps f:V\to W of vector spaces, forms a vector space.

The Hom construction thus creates a new A-module out of two existing ones. We have:

Proposition 3.

Let f:M \to M' and g:N\to N' be A-linear maps of A-modules. These induce A-linear maps

\begin{aligned} f^* : \mathrm{Hom}_A(M', N) \longrightarrow \mathrm{Hom} _A(M, N), \quad &h \mapsto h\circ f\\ g_* : \mathrm{Hom}_A(M, N) \longrightarrow \mathrm{Hom}_A(M, N'), \quad &h\mapsto g\circ h.\end{aligned}

Proof. Easy exercise. ♦


Suppose M = A/aA for some a\in A, considered as an A-module (not a ring!). Then \mathrm{Hom}_A(M, N) corresponds to the submodule \{n\in N: a n = 0\} of N. Indeed any A-linear map f : A/aA \to N is determined by the image f(1 + aA) \in N. This can be any element n\in N as long as a n =0.

[We usually say n is annihilated by a\in A. This will be covered in greater detail later.]

E.g. if A = \mathbb Z and M = \mathbb Z/2\mathbb Z, then \mathrm{Hom}_A(M, N) picks out the 2-torsion subgroup of N.

Now if g : N\to N' is an A-linear map, the above map g_* can be interpreted as follows:


where the composed map is simply g restricted to \{n : N : an = 0\}.

The upshot is: the process of taking the a-torsion elements of a module can be expressed by Hom. We write this process as N \mapsto \mathrm{Hom}_A(A/aA, N), or simply \mathrm{Hom}_A(A/aA, -). All these can be formalized in the language of category theory.


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