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.

Definition.

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).

Definition.

A module M is said to be finitely generated (or just finite) if $M = (S)$ for some finite subset S of M.

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

warning 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 $\mathfrak a \subseteq A$ 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).

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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.

Definition.

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.

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Two More Isomorphism Theorems

Just like the case of group theory, we have the following.

Theorem.

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

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

Note

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

Proof

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. ♦

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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.

submodules_correspondence

[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.

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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. ♦

Example

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:

equation_hom_1

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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Commutative Algebra 8

Generated Submodule

Linear Dependence

Two More Isomorphism Theorems

Correspondence of Submodules

Hom Module

Example

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Commutative Algebra 8

Generated Submodule

Linear Dependence

Two More Isomorphism Theorems

Correspondence of Submodules

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Example

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