Elementary Module Theory (II)

Having defined submodules, let’s proceed to quotient modules. Unlike the case of groups and rings, any submodule can give a quotient module without any additional condition imposed.

Definition. Let N be a submodule of M. By definition, it’s an additive subgroup which is normal since (M, +) is abelian. So take the group quotient M/N and define a module structure on it via the scalar multiplication:

$R\times M/N \to M/N,\quad (r, m+N) \mapsto (rm)+N.$

This is called the module quotient.

The only worrying aspect is that scalar multiplication is well-defined: i.e. suppose m+N = m’+N; is it true that (rm)+N = (rm’)+N then? Well:

$\begin{aligned}m+N = m'+N &\implies m-m'\in N \implies r(m-m')\in N\\ &\implies rm-rm'\in N \implies (rm)+N = (rm')+N.\end{aligned}$

The remaining axioms of a module are then easily verified.

Examples

Clearly, M/{0} = M and M/M = {0}.
Let $R = M_n(\mathbf{Z})$ be the ring of n × n matrices with integer entries. Then $M:=\mathbf{Z}^n$ is a module and $N :=2M = \{2\mathbf{v} : \mathbf{v} \in M\}$ is a submodule. The resulting quotient is isomorphic to $(\mathbf{Z}/2)^n.$
Recall that if R = Z, the class of R-modules is precisely the class of abelian groups. The submodules are precisely the subgroups and quotient modules are the quotient groups.
Let $I\subseteq R$ be an ideal of R and M be an R-module. The module quotient M/IM is not only an R-module, but an (R/I)-module as well. Indeed, if $r+I\in R/I$ then multiplying with $m+IM \in M/IM$ gives $rm+IM\in M/IM.$ This is well-defined since if $r+I = r'+I$ and $m+IM = m'+IM$ , then we have $r-r'\in I$ and $m-m'\in IM$ which gives us $rm - r'm' = (r-r')m +r'(m-m')\in IM$ and so rm + I = r’m’ + I. Thus scalar multiplication by R/I is well-defined and it’s easy to show that M/IM is an R/I-module.

Homomorphisms

The definition is straightforward.

Definition. Let M and N be R-modules. A module homomorphism is a function f : M → N such that:

$f(m+m') = f(m) + f(m')$ and $f(rm) = r\cdot f(m)$

for any $m, m'\in M$ and $r\in R.$

Notice that since f is a homomorphism of the underlying additive groups, we automatically have f(0) = 0 and f(-m) = –f(m). As in the case of groups, an injective (resp. surjective / bijective) homomorphism is called a monomorphism (resp. epimorphism / isomorphism).

The following properties are hardly surprising:

Proposition. Let $f : M\to N$ be a homomorphism of R-modules.

If $M'\subseteq M$ is a submodule, then f(M’) is a submodule of N.

If $N'\subseteq N$ is a submodule, then f^-1(N’) is a submodule of M.

In other words, the “push-forward” or “pull-back” of a submodule is also a submodule.

Proof.

The proof is easy: we know that f(M’) is an additive subgroup of N. Next, if $n\in f(M')$ and $r\in R$ , then we can write n = f(m) for some m in M’. This gives rn = r·f(m) = f(rm), which is in f(M’).

Likewise, f^-1(N’) is an additive subgroup of M. If $m\in f^{-1}(N')$ and $r\in R$ , then f(m) lies in N’ so f(rm) = r·f(m) which lies in N’. Hence $rm\in f^{-1}(N')$ and we’re done. ♦

In particular, we have the following special cases.

Definition. For a module homomorphism $f:M\to N$ , the pullback $f^{-1}(0)$ is called the kernel of f; this is a submodule of M, denoted ker(f).

Likewise, $f(M)$ is called the image of f; this is a submodule of N, denoted im(f).

Finally the cokernel of f, denoted coker(f), is the quotient $N/\text{im}(f)$ .

[ Note: the cokernel is kind of unusual since we didn’t encounter it in the case of groups and rings. Its primary purpose is as a “dual” to the kernel. Roughly, the kernel satisfies some universal property; if we reverse the arrows around the resulting universal property describes the cokernel. ]

As before, we have the standard three isomorphism theorems.

First Isomorphism Theorem. Let $f:M\to N$ be a homomorphism of R-modules. Then $M/\text{ker}(f) \cong \text{im}(f).$

Proof.

Construct a map $g:M/\text{ker}(f) \to \text{im}(f)$ which takes $m + \text{ker}(f)$ to f(m). Let’s show that g is injective and well-defined. Indeed:

$m + \text{ker}(f) = m'+\text{ker}(f) \iff m-m'\in \text{ker}(f) \iff f(m-m')=0 \iff f(m) = f(m').$

Finally, g is surjective by the definition of im(f), so we’re done. ♦

Armed with the first isomorphism theorem, the remaining two are a piece of cake.

Second Isomorphism Theorem. Let N, N’ be submodules of M. Then $N/(N\cap N') \cong (N+N')/N'.$

Third Isomorphism Theorem. Let $P\subseteq N\subseteq M$ be submodules. Then N/P is a submodule of M/P and $(M/P)/(N/P) \cong M/N.$

Proof.

For the first result, map N → (N+N’) → (N+N’)/N’ by composing the inclusion $N\subseteq N+N'$ with projection $N+N' \to (N+N')/N'.$ Apply the first isomorphism theorem. The kernel of this map is $\{n\in N : n+N'=0\} = N\cap N'$ while the image is the whole of (N+N’)/N’ since any element (n+n’)+N’ (for $n\in N$ and $n'\in N'$ ) is basically just n+N’.

For the second result, map M/P → M/N by taking m+P to m+N. The map is well-defined since if m+P = m’+P, then m–m’ lies in P and hence N, so m+N = m’+N also. The map is obviously surjective. Finally, the kernel is

$\{m+P\in M/P : m+N = 0\} = \{m+P : m\in N\} = P/N.$

Now apply the first isomorphism theorem. ♦

Let $N\subseteq M$ . As in the case of groups and rings, there’s a correspondence between submodules of M containing N, as well as submodules of M/N.

Theorem. Let N be a submodule of M. We have a bijection:

$\{ \text{ submodule } P \subseteq M : P \supseteq N\} \leftrightarrow \{\text{ submodule } Q \subseteq M/N\}.$

The correspondence preserves inclusion.

Sketch of Proof.

Let f : M → M/N be the projection map.

Any submodule P of M gives a submodule f(P) of M/N.
On the other hand, any submodule Q of M/N gives a submodule f-1(Q) of M containing N.

One then checks that $f(f^{-1}(Q)) = Q$ always holds, and that if P contains N, then we have $f^{-1}(f(P)) = P$ . ♦

Exercise.

Prove that in the above correspondence, the intersection (resp. sum) of submodules on one side corresponds to the intersection (resp. sum) of submodules on another.

[ Hint for exercise: avoid “picking elements” from the submodules if possible; instead argue that the intersection or sum satisfies some unique property pertaining to inclusion of submodules.]

Direct Sum and Direct Product

It’s clear that given any two R-modules M and N, we can form the product M × N and let R act on it component-wise. On the other hand, with infinitely many modules, there’re two common ways to construct this “product”.

Definition. Consider any class of R-modules $\{M_i\}$ .

The direct product is the set-theoretic product $\prod_i M_i$ . It’s given the structure of an R-module, by letting $r\in R$ act on it component-wise.

The direct sum $\oplus_i M_i$ is the subset of $(m_i) \in \prod_i M_i$ for which only finitely many m_i‘s are non-zero. This is clearly a submodule of the direct product.

For example, suppose R = Z and M_n = Z for n = 1, 2, 3, … .The direct product of the M_n‘s is the set of all sequences of integers, while the direct sum is the set of sequences of integers in which only finitely many terms are non-zero.

The difference may seem subtle, but these two modules turn out to be quite different in their behaviour. In fact, one can even say they’re dual to each other, thanks to their universal properties.

Universal Property of Direct Product. For each index j, let $\pi_j : \prod_i M_i \to M_j$ be the projection map. If N is a module and $f_j : N\to M_j$ is any collection of module homomorphisms, then there’s a unique $\phi : N\to \prod_i M_i$ such that:

$\pi_j\circ \phi = f_j : N\to M_j$ for all j.

Universal Property of Direct Sum. For each index j, let $\iota_j : M_j \to \oplus_i M_i$ be the natural inclusion map which takes $m\in M_j$ to the tuple $(m_i)$ which comprises of all zeros except $m_j = m.$ If N is a module and $g_j : M_j \to N$ is any collection of module homomorphisms, then there’s a unique $\psi : \oplus_i M_i \to N$ such that:

$\psi\circ \iota_j = g_j : M_j\to N$ for all j.

Sketch of Proof

For direct product, the collection of maps $f_j : N\to M_j$ gives rise to the map $\phi : N\to \prod_i M_i$ by taking:

$\phi(n) := (f_i(n))$ for all $n\in N$ .

One easily checks that this is a module homomorphism and that it is the unique map satisfies the desired requirement.

For direct sum, the maps $g_j : M_j \to N$ give $\psi : \oplus_i M_i \to N$ by taking

$\psi((m_i)) := \sum_i g_i(m_i)$ for any $(m_i) \in \oplus_i M_i.$

This is well-defined since there are only finitely many non-zero m_i‘s so the RHS is a finite sum. ♦

Exercise

Consider the case where R=Z and $M_1 = M_2 = M_3 = \ldots = \mathbf{Z}$ . Explain why $\prod_i M_i$ doesn’t satisfy the universal property for direct sum and why $\oplus_i M_i$ doesn’t satisfy that of direct product. [ Hint: in one case, the induced map fails to exist; in the other, the induced map is not unique. ]