# Modules

Having dipped our toes into algebraic geometry, we are back in commutative algebra.

Next we would like to introduce “linear algebra” over a ring A. Most of the proofs should pose no difficulty to the reader so we will skip them. We recommend that the reader prove these properties at least once.

Definition.

An A-module is an abelian group $(M, +)$ together with a map $A\times M \to M$, written as $(a,m) \mapsto am$, such that the following hold for all $a, a' \in A$ and $m, m' \in M$.

• $1_A \cdot m = m$.
• $(a+a')m = am + a'm \in M$.
• $(aa')m = a(a'm) \in M$.
• $a(m + m') = am + am' \in M$.

### Alternative Look

If we fix $a\in A$, we obtain a “multiplication-by-a” map defined by $\mu_a : M\to M, m\mapsto am$. The last property says $\mu_a : M\to M$ is an (additive) homomorphism so we get a map $A \to \mathrm{End}(A)$, where $\mathrm{End}(M)$ is the set of all group homomorphisms $(M, +) \to (M, +)$. Now $\mathrm{End}(M)$ is not just any old set – it has a natural structure of a (non-commutative) ring if we introduce the following definitions: $\phi, \psi \in \mathrm{End}(M) \implies \begin{cases} (\phi + \psi) : m \mapsto \phi(m) + \psi(m),\\ (\phi\psi) : m \mapsto \phi(\psi(m)). \end{cases}$

In words, we say “addition is defined on the image and product is given by composition”.

Now that $\mathrm{End}(M)$ has a ring structure, the first three axioms of the definition says $A \to \mathrm{End}(A), a\mapsto \mu_a$ is a ring homomorphism.

### Easy Properties

Immediately, we can show the following: $a\in A, m\in M \implies \begin{cases} 0_A\cdot m = a\cdot 0_M = 0_M,\\ (-a)m = a(-m) = -(am).\end{cases}$

E.g. a(-m) = -(am) because $\mu_a : M\to M$ is an additive homomorphism.

### Examples

1. Since the homomorphism $\mathbb Z \to \mathrm{End}(M)$ is unique, every abelian group is automatically a $\mathbb Z$-module.

2. A module over a field k is the same as a vector space over k.

3. Any ring A is a module over itself. More generally if $A\subseteq B$ is a subring, then B is an A-module.

4. If M and M’ are A-modules, then $M\times M'$ becomes an A-module via $(m_1, m_1') + (m_2, m_2') = (m_1 + m_2, m_1' + m_2'), \quad a(m, m') = (am, am')$.

5. From examples 3 and 4, we get $M = A^n = \{ (a_1, \ldots, a_n) : a_1, \ldots, a_n \in A\}$ with $(a_1, \ldots, a_n) + (a_1', \ldots, a_n') = (a_1 + a_1', \ldots, a_n + a_n'), \quad a\cdot (a_1, \ldots, a_n) := (aa_1, \ldots, aa_n)$.

6. Let $A = B \times B'$ be a product of two rings. Then any B-module M gives an A-module $M\times \{0\}$; similarly, any B’-module M’ gives an A-module $\{0\} \times M'$.

7. More generally for $A = B\times B'$, any B-module N and B’-module N’ gives an A-module $M = N\times N'$. # Basic Constructions

Just as we have subspaces of vector spaces, we have:

Definition.

If M is an A-module, a A-submodule is a subgroup N of $(M, +)$ such that for all $a\in A, m\in N$ we have $am\in N$.

And this gives:

Proposition 1.

Let N be an A-submodule of M. The quotient of $(M, +)$ by the subgroup N has a natural structure of an A-module, given by: $A \times (M/N) \to M/N, \quad (a, m + N) \mapsto am + N$.

Thus we obtain the quotient module $M/N$.

Proof. Easy exercise. ♦

Next, we have the linear maps.

Definition.

Let M, N be A-modules. A module homomorphism is a function $f:M\to N$ such that $a\in A, m, m'\in M \implies \begin{cases} f(m + m') = f(m) + f(m'),\\ f(am) = a\cdot f(m).\end{cases}$

We will also say f is an A-linear map (or just linear map if the base ring is implicit).

A monomorphism (resp. epimorphismisomorphism) of modules is an injective (resp. surjective, bijective) homorphism of modules.

Immediately we have the following.

First Isomorphism Theorem.

Let $f:M\to N$ be a linear map of A-modules.

• The kernel of f, $\mathrm{ker} f := \{x \in M : f(x) = 0\}$, is a submodule of M.
• The image of f, $\mathrm{im} f := \{f(x) \in N : x\in M\}$, is a submodule of N.
• The map f induces an isomorphism of modules $\overline f : M/\mathrm{ker} f \longrightarrow \mathrm{im} f, \quad m + \mathrm{ker} f \mapsto f(m).$

Proof. Easy exercise. ♦

Exercise A

Let $A = B\times B'$ be a product of two rings. For a B-module N and B’-module N’, take the A-module $M = N \times N'$. Must every submodule of M be of the form $N_1 \times N_1'$ for B-submodule $N_1 \subseteq N$ and B’-submodule $N_1' \subseteq N'$? # Ideals as Modules

Recall that every non-trivial ring A comes with two modules.

• The zero module is the trivial group (0, +), and $a\cdot 0 = 0$ for all $a\in A$.
• The ring A is a module over itself, where $A\times M\to M$ is given by ring product $A\times A \to A$.

Lemma 1.

For any ring A, a subset of A is a submodule if and only if it is an ideal of A.

Proof. Easy exercise. ♦

## Some Notes

Although the lemma is a trivial observation, it highlights an important philosophy: modules can be considered as a generalization of ideals. For an ideal $\mathfrak a\subseteq A$, it is geometrically more natural to associate it with $A/\mathfrak a$ as a module, although the reasons are not clear for now.

Note that an ambiguity may arise if we were sloppy with words: for $A/\mathfrak a$, as a module, could mean the ring $A/\mathfrak a$ as a module over itself, or the quotient A-module of $A$ by the submodule $\mathfrak a$. We will try to be extra careful in stating the base ring. Often, the base ring is used as a subscript in notation, e.g. $C_A(M)$ means a certain construction from M by regarding it as an A-module.

For readers with some experience with differentiable manifolds, another geometrical application of modules is as follows. Recall that closed sets V correspond with their coordinate ring k[V]. Modules over k[V] then correspond to “vector bundles” over the set V. Of special interest are those bundles which are “locally free”, i.e. close to each point the vector bundle restricts to a trivial one. # Operations on Submodules

Finally, the operations on ideals can be generalized.

Proposition 2.

Let $(N_i)$ be a collection of submodules of M, over a ring A.

• $\cap N_i$ is a submodule of M.
• Let $\sum_i N_i$ be the set of all finite sums $m_{1} + \ldots + m_{k} \in M$, where $m_{1} \in N_{i_1}, \ldots, m_k \in N_{i_k}$. This is a submodule of M.
• If $\mathfrak a$ is an ideal of A, let $\mathfrak a M$ be the set of all finite sums $a_1 m_1 + \ldots + a_k m_k$ for $a_i \in \mathfrak a$ and $m_i \in M$. This is a submodule of M.

Proof. Easy exercise. ♦

Exercise B

An A-module is said to be simple if it is non-zero and its only submodules are 0 and itself. Prove that a simple A-module must be isomorphic to $A/\mathfrak m$ for some maximal ideal $\mathfrak m\subset A$.

Note

When MA and each $N_i = \mathfrak a_i$ is an ideal of A, these definitions are consistent with the earlier ones for ideals. Thus we can think of them as generalizations of the corresponding operations on ideals.

Note that product of submodules is not defined since we cannot multiply elements in a module. But as we shall see later, one can forcibly define a “product space” P for modules M and N so that $m\in M$ and $n\in N$ multiply to give an element in P. That leads us to tensor products.

The next result is important, so we will state it separately.

Proposition 3.

If M is an A-module and $\mathfrak a$ is an ideal of A, then $M/\mathfrak a M$ has a canonical structure of an $(A/\mathfrak a)$-module, via $(A/\mathfrak a) \times (M/\mathfrak a M) \longrightarrow (M/\mathfrak a M), \quad (a + \mathfrak a, m + \mathfrak a M) \mapsto am + \mathfrak aM$.

Proof. Easy exercise. ♦

Proposition 3 shows a “canonical” way of constructing a module over $A/\mathfrak a$ given a module over A. We will state more precisely what canonicity means in later articles, by universal properties. This entry was posted in Advanced Algebra and tagged , , , , , . Bookmark the permalink.