Modules

Now, for an arbitrary ring \(A\), we define an \(A\)-module and examine its properties. More advanced material on \(A\)-modules can be found in the Multilinear Algebra category.

Definition 1 In the symmetric monoidal category \((\Ab,\otimes, \mathbb{Z})\), fix a monoid object \((A,\cdot, 1)\). Then \(M\in\Ab\) equipped with a left \(A\)-action in the sense of §Group Action, ⁋Definition 1 is called a left \(A\)-module, and one equipped with a right \(A\)-action is called a right \(A\)-module.

Since a monoid object in the monoidal category \((\Ab,\otimes, \mathbb{Z})\) is a ring \(A\), if we rewrite the above definition in this case, \(M\) being a left \(A\)-module means that \(M\) is an abelian group under addition, and is given an action of \(A\), \(\cdot:A\times M \rightarrow M\), satisfying the following conditions.

1. For any \(\alpha\in A\) and \(x,y\in M\), we have \(\alpha\cdot(x+y)=\alpha\cdot x+\alpha\cdot y\).
2. For any \(\alpha,\beta\in A\) and \(x\in M\), we have \((\alpha+\beta)\cdot x=\alpha\cdot x+\beta\cdot x\).
3. For any \(\alpha,\beta\in A\) and \(x\in M\), we have \((\alpha\beta)\cdot x=\alpha\cdot(\beta\cdot x)\).
4. For any \(x\in M\), we have \(1\cdot x=x\).

Here, instead of viewing \(\cdot\) as a homomorphism from \(A\otimes M\) to \(M\), we have explicitly written the bilinearity condition in conditions 1 and 2. On the other hand, just as a right action was the same as a left action via the opposite in §Group Action, a right \(A\)-module is the same as a left \(A^\op\)-module. Therefore, unless stated otherwise, an \(A\)-module will mean a left \(A\)-module from now on.

Definition 2 For an \(A\)-module \(M\) and a family \((x_i)_{i\in I}\) of its elements, a linear combination of these refers to an element that appears in the form

\[\sum_{i\in I} \alpha_i x_i,\qquad\text{$(\alpha_i)$ finitely supported}\]

If every element of \(M\) can be expressed as a linear combination of elements of this family, we say that \((x_i)_{i\in I}\) generates \(M\), and if there exists a finite family that generates \(M\), we call \(M\) a finitely generated \(A\)-module.

Submodules and Quotient Modules

Definition 3 For an arbitrary \(A\)-module \(M\) and its subset \(N\), if restricting the addition of \(M\) and its \(A\)-action to \(N\) gives \(N\) an \(A\)-module structure, we call this a submodule.

Meanwhile, for an \(A\)-module \(M\) and any of \(M\)’s submodules \(N\), the quotient group \(M/N\) is well-defined. On the other hand, considering the action of \(A\) defined on it, since \(N\) is closed under the action of \(A\) anyway, there is no problem in giving \(M/N\) an \(A\)-module structure.

Definition 4 For an arbitrary \(A\)-module \(M\) and its submodule \(N\), we call \(M/N\) a quotient module.

Example 5 For a ring \(A\), the multiplication map \(\mu:A\otimes A \rightarrow A\) satisfies precisely all the properties that an action must satisfy. Therefore, any ring is always a module over itself. If we view \(A\) as a left \(A\)-module, the submodules of \(A\) are precisely the left ideals of \(A\), and similarly, if we view \(A\) as a right \(A\)-module, the submodules of \(A\) are precisely the right ideals of \(A\).

On the other hand, if we reconsider why we only considered two-sided ideals when defining quotient rings in §Quotient Rings and Ring Isomorphisms, we can see that for a left ideal \(\mathfrak{a}\), the quotient module \(A/\mathfrak{a}\) is still a left \(A\)-module (even though it does not have a ring structure), and the same holds for right ideals.

Linear Maps

Definition 6 For two \(A\)-modules \(M\), \(N\), a map from \(M\) to \(N\) being an \(A\)-linear map means that the following two conditions

\[u(x+y)=u(x)+u(y),\qquad u(\alpha x)=\alpha u(x)\]

are satisfied for all \(x,y\in M\) and \(\alpha\in A\) by a function \(u\). When there is no risk of confusion, this is also simply written as a linear map.

Proposition 7 The composition of linear maps is a linear map. Moreover, a bijective linear map is always an isomorphism.

Proof

That the composition is linear is obtained by directly substituting into the definition. Meanwhile, for a bijective linear map \(u:M\rightarrow N\), for any \(x,y\in N\) and \(\alpha\in A\) we have \(u(u^{-1}(x)+u^{-1}(y))=x+y\) and \(u(\alpha u^{-1}(x))=\alpha x\), so from the fact that \(u\) is injective we obtain \(u^{-1}(x+y)=u^{-1}(x)+u^{-1}(y)\) and \(u^{-1}(\alpha x)=\alpha u^{-1}(x)\). That is, \(u^{-1}\) is also a linear map, and therefore \(u\) is an isomorphism.

We write the category of left \(A\)-modules and \(A\)-linear maps as \(\lMod{A}\). The category of right \(A\)-modules and \(A\)-linear maps is written as \(\rMod{A}\). Also, we sometimes write \(\Hom_\lMod{A}(M,N)\) simply as \(\Hom_A(M,N)\). These have full subcategories consisting of finitely generated \(A\)-modules, \(\lmod{A}\) and \(\rmod{A}\). The zero object of these four categories is \(\{0\}\).

Meanwhile, one of the special properties of \(\lMod{A}\) is that \(\Hom_{\lMod{A}}(M,N)\) has the structure of an abelian group. Furthermore, the following holds.

Proposition 8 Any \(\Hom_{\lMod{A}}(M,N)\) is an abelian group. Furthermore, for any \(A\)-linear map \(u:M \rightarrow M'\),

\[\Hom_{\lMod{A}}(u, N):\Hom_{\lMod{A}}(M',N)\rightarrow \Hom_{\lMod{A}}(M,N)\]

is a homomorphism between abelian groups.

Proof

In \(\Hom_{\lMod{A}}(M,N)\), for any two elements \(v,w\in\Hom_{\lMod{A}}(M,N)\), the sum \(v+w\in\Hom_{\lMod{A}}(M,N)\) is defined by the following formula

\[(v+w)(x)=v(x)+w(x)\qquad\text{for all $x\in M$}\]

and one must prove that this is indeed an \(A\)-linear map, but this is clear.

Now, for \(v,w\in\Hom_{\lMod{A}}(M',N)\) and \(x\in M\), that \(\Hom_{\lMod{A}}(u,N)\) is a homomorphism between abelian groups is clear from the following equation:

\[\left(\Hom_{\lMod{A}}(u, N)(v+w)\right)(x)=(v+w)(u(x))=v(u(x))+w(u(x))=\left(\Hom_{\lMod{A}}(u,N)(v)\right)(x)+\left(\Hom_{\lMod{A}}(u,N)(w)\right)(x)\]

A similar theorem holds, for an \(A\)-linear map \(u:N \rightarrow N'\), regarding \(\Hom_{\lMod{A}}(M, u)\), and also for right \(A\)-modules. If \(A\) were a commutative ring, we know that for any \(u:M \rightarrow N\),

\[(\alpha\cdot u)(x):=\alpha\cdot u(x)\qquad\text{for all $x\in M$}\]

defines an operation \(\cdot\) giving \(\Hom_{\lMod{A}}(M,N)\) an \(A\)-module structure; but if \(A\) is not commutative, then for any \(\beta\in A\) we have

\[(\alpha\cdot u)(\beta x)=\alpha\cdot u(\beta x)=\alpha\beta u(x)\]

and since there is no clear way to turn this into \(\beta(\alpha u(x))=\beta\cdot(\alpha\cdot u)(x)\), \(\Hom_{\lMod{A}}(M,N)\) does not generally have an \(A\)-module structure. Instead, if \(A\) is commutative, one can even show that the abelian group homomorphism in Proposition 8 is an \(A\)-linear map.

Definition 9 For an \(A\)-linear map \(u:M \rightarrow N\), we define the kernel and image of \(u\) respectively by

\[\ker u=\{x\in M\mid u(x)=0\},\qquad \im u=\{u(x)\in N\mid x\in M\}\]

The following is the isomorphism theorem we have always used, and its proof also proceeds in exactly the same way as that treated in §Group Isomorphisms; one only needs to verify additionally that the induced group homomorphism is actually \(A\)-linear, so we do not write it out separately.

Theorem 10 Let an \(A\)-linear map \(u:M \rightarrow N\) be given.

1. \(\ker u\) is a submodule of \(M\), and \(x+\ker u \mapsto u(x)\) defines a well-defined isomorphism \(M/\ker u \rightarrow \im u\).
2. For two submodules of \(M\), \(M',M''\), both \(M'+M''\) and \(M'\cap M''\) are submodules of \(M\), and the isomorphism \((M'+M'')/M''\cong M'/(M'\cap M'')\) holds.
3. If two submodules of \(M\), \(M',M''\), satisfy \(M''\subseteq M'\), then \(M'/M''\) is a submodule of \(M/M''\) and \((M/M'')/(M'/M'')\cong M/M'\) holds.
4. For a submodule of \(M\), \(M'\), there exists an inclusion-preserving bijection between the set of submodules of \(M/M'\) and the set of submodules containing \(M'\) in \(M\).

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References

[Bou] Bourbaki, N. Algebra I. Elements of Mathematics. Springer. 1998.

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