Modules

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

We now define an $A$-module for an arbitrary ring $A$ and examine its properties. More advanced material on $A$-modules can be found in the Multilinear Algebra category.

Definition 1 Fix a monoid object $(A,\cdot, 1)$ in the symmetric monoidal category $(\Ab,\otimes, \mathbb{Z})$. Then an object $M\in\Ab$ equipped with a left $A$-action in the sense of §Group Actions, ⁋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 precisely a ring $A$, let us rewrite the above definition in this case: $M$ is a left $A$-module means that $M$ is an abelian group under addition, and we are given an action $\cdot:A\times M \rightarrow M$ of $A$ satisfying the following conditions.

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

Here, rather than viewing $\cdot$ as a homomorphism $A\otimes M \rightarrow M$, we have explicitly written out the bilinearity conditions in (1) and (2). On the other hand, just as a right action becomes a left action via the opposite ring in §Group Actions, a right $A$-module is the same as a left $A^\op$-module. Henceforth, unless stated otherwise, an $A$-module will mean a left $A$-module.

Definition 2 For an $A$-module $M$ and a family of its elements $(x_i)_{i\in I}$, a linear combination of these elements is an element of 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 from this family, we say that $(x_i)_{i\in I}$ generates $M$, and if there exists a finite family generating $M$, we call $M$ a finitely generated $A$-module.

Submodules and Quotient Modules

Definition 3 For an arbitrary $A$-module $M$ and a subset $N$ of $M$, if restricting the addition and $A$-action of $M$ to $N$ yields an $A$-module structure on $N$, we call $N$ a submodule of $M$.

Now, for an $A$-module $M$ and any submodule $N$ of $M$, the quotient group $M/N$ is well defined. Moreover, considering the induced action of $A$ on it, since $N$ is closed under the action of $A$, there is no obstruction to 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 The multiplication map $\mu:A\otimes A \rightarrow A$ of a ring $A$ satisfies exactly the properties that an action must satisfy. Therefore any ring is always a module over itself. If we regard $A$ as a left $A$-module, then the submodules of $A$ are precisely the left ideals of $A$, and similarly, if we regard $A$ as a right $A$-module, then the submodules of $A$ are precisely the right ideals of $A$.

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

Linear Maps

Definition 6 For two $A$-modules $M$, $N$, an $A$-linear map from $M$ to $N$ is a function $u$ satisfying the following two conditions

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

for all $x,y\in M$ and $\alpha\in A$. When there is no risk of confusion, we may simply call this 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 composition is linear follows immediately by substituting into the definition. Now, for a bijective linear map $u:M\rightarrow N$ and 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 injectivity of $u$ we obtain $u^{-1}(x+y)=u^{-1}(x)+u^{-1}(y)$ and $u^{-1}(\alpha x)=\alpha u^{-1}(x)$. Thus $u^{-1}$ is also a linear map, and therefore $u$ is an isomorphism.

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

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

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

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

is a homomorphism of abelian groups.

Proof

The sum $v+w\in\Hom_{\lMod{A}}(M,N)$ of two arbitrary elements $v,w\in\Hom_{\lMod{A}}(M,N)$ is defined by the formula

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

and one must verify that this is indeed an $A$-linear map, but this is immediate.

Now for $v,w\in\Hom_{\lMod{A}}(M',N)$ and $x\in M$, that $\Hom_{\lMod{A}}(u,N)$ is a homomorphism of abelian groups follows immediately from the equality

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

A similar statement holds for $\Hom_{\lMod{A}}(M, u)$ with respect to an $A$-linear map $u:N \rightarrow N'$, and also for right $A$-modules. If $A$ were a commutative ring, then for any $u:M \rightarrow N$ the operation defined by

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

would give $\Hom_{\lMod{A}}(M,N)$ an $A$-module structure; but if $A$ is not commutative, then for arbitrary $\beta\in A$ we have

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

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

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 proceeds in exactly the same way as in §Group Homomorphisms; the only additional check needed is that the induced group homomorphisms are actually $A$-linear, so we do not write it out separately.

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

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

References

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

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Modules

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