This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
Now we define the \(A\)-module for an arbitrary ring \(A\), and study its properties. More advanced topics 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 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. That \(M\) is a left \(A\)-module means that \(M\) is an abelian group under addition, and an action \(\cdot:A\times M \rightarrow M\) of \(A\) is given 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, instead of viewing \(\cdot\) as a homomorphism from \(A\otimes M\) to \(M\), we have explicitly written out the bilinearity conditions in (1) and (2).
Definition 2 For an \(A\)-module \(M\) and a family \((x_i)_{i\in I}\) of its elements, 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 there exists a family \((x_i)_{i\in I}\) of elements of \(M\) such that 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 \(M\). If this family \((x_i)_{i\in I}\) can be chosen to be finite, 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 it, if \(N\) itself carries an \(A\)-module structure, we call it a submodule.
Meanwhile, for an \(A\)-module \(M\) and any submodule \(N\) of \(M\), the quotient group \(M/N\) is well-defined. Moreover, 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 The multiplication map \(\mu:A\otimes A \rightarrow A\) of a ring \(A\) satisfies exactly all the properties that an action must satisfy. Thus any ring is always a module over itself. If we view \(A\) as a left \(A\)-module, then the submodules of \(A\) are precisely the left ideals of \(A\), and similarly, if we view \(A\) 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, we can verify 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. Also, a bijective linear map is always an isomorphism.
Proof
Obvious.
We write \(\lMod{A}\) for the category of left \(A\)-modules and \(A\)-linear maps. The category of right \(A\)-modules and \(A\)-linear maps is written \(\rMod{A}\). Also, we sometimes write \(\Hom_A(M,N)\) for short instead of \(\Hom_\lMod{A}(M,N)\). 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 \(\Hom_{\lMod{A}}(M,N)\) has the structure of an abelian group. Moreover, the following holds.
Proposition 8 For any \(\Hom_{\lMod{A}}(M,N)\) is an abelian group. Moreover, 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 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 although one must prove that this is actually an \(A\)-linear map, this is obvious.
That \(\Hom_{\lMod{A}}(u,N)\) is a homomorphism of abelian groups follows from the identity
\[\left(\Hom_{\lMod{A}}(u, N)(v+w)\right)(x)=(v+w)(u(x))=v(u(x))+w(u(x))=\left(\Hom_{\lMod{A}}(u,N)(v)\right)(x)+\left(\Hom_{\lMod{A}}(u,N)(w)\right)(x)\]which is obvious.
A similar statement holds for \(\Hom_{\lMod{A}}(M, f)\), and also for right \(A\)-modules. If \(A\) were a commutative ring, then for any \(u:M \rightarrow N\), the operation defined by
\[(\alpha\cdot u)(x):=\alpha\cdot u(x)\qquad\text{for all $x\in M$}\]gives \(\Hom_{\lMod{A}}(M,N)\) 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 have an \(A\)-module structure. Instead, if \(A\) is commutative, one can also show that the abelian group homomorphism in Proposition 8 is in fact an \(A\)-linear map.
Definition 9 For an \(A\)-linear map \(u:M \rightarrow N\), we define the kernel and image of \(u\) respectively as
\[\ker u=\{x\in M\mid u(x)=0\},\qquad \im u=\{u(x)\in N\mid x\in M\}\]The following is the isomorphism theorem that we have always used, and since its proof is the same as before, 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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