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.
Direct Products and Direct Sums of Modules
The category \(\lMod{A}\) is a bicomplete category. To show this, we need to construct arbitrary products and coproducts in \(\lMod{A}\), and for this it suffices to show that there is a natural \(A\)-action on the product and coproduct in \(\Ab\).
Let a family of \(A\)-modules \((M_i)_{i\in I}\) be given. Then the action on \(\prod M_i\) is defined by first defining \(A\otimes\left(\prod M_i\right) \rightarrow M_i\) through the formula
\[A\otimes\left(\prod_{i\in I}M_i\right)\overset{\id_A\otimes\pr_i}{\longrightarrow} A\otimes M_i \overset{\rho_i}{\longrightarrow} M_i\]and then using the universal property of the product in \(\Ab\) to obtain \(A\otimes\left(\prod M_i\right) \rightarrow \prod M_i\), and verifying that this satisfies the axioms for an action.
For the coproduct, since \(A\otimes-\) is a left adjoint from \(\Ab\) to \(\Ab\), it preserves colimits, and therefore the action on \(\bigoplus M_i\) is defined via
\[A\otimes\left(\bigoplus_{i\in I} M_i\right)\cong\bigoplus_{i\in I}(A\otimes M_i)\overset{\bigoplus \rho_i}{\longrightarrow} \bigoplus_{i\in I}M_i\]For equalizers and coequalizers, given two module homomorphisms \(u,v:M \rightarrow N\), we can define
\[\Eq(u,v)=\{x\in M\mid u(x)=v(x)\}\]and
\[\CoEq(u,v)=N/N',\qquad N'=\langle u(x)-v(x)\rangle\rangle\]Thus the following holds.
Theorem 1 \(\lMod{A}\) is a bicomplete category; in particular, the product of a family of \(A\)-modules \((M_i)\) is their direct product, and the coproduct is their direct sum.
Then the direct product preserves kernels, and the direct sum preserves cokernels. ([Category Theory] §Limits, ⁋Proposition 10) Additionally, they also satisfy the following proposition.
Proposition 2 Let two families of \(A\)-modules \((M_i)_{i\in I},(N_i)_{i\in I}\) and linear maps between them \(u_i: M_i \rightarrow N_i\) be given, and consider the induced maps \(\bigoplus u_i:\bigoplus M_i \rightarrow \bigoplus N_i\) and \(\prod u_i: \prod M_i \rightarrow \prod N_i\). Then the following hold.
- If each \(u_i\) is surjective, then \(\prod u_i\) is also surjective, and the converse holds as well.
- If each \(u_i\) is injective, then \(\bigoplus u_i\) is also injective, and the converse holds as well.
The proof is obtained by writing out \(\prod u_i\) and \(\bigoplus u_i\) directly in coordinates. In particular, from this proposition we see that the direct product also preserves cokernels, and the direct sum also preserves kernels.
Earlier we saw that for any \(M,N\in\lMod{A}\), the set \(\Hom_{\lMod{A}}(M,N)\) has the structure of an abelian group. It is not difficult to check that this addition behaves well with respect to composition, and that the category \(\lMod{A}\) is an additive category with the zero module \(0\) as zero object. ([Category Theory] §Abelian Categories, ⁋Definition 1)
Moreover, \(\lMod{A}\) is an abelian category. ([Category Theory] §Abelian Categories, ⁋Definition 7) To verify this, one checks that any monomorphism \(u:M \rightarrow N\) is the kernel of its cokernel \(N \rightarrow N/M\), and any epimorphism \(v:M \rightarrow N\) is the cokernel of its kernel \(\ker v\), namely \(M \rightarrow M/\ker v\).
Free Modules
In §Modules, ⁋Example 5 we saw that a ring \(A\) carries the structure of an \(A\)-module. Then any \(A\)-module homomorphism \(u:A \rightarrow M\) is uniquely determined by \(u(1)\), since for any \(\alpha\in A\),
\[u(\alpha)=u(\alpha\cdot 1)=\alpha\cdot u(1)\]In other words, the following isomorphism holds:
\[\Hom_A(A, M)\cong\Hom_\Set(\ast, U(M))\]Here \(U:\lMod{A} \rightarrow \Set\) is the forgetful functor. Thus \(A\) can be said to be a representation of the forgetful functor \(U\).
On the other hand, since we have verified that \(\lMod{A}\) has coproducts \(\bigoplus\), if the left adjoint \(F: \Set \rightarrow \lMod{A}\) of \(U\) exists, then the formula
\[F(X)=F\left(\coprod_{x\in X} \{x\}\right)\cong\bigoplus_{x\in X} F(\{x\})\]must hold, and using the above representation we know that we must define \(F(X)=\bigoplus_{x\in X}Ax\). That is, the following holds.
Proposition 3 For the forgetful functor \(U:\lMod{A} \rightarrow\Set\) and the free functor \(F:\Set \rightarrow\lMod{A}\) defined above, the adjunction \(F\dashv U\) exists.
For any set \(X\), \(A\)-modules isomorphic to \(F(X)\) are called free \(A\)-modules.
Tensor Products of Modules
We can also define the tensor product of \(A\)-modules. We begin with the following definition.
Definition 4 Let a ring \(A\), a right \(A\)-module \(M\), and a left \(A\)-module \(N\) be given. Then a function \(f:M\times N \rightarrow L\) is called \(A\)-balanced if \(f\) is bilinear as a map of abelian groups and additionally satisfies the identity
\[f(x\alpha, y)=f(x,\alpha y)\]For fixed \(M\in\obj(\rMod{A}),N\in\obj(\lMod{A})\), define the set \(\Balan_A(M,N;L)\) by
\[\Balan_A(M,N;L)=\{\text{$A$-balanced maps from $M\times N$ to $L$}\}\]Then the following theorem holds.
Theorem 5 The functor \(\Balan_A(M,N;-):\lMod{\mathbb{Z}}=\Ab\rightarrow\Set\) is a representable functor.
Proof
Define the subgroup \(M'\) of the free abelian group \(F(M\times N)\) by
\[M'=\left\langle (x, y_1+y_2)-(x,y_1)-(x,y_2), (x_1+x_2,y)-(x_1,y)-(x_2,y), (\alpha x,y)-(x,\alpha y)\right\rangle\]Then by the universal property of the free group, for any function \(f:M\times N \rightarrow L\) there exists a group homomorphism \(\hat{f}:F(M\times N)\rightarrow L\), and if \(f\) is \(A\)-balanced, the kernel of this \(\hat{f}\) contains \(M'\), so \(\hat{f}\) defines a group homomorphism from \(F(M\times N)/M'\) to \(L\).
The naturality of the isomorphism \(\Balan_A(M,N;L)\cong\Hom_\Ab(F(M\times N)/M',L)\) also needs to be verified, but it is a straightforward computation so we omit it.
We denote the representation thus obtained by \(M\otimes_AN\). Then the following holds.
Theorem 6 (\(\otimes\dashv\Hom\)) The adjunction
\[\Hom_\mathbb{Z}(M\otimes_A N, L)\cong\Hom_{\rMod{A}}(M,\Hom_\mathbb{Z}(N, L))\cong\Hom_{\lMod{A}}(N,\Hom_\mathbb{Z}(M, L))\]exists.
Therefore \(\otimes\) commutes with colimits, and \(\Hom\) commutes with limits. In particular, we obtain the following isomorphisms of abelian groups:
\[M\otimes_A\left(\bigoplus_{i\in I} N_i\right)\cong \bigoplus_{i\in I} M\otimes_AN_i,\qquad \left(\bigoplus_{i\in I} M_i\right)\otimes_A N\cong\bigoplus_{i\in I} M_i\otimes_AN\tag{1}\]and
\[\Hom_{\lMod{A}}\left(M,\prod_{i\in I} N_i\right)\cong\prod_{i\in I}\Hom_{\lMod{A}}(M, N_i),\qquad \Hom_{\lMod{A}}\left(\bigoplus_{i\in I} M_i, N\right)\cong \prod_{i\in I}\Hom_{\lMod{A}}(M_i,N)\tag{2}\]In the special case where \(A=\mathbb{Z}\), these recover the contents of §Abelian Groups, §§Tensor Product; the above isomorphisms were omitted in that post for reasons of length.
Tensor Products of Modules over Commutative Rings
The \(M\otimes_A N\) defined above does not have an \(A\)-module structure. If we try to define an action of \(A\) on \(M\otimes_A N\), it would be natural to set the element
\[(x\alpha)\otimes_A y=x\otimes_A(\alpha y)\]equal to \(\alpha(x\otimes_Ay)\), but computing \((\alpha\beta)(x\otimes_Ay)\) would yield
\[(x\alpha\beta)\otimes_A y,\qquad x\otimes_A(\alpha\beta y)\]which would be different elements. In the definition of the tensor product, taking \(M\) as a right module and \(N\) as a left module is also for a similar reason.
If \(M\) has not only a right \(A\)-module structure but also a compatible left \(B\)-module structure, we call \(M\) a \((B,A)\)-bimodule. That is, for any \(\alpha\in A\), \(\beta\in B\), \(x\in M\), the identity
\[(\alpha\cdot_A x)\cdot_B\beta=\alpha\cdot_A(x\cdot_B\beta)\]must hold. Then one can verify that the formula
\[\beta(x\otimes_A y)=(\beta x)\otimes_Ay\]gives a left \(B\)-module structure on \(M\otimes_AN\).
We are mostly interested in the case where \(A\) is a commutative ring. Then any left \(A\)-module is also a right \(A\)-module, and vice versa. Moreover, viewing any left \(A\)-module as a right \(A\)-module in this way, these two structures form an \((A,A)\)-bimodule structure. Therefore there is a natural \(A\)-action on \(M\otimes_AN\) given by
\[\alpha(x\otimes_Ay)=(\alpha x)\otimes_Ay=x\otimes_A(\alpha y)\]This is again a representation of an appropriate functor.
Definition 7 Let a commutative ring \(A\) and three \(A\)-modules \(M,N,L\) be given. Then a function \(f:M\times N \rightarrow L\) is called \(A\)-bilinear if \(f\) is bilinear as a map of abelian groups and additionally satisfies the identity
\[\alpha f(x,y)=f(\alpha x,y)=f(x,\alpha y)\]Define the set \(\Bilin_A(M,N;L)\) by
\[\Bilin_A(M,N;L)=\{\text{$A$-bilinear maps from $M\times N$ to $L$}\}\]Proposition 8 The functor \(\Bilin_A(M,N;-):\lMod{A}\rightarrow\Set\) is a representable functor, and its representation is the \(A\)-module \(M\otimes_AN\) defined above.
On the other hand, if \(A\) is a general ring then \(\Hom_{\lMod{A}}(M,M')\) does not have an \(A\)-module structure, but if \(A\) is a commutative ring then \(\Hom_{\lMod{A}}(M,M')\) does have an \(A\)-module structure. That is, \(\Hom_A\) is an internal \(\Hom\), and therefore the adjunction of Theorem 6 can be refined to prove the following.
Theorem 9 For a commutative ring \(A\), the adjunction
\[\Hom_A(M\otimes_AN, L)\cong\Hom_A(M,\Hom_A(N,L))\cong\Hom_A(N,\Hom_A(M,L))\]exists.
In particular, the above formulas (1) and (2) become isomorphisms of \(A\)-modules. Also, one can verify that \((\lMod{A},\otimes_A,A)\) is a symmetric monoidal category.
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