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 \(\lMod{A}\), which amounts to showing that there exist natural \(A\)-actions on the product and coproduct in \(\Ab\).
Let a family \((M_i)_{i\in I}\) of \(A\)-modules 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 composition
\[A\otimes\left(\prod_{i\in I}M_i\right)\overset{\id_A\otimes\pr_i}{\longrightarrow} A\otimes M_i \overset{\rho_i}{\longrightarrow} M_i\]and then using the universal property of products in \(\Ab\) to construct \(A\otimes\left(\prod M_i\right) \rightarrow \prod M_i\), and showing that this satisfies the action conditions.
For coproducts, since \(A\otimes-\) is a left adjoint from \(\Ab\) to \(\Ab\), it preserves colimits, and thus the action on \(\bigoplus M_i\) is defined through
\[A\otimes\left(\bigoplus_{i\in I} M_i\right)\cong\bigoplus_{i\in I}(A\otimes M_i)\overset{\bigoplus \rho_i}{\longrightarrow} \bigoplus_{i\in I}M_i\]For equalizers and coequalizers, for two module homomorphisms \(u,v:M \rightarrow N\), we 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, and in particular, the product of a family \((M_i)\) of \(A\)-modules is given by their direct product, and the coproduct is given by their direct sum.
Then direct products preserve kernels and direct sums preserve cokernels. ([Category Theory] §Limits, ⁋Proposition 10) Additionally, these satisfy the following proposition.
Proposition 2 Let two families \((M_i)_{i\in I},(N_i)_{i\in I}\) of \(A\)-modules and linear maps \(u_i: M_i \rightarrow N_i\) between them be given, and consider the induced functions \(\bigoplus u_i:\bigoplus M_i \rightarrow \bigoplus N_i\) and \(\prod u_i: \prod M_i \rightarrow \prod N_i\). Then the following holds:
- If each \(u_i\) is surjective, then \(\prod u_i\) is also surjective, and conversely.
- If each \(u_i\) is injective, then \(\bigoplus u_i\) is also injective, and conversely.
The proof is obtained by writing \(\prod u_i\) and \(\bigoplus u_i\) explicitly in terms of coordinates. In particular, from this proposition, we see that direct products also preserve cokernels and direct sums also preserve kernels.
Earlier, we saw that for any \(M,N\in\lMod{A}\), the set \(\Hom_{\lMod{A}}(M,N)\) is an abelian group. It is not difficult to verify 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 its zero object. ([Category Theory] §Abelian Categories, ⁋Definition 1)
Moreover, \(\lMod{A}\) is an abelian category. ([Category Theory] §Abelian Categories, ⁋Definition 7) To verify this, we need to check 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 \(M \rightarrow M/\ker v\).
Free Modules
In §Modules, ⁋Example 5, we saw that a ring \(A\) carries an \(A\)-module structure. Then any \(A\)-module homomorphism \(u:A \rightarrow M\) is uniquely determined by \(u(1)\). For any \(\alpha\in A\),
\[u(\alpha)=u(\alpha\cdot 1)=\alpha\cdot u(1)\]In other words, the isomorphism
\[\Hom_A(A, M)\cong\Hom_\Set(\ast, U(M))\]holds. Here \(U:\lMod{A} \rightarrow \Set\) is the forgetful functor. Thus, \(A\) can be called a representation of the forgetful functor \(U\).
On the other hand, since we have verified that \(\lMod{R}\) has coproducts \(\bigoplus\), if a left adjoint \(F: \Set \rightarrow \lMod{A}\) to \(U\) exists, then the equation
\[F(X)=F\left(\coprod_{x\in X} \{x\}\right)\cong\bigoplus_{x\in X} F(\{x\})\]must hold, and using the above representation, we see that we should define \(F(X)=\bigoplus_{x\in X}Ax\). Thus, 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, an adjunction \(F\dashv U\) exists.
For any set \(X\), \(A\)-modules isomorphic to \(F(X)\) are called free \(A\)-modules.
Tensor Products of Modules
On the other hand, we can also define the tensor product of \(A\)-modules. First, 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 for any abelian group \(L\), a function \(f:M\times N \rightarrow L\) is called \(A\)-balanced if \(f\) is bilinear as a function between abelian groups, and additionally the equation
\[f(x\alpha, y)=f(x,\alpha y)\]holds.
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 free groups, whenever a function \(f:M\times N \rightarrow L\) is given, there exists a group homomorphism \(\hat{f}:F(M\times N)\rightarrow L\), and if \(f\) is \(A\)-balanced, then the kernel of \(\hat{f}\) contains \(M'\), so \(\hat{f}\) defines a group homomorphism from \(F(M\times N)/M'\) to \(L\).
The naturality of the isomorphism \(\Balan_A(M,N;L)\cong\Hom_\Ab(F(M\times N)/M',L)\) requires additional verification, but this is a straightforward computation, so we omit it.
The representation thus obtained is denoted by \(M\otimes_AN\). Then the following holds.
Theorem 6 (\(\otimes\dashv\Hom\)) There exists an adjunction
\[\Hom_\mathbb{Z}(M\otimes_A N, L)\cong\Hom_{\rMod{A}}(M,\Hom_\mathbb{Z}(N, L))\cong\Hom_{\lMod{A}}(N,\Hom_\mathbb{Z}(M, L))\]Therefore, \(\otimes\) commutes with colimits, and \(\Hom\) commutes with limits. In particular, the following isomorphisms between 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
\[\Hom_{\lMod{A}}\left(M,\prod_{i\in I} N_i\right)\cong\prod_{i\in I}\Hom_{\lMod{A}}(M, N_i),\qquad \Hom_{\lMod{A}}\left(\bigoplus_{i\in I} M_i, N\right)\cong \prod_{i\in I}\Hom_{\lMod{A}}(M_i,N)\tag{2}\]are obtained. In particular, when \(A=\mathbb{Z}\), we recover the contents of §Abelian Groups, §§Tensor Products; the isomorphisms above were not written in that article due to space constraints.
Tensor Products of Modules over Commutative Rings
The \(M\otimes_A N\) defined above does not carry an \(A\)-module structure. If we try to define an \(A\)-action on \(M\otimes_A N\), it would be natural to define the element
\[(x\alpha)\otimes_A y=x\otimes_A(\alpha y)\]as \(\alpha(x\otimes_Ay)\), but then computing \((\alpha\beta)(x\otimes_Ay)\) would give
\[(x\alpha\beta)\otimes_A y,\qquad x\otimes_A(\alpha\beta y)\]which would be different elements. This is also why, in the definition of the tensor product, \(M\) is taken as a right module and \(N\) as a left module.
If \(M\) has not only a right \(A\)-module structure but also a compatible left \(B\)-module structure, then \(M\) is called a \((B,A)\)-bimodule. That is, for any \(\alpha\in A\), \(\beta\in B\), and \(x\in M\), the equation
\[(\alpha\cdot_A x)\cdot_B\beta=\alpha\cdot_A(x\cdot_B\beta)\]must hold. Then one can verify that the equation
\[\beta(x\otimes_A y)=(\beta x)\otimes_Ay\]gives \(M\otimes_AN\) a left \(B\)-module structure.
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, regarding any left \(A\)-module as a right \(A\)-module, these two structures form an \((A,A)\)-bimodule structure. Therefore, there exists a natural \(A\)-action on \(M\otimes_AN\):
\[\alpha(x\otimes_Ay)=(\alpha x)\otimes_Ay=x\otimes_A(\alpha y)\]This also becomes a representation of an appropriate functor.
Definition 7 Let a commutative ring \(A\) and three \(A\)-modules \(M,N,L\) be given. A function \(f:M\times N \rightarrow L\) is called \(A\)-bilinear if \(f\) is bilinear as a function between abelian groups, and additionally the equation
\[\alpha f(x,y)=f(\alpha x,y)=f(x,\alpha y)\]holds.
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 carry an \(A\)-module structure, but if \(A\) is a commutative ring, then \(\Hom_{\lMod{A}}(M,M')\) also has an \(A\)-module structure. That is, \(\Hom_A\) is an internal \(\Hom\), and thus we can refine the adjunction in Theorem 6 to prove the following.
Theorem 9 For a commutative ring \(A\), there exists an adjunction
\[\Hom_A(M\otimes_AN, L)\cong\Hom_A(M,\Hom_A(N,L))\cong\Hom_A(N,\Hom_A(M,L))\]In particular, the isomorphisms (1) and (2) above become isomorphisms between \(A\)-modules. Also, one can verify that \((\lMod{A},\otimes_A,A)\) becomes a symmetric monoidal category.
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