대수적 구조
Direct Products, Direct Sums, and Tensor Products of Modules
Products, coproducts, and tensor products in the module category
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.
Direct Products and Direct Sums of Modules
The category \(\lMod{A}\) is bicomplete. To show this, we must construct arbitrary products and coproducts in \(\lMod{A}\), and for this it suffices to exhibit natural \(A\)-actions on the product and coproduct taken in \(\Ab\).
Let \((M_i)_{i\in I}\) be a family of \(A\)-modules. Then the action on \(\prod M_i\) is given by the formula
\[A\otimes\left(\prod_{i\in I}M_i\right)\overset{\id_A\otimes\pr_i}{\longrightarrow} A\otimes M_i \overset{\rho_i}{\longrightarrow} M_i\]which defines \(A\otimes\left(\prod M_i\right) \rightarrow M_i\), and then by the universal property of the product in \(\Ab\) we obtain \(A\otimes\left(\prod M_i\right) \rightarrow \prod M_i\) and verify that this satisfies the action axioms.
For the coproduct, since \(A\otimes-\) is a left adjoint from \(\Ab\) to \(\Ab\), it preserves colimits, and therefore
\[A\otimes\left(\bigoplus_{i\in I} M_i\right)\cong\bigoplus_{i\in I}(A\otimes M_i)\overset{\bigoplus \rho_i}{\longrightarrow} \bigoplus_{i\in I}M_i\]defines an action on \(\bigoplus M_i\). For equalizers and coequalizers, given two module homomorphisms \(u,v:M \rightarrow N\), we have
\[\Eq(u,v)=\{x\in M\mid u(x)=v(x)\}\]and
\[\CoEq(u,v)=N/N',\qquad N'=\langle u(x)-v(x)\rangle\]That is, the following holds.
Theorem 1 \(\lMod{A}\) is a bicomplete category; in particular, the product of a family \((M_i)\) of \(A\)-modules is their direct product, and the coproduct is their direct sum.
Thus the direct product preserves kernels, and the direct sum preserves cokernels. ([Category Theory] §Limits, ⁋Proposition 10) Moreover, they satisfy the following further property.
Proposition 2 Let \((M_i)_{i\in I},(N_i)_{i\in I}\) be two families of \(A\)-modules and let \(u_i: M_i \rightarrow N_i\) be linear maps between them; 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 conversely.
- If each \(u_i\) is injective, then \(\bigoplus u_i\) is also injective, and conversely.
The proof follows by writing out \(\prod u_i\) and \(\bigoplus u_i\) coordinatewise. In particular, from this proposition we see that the direct product also preserves cokernels, and the direct sum also preserves kernels.
We have already observed that for arbitrary \(M,N\in\lMod{A}\), the set \(\Hom_{\lMod{A}}(M,N)\) is an abelian group. It is easy to check that this addition is compatible with 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 3) 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 observed that a ring \(A\) carries the structure of an \(A\)-module. Hence 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 following isomorphism
\[\Hom_A(A, M)\cong\Hom_\Set(\ast, U(M))\]holds. Here \(U:\lMod{A} \rightarrow \Set\) is the forgetful functor. That is, \(A\) is 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 representation above we know that we must define \(F(X)=\bigoplus_{x\in X}Ax\). Conversely, defining \(F(X)\) in this way and, for a function \(u:X\rightarrow Y\), defining \(F(u)\) to be the linear map sending each generator to its image, the universal property of the coproduct together with the representation above yield, for any \(A\)-module \(M\), the isomorphism
\[\Hom_A\biggl(\bigoplus_{x\in X}Ax,M\biggr)\cong\prod_{x\in X}\Hom_A(A,M)\cong\prod_{x\in X}U(M)\cong\Hom_\Set(X,U(M))\]Since each correspondence here is given only by composition of linear maps and of functions, it is natural in both \(X\) and \(M\), and therefore we obtain the following.
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 \(X\), an \(A\)-module isomorphic to \(F(X)\) is called a free \(A\)-module.
Tensor Products of Modules
Meanwhile, 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
\[f(x\alpha, y)=f(x,\alpha y)\]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
\[M'=\left\langle (x, y_1+y_2)-(x,y_1)-(x,y_2), (x_1+x_2,y)-(x_1,y)-(x_2,y), (x\alpha,y)-(x,\alpha y)\right\rangle\]Then by the universal property of the free abelian 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 then 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 \(\Balan_A(M,N;L)\cong\Hom_\Ab(F(M\times N)/M',L)\) still needs to be checked, but this is a straightforward computation so we omit it.
We write the representation thus obtained as \(M\otimes_AN\). 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.
The proof is obtained by associating to an \(A\)-balanced map \(f:M\times N\rightarrow L\) the map \(x\mapsto f(x,-)\), which gives \(\tilde{f}(x\alpha)=\tilde{f}(x)\alpha\) and hence \(\Balan_A(M,N;L)\cong\Hom_{\rMod{A}}(M,\Hom_\mathbb{Z}(N,L))\); fixing the second variable similarly yields \(\Hom_{\lMod{A}}(N,\Hom_\mathbb{Z}(M,L))\).
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 \(A=\mathbb{Z}\), this recovers the contents of [§Abelian Groups, §§Tensor Products]; the isomorphisms above were omitted in that article for reasons of length.
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 action of \(A\) on \(M\otimes_A N\), it would be natural to set
\[(x\alpha)\otimes_A y=x\otimes_A(\alpha y)\]equal to \(\alpha(x\otimes_Ay)\), but computing \((\alpha\beta)(x\otimes_Ay)\) and \(\alpha(\beta(x\otimes_Ay))\) yields
\[(x\alpha\beta)\otimes_A y,\qquad (x\beta\alpha)\otimes_A y\]which would be different elements. The reason \(M\) is taken as a right module and \(N\) as a left module in the definition of the tensor product is similar.
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 equation
\[(\beta\cdot_B x)\cdot_A\alpha=\beta\cdot_B(x\cdot_A\alpha)\]must hold. Then the formula
\[\beta(x\otimes_A y)=(\beta x)\otimes_Ay\]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, 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 \(M\otimes_AN\)
\[\alpha(x\otimes_Ay)=(\alpha x)\otimes_Ay=x\otimes_A(\alpha y)\]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
\[\alpha f(x,y)=f(\alpha x,y)=f(x,\alpha y)\]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.
Since an \(A\)-bilinear map is in particular \(A\)-balanced, the correspondence of Theorem 5 applies directly, and under this correspondence \(f\) being \(A\)-bilinear is equivalent to \(\hat{f}\) being \(A\)-linear. Indeed,
\[\hat{f}(\alpha(x\otimes_Ay))=\hat{f}((\alpha x)\otimes_Ay)=f(\alpha x,y)=\alpha f(x,y)=\alpha\hat{f}(x\otimes_Ay)\]and since \(M\otimes_AN\) is generated by elements of the form \(x\otimes_Ay\), the converse also holds. That is, \(\Bilin_A(M,N;L)\cong\Hom_A(M\otimes_AN,L)\).
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 carry an \(A\)-module structure. That is, \(\Hom_A\) is an internal \(\Hom\), and therefore we can refine the adjunction of Theorem 6 (\(\otimes\dashv\Hom\)) 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 formulas (1), (2) above become isomorphisms of \(A\)-modules. Also, one can verify that \((\lMod{A},\otimes_A,A)\) is a symmetric monoidal category.
References
[Bou] Bourbaki, N. Algebra I. Elements of Mathematics. Springer. 1998.
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