Direct Products, Direct Sums, and Tensor Products of Modules

Direct Products and Direct Sums of Modules

The category \(\lMod{A}\) is a bicomplete category. To show this, we must construct arbitrary products and coproducts in \(\lMod{A}\), and to do so, it suffices to show that natural \(A\)-actions exist on the product and coproduct in \(\Ab\).

Let \((M_i)_{i\in I}\) be a family of \(A\)-modules. Then, for the action on \(\prod M_i\), it suffices to define \(A\otimes\left(\prod M_i\right) \rightarrow M_i\) via 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 use the universal property of the product in \(\Ab\) to construct \(A\otimes\left(\prod M_i\right) \rightarrow \prod M_i\) and show that this satisfies the conditions for an action.

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 the action on \(\bigoplus M_i\). For equalizers and coequalizers, given two module homomorphisms \(u,v:M \rightarrow N\), they can be defined by

\[\Eq(u,v)=\{x\in M\mid u(x)=v(x)\}\]

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.

Then the direct product preserves kernels, and the direct sum preserves cokernels. ([Category Theory] §Limits, ⁋Proposition 10) In addition, they also satisfy the following proposition.

Proposition 2 Suppose that two families of \(A\)-modules \((M_i)_{i\in I},(N_i)_{i\in I}\) and linear maps \(u_i: M_i \rightarrow N_i\) between them are given, and consider the functions \(\bigoplus u_i:\bigoplus M_i \rightarrow \bigoplus N_i\) and \(\prod u_i: \prod M_i \rightarrow \prod N_i\) induced by them. Then the following hold.

1. If each \(u_i\) is surjective, then \(\prod u_i\) is also surjective, and conversely.
2. If each \(u_i\) is injective, then \(\bigoplus u_i\) is also injective, and conversely.

The proof of this is obtained by writing out \(\prod u_i\) and \(\bigoplus u_i\) directly coordinatewise. In particular, from this proposition we see that the direct product also preserves cokernels, and the direct sum also preserves kernels.

Previously, we observed that for any \(M,N\in\lMod{A}\), \(\Hom_{\lMod{A}}(M,N)\) is an abelian group. It is easy 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 a 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 only needs to check that any monomorphism \(u:M \rightarrow N\) is equal to the kernel of its cokernel \(N \rightarrow N/M\), and any epimorphism \(v:M \rightarrow N\) is equal to the cokernel of its kernel \(\ker v\), namely \(M \rightarrow M/\ker v\).

Free module

In §Modules, ⁋Example 5, we observed that the ring \(A\) has the structure of an \(A\)-module. Then any \(A\)-module homomorphism \(u:A \rightarrow M\) is uniquely determined by \(u(1)\). This is because for any \(\alpha\in A\),

\[u(\alpha)=u(\alpha\cdot 1)=\alpha\cdot u(1)\]

holds. 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, \(A\) can be said to be a representation of the forgetful functor \(U\).

On the other hand, since we previously verified that \(\lMod{A}\) has the coproduct \(\bigoplus\), if \(U\) has a left adjoint \(F: \Set \rightarrow \lMod{A}\), 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 representation above we see that we must define \(F(X)=\bigoplus_{x\in X}Ax\). Conversely, defining \(F(X)\) in this way and, for a function \(u:X\rightarrow Y\), setting \(F(u)\) to be the linear map that maps each generator, from the universal property of the coproduct and the representation above we obtain, for any \(A\)-module \(M\), the following 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 the composition of linear maps and the composition 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, there exists an adjunction \(F\dashv U\).

For any set \(X\), if isomorphic to \(F(X)\), an \(A\)-module is called a free \(A\)-module.

Tensor Product of Modules

Meanwhile, we can also define the tensor product of \(A\)-modules. First, we begin with the following definition.

Definition 4 Suppose that a ring \(A\), a right \(A\)-module \(M\), and a left \(A\)-module \(N\) are given. Then, for any abelian group \(L\), a function \(f:M\times N \rightarrow L\) is said to be \(A\)-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

Let us 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, 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, 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 shown, but as it is a straightforward calculation, we omit it.

We write the representation obtained in this way as \(M\otimes_AN\). Then the following holds.

Theorem 6 (\(\otimes\dashv\Hom\)) There exists an adjunction

\[\Hom_\mathbb{Z}(M\otimes_A N, L)\cong\Hom_{\rMod{A}}(M,\Hom_\mathbb{Z}(N, L))\cong\Hom_{\lMod{A}}(N,\Hom_\mathbb{Z}(M, L))\]

The proof of this 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 yields \(\Balan_A(M,N;L)\cong\Hom_{\rMod{A}}(M,\Hom_\mathbb{Z}(N,L))\); 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 where \(A=\mathbb{Z}\), this recovers the contents of §Abelian Groups, §§Tensor Product; the above isomorphisms are ones that 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 have an \(A\)-module structure. This is because, if we think about defining an action of \(A\) 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 the elements

\[(x\alpha\beta)\otimes_A y,\qquad (x\beta\alpha)\otimes_A y\]

obtained by computing \((\alpha\beta)(x\otimes_Ay)\) and \(\alpha(\beta(x\otimes_Ay))\) will be different elements. It is also for a similar reason that 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, 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 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 \(M\otimes_AN\):

\[\alpha(x\otimes_Ay)=(\alpha x)\otimes_Ay=x\otimes_A(\alpha y)\]

This is also 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 function between abelian groups and, additionally, the following equation

\[\alpha f(x,y)=f(\alpha x,y)=f(x,\alpha y)\]

holds.

Define the set \(\Bilin_A(M,N;L)\) by the following equation:

\[\Bilin_A(M,N;L)=\{\text{$A$-bilinear maps from $M\times N$ to $L$}\}\]

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 \(\Hom_{\lMod{A}}(M,M')\) did not have an \(A\)-module structure, but if \(A\) is a commutative ring then an \(A\)-module structure also exists on \(\Hom_{\lMod{A}}(M,M')\). That is, \(\Hom_A\) is an internal \(\Hom\), and therefore we can further 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 both become isomorphisms between \(A\)-modules. Also, one can verify that \((\lMod{A},\otimes_A,A)\) is a symmetric monoidal category.

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References

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