Direct Products, Direct Sums, and Tensor Products of Modules

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 \(\lMod{A}\); for this it suffices to show that there is a natural \(A\)-action 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 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 gives \(A\otimes\left(\prod M_i\right) \rightarrow M_i\), and then using the universal property of the product in \(\Ab\) we obtain \(A\otimes\left(\prod M_i\right) \rightarrow \prod M_i\); one checks that this satisfies the axioms of an action.

For the coproduct, since \(A\otimes-\) is a left adjoint from \(\Ab\) to \(\Ab\), it preserves colimits, and thus through

\[A\otimes\left(\bigoplus_{i\in I} M_i\right)\cong\bigoplus_{i\in I}(A\otimes M_i)\overset{\bigoplus \rho_i}{\longrightarrow} \bigoplus_{i\in I}M_i\]

an action on \(\bigoplus M_i\) is defined. 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

\[\CoEq(u,v)=N/N',\qquad N'=\langle u(x)-v(x)\rangle\rangle\]

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. (§Limits, ⁋Proposition 10) Additionally, they 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 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.

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 coordinate-wise. In particular, from this proposition we see that the direct product also preserves cokernels, and the direct sum also preserves kernels.

Earlier we observed that for arbitrary \(M,N\in\lMod{A}\), the set \(\Hom_{\lMod{A}}(M,N)\) becomes an abelian group. It is not difficult 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 a zero object. (§Abelian Categories, ⁋Definition 1)

Moreover, \(\lMod{A}\) is an abelian category. (§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. Then any \(A\)-module homomorphism \(u:A \rightarrow M\) is uniquely determined by \(u(1)\). For any \(\alpha\in A\),

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

because of this. 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 have verified that \(\lMod{A}\) has coproducts \(\bigoplus\), if the left adjoint \(F: \Set \rightarrow \lMod{A}\) of \(U\) exists, 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 \(F(X)=\bigoplus_{x\in X}Ax\). 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 \(X\), we call \(A\)-modules isomorphic to \(F(X)\) 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 for any abelian group \(L\), a function \(f:M\times N \rightarrow L\) is called \(A\)-balanced if \(f\) is bilinear as a map of abelian groups and additionally the formula

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

holds.

For fixed \(M\in\obj(\rMod{A}),N\in\obj(\lMod{A})\), define the set \(\Balan_A(M,N;L)\) by the formula

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

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, 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 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 \(\Balan_A(M,N;L)\cong\Hom_\Ab(F(M\times N)/M',L)\) should additionally be shown, but it is a simple 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.

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 \(A=\mathbb{Z}\), we recover the contents of §Abelian Groups, §§Tensor Products; the isomorphisms above 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 carry an \(A\)-module structure. If we think of 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 reason this fails is that computing \((\alpha\beta)(x\otimes_Ay)\) yields

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

which would be different elements. In the definition of the tensor product, the reason \(M\) is taken as a right module and \(N\) as a left module 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 formula

\[(\alpha\cdot_A x)\cdot_B\beta=\alpha\cdot_A(x\cdot_B\beta)\]

must hold. Then one can check 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 conversely. 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 the formula

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

holds.

Define the set \(\Bilin_A(M,N;L)\) by the formula

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

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 there is an \(A\)-module structure on \(\Hom_{\lMod{A}}(M,M')\) as well. 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 check that \((\lMod{A},\otimes_A,A)\) is a symmetric monoidal category.