Tensor Products of Modules

Regard $A$ as a $\mathbb{Z}$-module by integer multiplication in its additive group. A group homomorphism between abelian groups is automatically $\mathbb{Z}$-linear: additivity gives $f(na)=nf(a)$ for $n\ge 0$, and $f(-a)=-f(a)$ gives the formula for negative integers. Thus $\mathbb{Z}$-module homomorphisms and group homomorphisms between these underlying additive groups are the same maps.

If $h:M{\otimes }_{R}N\to A$ is a group homomorphism, then $(m,n)\mapsto h(m\otimes n)$ is balanced because the elementary tensors satisfy all three relations in [L1].

Apply [L4] at $S=\mathbb{Z}$ and $X=M\times N$ to extend the set map $b$ uniquely to a $\mathbb{Z}$-linear map $\tilde{b}:F\to A$ satisfying $\tilde{b}(e_{(m,n)})=b(m,n)$.

By [L2], $\tilde{b}$ sends each generator of $H$ to zero: the two additivity generators map respectively to $b(m+m^{\mathrm{\prime }},n)-b(m,n)-b(m^{\mathrm{\prime }},n)$ and $b(m,n+n^{\mathrm{\prime }})-b(m,n)-b(m,n^{\mathrm{\prime }})$, while the balance generator maps to $b(mr,n)-b(m,rn)$. Hence $H\subseteq \ker \tilde{b}$.

By [L5], $\tilde{b}$ factors uniquely through a group homomorphism $\bar{b}:F\mathrm{/}H=M{\otimes }_{R}N\to A$, and $\bar{b}(m\otimes n)=\tilde{b}(e_{(m,n)})=b(m,n)$.

The operations in steps 4.1 and 1.2 are inverse: starting from $b$ recovers its values on every pair, while starting from $h$ produces a homomorphism agreeing with $h$ on every elementary tensor, and those tensors generate $M{\otimes }_{R}N$. This proves both the asserted uniqueness and the displayed bijection.

No selection is made in the construction. If $M=0$ or $N=0$, every balanced map out of $M\times N$ is zero and the relations make every elementary tensor zero, so the same proof gives $M{\otimes }_{R}N=0$; the zero ring is covered as well. Since a module contains its zero element, $M\times N$ is never empty, so there is no separate empty-domain case.

Apply [L1] for $(T,\tau )$ to the balanced map ${\tau }^{\mathrm{\prime }}$ and for $(T^{\mathrm{\prime }},{\tau }^{\mathrm{\prime }})$ to $\tau$; this gives unique homomorphisms $u:T\to T^{\mathrm{\prime }}$ and $v:T^{\mathrm{\prime }}\to T$ with $u\tau ={\tau }^{\mathrm{\prime }}$ and $v{\tau }^{\mathrm{\prime }}=\tau$.

Both $v\circ u$ and ${\mathrm{id}}_T$ compose with $\tau$ to give $\tau$, so uniqueness in [L1] gives $v\circ u={\mathrm{id}}_T$; similarly $u\circ v={\mathrm{id}}_{T^{\mathrm{\prime }}}$.

Thus $u$ is an isomorphism with inverse $v$, and the same uniqueness clause shows that no other isomorphism carrying $\tau$ to ${\tau }^{\mathrm{\prime }}$ exists.

The pairing $(m,n)\mapsto f(m)\otimes g(n)$ is additive in both variables by [L2], and $(f(mr))\otimes g(n)=(f(m)r)\otimes g(n)=f(m)\otimes (rg(n))=f(m)\otimes g(rn)$, so it is balanced.

By [L1] the pairing of step 1.1 induces a unique homomorphism $f\otimes g$ with the stated formula.

The maps ${\mathrm{id}}_M\otimes {\mathrm{id}}_N$ and ${\mathrm{id}}_{M{\otimes }_{R}N}$ agree on every elementary tensor, so uniqueness in [L1] makes them equal.

The two sides of the composition formula both send $m\otimes n$ to $f^{\mathrm{\prime }}(f(m))\otimes g^{\mathrm{\prime }}(g(n))$, so uniqueness in [L1] makes them equal.

Steps 2.1, 3.1 and 3.2 prove existence, uniqueness, identity preservation, and composition preservation.

If $q$ is balanced, [L1] supplies a unique group homomorphism $Q$ satisfying $Q(m\otimes n)=q(m,n)$.

Conversely, if such a homomorphism $Q$ exists, composing it with the elementary-tensor map gives $q$; [L2] and additivity of $Q$ show that $q$ is additive in each variable and satisfies $q(mr,n)=q(m,rn)$, so $q$ is balanced.

For $R=M=N=\mathbb{Z}$, which is permitted by [L3], balance gives $2\otimes 1=1\otimes 2$ by [L2]. The function $q(m,n)=m$ assigns $2$ to $(2,1)$ and $1$ to $(1,2)$, so it is not balanced and no homomorphism can have the proposed elementary-tensor values.

Steps 1.1 and 1.2 prove the equivalence and uniqueness, while step 1.3 verifies the asserted failure.

For fixed $s\in S$, the pairing $(m,n)\mapsto m\otimes (ns)$ is additive in both variables and is balanced because $(mr)\otimes (ns)=m\otimes r(ns)=m\otimes (rn)s$ by [L1].

The left-handed construction is identical: for fixed $s\in S$, the pairing $(m^{\mathrm{\prime }},n^{\mathrm{\prime }})\mapsto (s{m}^{\mathrm{\prime }})\otimes n^{\mathrm{\prime }}$ is balanced by the commuting actions, and the induced maps satisfy the left module laws on elementary tensors.

By [L2] and [L3], step 1.1 induces an additive endomorphism $x\mapsto xs$ of $M{\otimes }_{R}N$ with $(m\otimes n)s=m\otimes (ns)$.

On every elementary tensor one has $(x(s+s^{\mathrm{\prime }}))=xs+x{s}^{\mathrm{\prime }}$, $(xs)s^{\mathrm{\prime }}=x(s{s}^{\mathrm{\prime }})$, and $x{1}_S=x$, using the right $S$-module laws of $N$. In each identity the two sides are additive maps that induce the same balanced pairing, so uniqueness in [L2] makes them equal on all $x$. Thus step 2.1 defines a right $S$-module structure.

When a left $S$-action and a right $T$-action are both present, $s((m\otimes n)t)=(sm)\otimes (nt)=(s(m\otimes n))t$ on elementary tensors, so the actions commute.

The formulas determine every action map on generators, so uniqueness follows from [L2].

For fixed $p\in P$, the pairing $(m,n)\mapsto m\otimes (n\otimes p)$ is $R$-balanced: $(mr,n)$ and $(m,rn)$ have the same image by the outer left $R$-action in [L1]. Thus [L2] and [L3] give a homomorphism $a_p:M{\otimes }_{R}N\to M{\otimes }_R(N{\otimes }_{S}P)$.

For fixed $m\in M$, the pairing $(n,p)\mapsto (m\otimes n)\otimes p$ is $S$-balanced, so [L2] and [L3] give $b_m:N{\otimes }_{S}P\to (M{\otimes }_{R}N){\otimes }_{S}P$.

The pairing $(x,p)\mapsto a_p(x)$ from $(M{\otimes }_{R}N)\times P$ is $S$-balanced. On generators, $a_p((m\otimes n)s)=m\otimes ((ns)\otimes p)=m\otimes (n\otimes sp)=a_{sp}(m\otimes n)$ by [L1], and additivity extends the equality to every $x$. Hence [L2] gives ${\alpha }_{M,N,P}$ with the displayed formula.

The pairing $(m,y)\mapsto b_m(y)$ is $R$-balanced. It is enough to check generators $y=n\otimes p$, where $b_{mr}(n\otimes p)=((mr)\otimes n)\otimes p=(m\otimes rn)\otimes p=b_m((rn)\otimes p)=b_m(r(n\otimes p))$ by [L1]. Therefore [L2] gives $\beta :M{\otimes }_R(N{\otimes }_{S}P)\to (M{\otimes }_{R}N){\otimes }_{S}P$.

The composite $\beta \alpha$ fixes every tensor $(m\otimes n)\otimes p$, and $\alpha \beta$ fixes every tensor $m\otimes (n\otimes p)$; successive applications of uniqueness in [L2] show that these composites are the identity maps.

The same elementary-tensor calculation shows compatibility with any outer actions, and replacing $m,n,p$ by their images under compatible homomorphisms proves naturality because both candidate composites agree on all elementary tensors.

Thus ${\alpha }_{M,N,P}$ is the asserted canonical natural isomorphism with inverse $\beta$.

For fixed $r\in R$, the pairing $(m,n)\mapsto (rm)\otimes n$ is additive in both variables and balanced: $(r(sm))\otimes n=((rs)m)\otimes n=((sr)m)\otimes n=(rm)\otimes (sn)$ by [L2].

By [L1] and [L3], step 1.1 induces an additive endomorphism $x\mapsto rx$ of $M{\otimes }_{R}N$ satisfying $r(m\otimes n)=(rm)\otimes n$. The tensor balance relation also gives $(rm)\otimes n=m\otimes (rn)$.

On elementary tensors, $(r+s)x=rx+sx$, $r(x+y)=rx+ry$, $(rs)x=r(sx)$, and $1_{R}x=x$ follow from the module axioms of $M$. In each identity the two additive maps induce the same balanced pairing, so uniqueness in [L1] makes the identity hold for every tensor.

Any $R$-module structure with the displayed formula has, for each $r$, a scalar-multiplication endomorphism inducing the same balanced pairing as step 2.1. Uniqueness in [L1] therefore makes the structure unique.

Steps 2.1, 3.1 and 3.2 give the asserted unique $R$-module structure and both elementary-tensor formulas.

The pairing $(m,n)\mapsto n\otimes m$ is balanced because $(rm,n)$ maps to $n\otimes (rm)=(rn)\otimes m$, which is also the image of $(m,rn)$; hence [L3] induces ${\sigma }_{M,N}$.

Regard all three modules as $(R,R)$-bimodules. Then [L2] supplies the displayed associativity isomorphism, and [L1] shows it is $R$-linear on elementary tensors.

Applying the same construction in the opposite order gives ${\sigma }_{N,M}$, and their composite fixes every $m\otimes n$; uniqueness in [L3] makes the composite the identity, so ${\sigma }_{M,N}$ is an isomorphism.

The map ${\sigma }_{M,N}$ is $R$-linear because $\sigma (r(m\otimes n))=\sigma ((rm)\otimes n)=n\otimes (rm)=r(n\otimes m)$.

Naturality of both maps follows by applying [L4]: after replacing the variables by their images under module homomorphisms, the two candidate composites agree on every elementary tensor, hence agree by [L3].

Steps 1.1 through 2.3 prove the two natural $R$-linear isomorphisms and the involutivity of symmetry.

The pairing $(r,n)\mapsto rn$ is additive in each variable and satisfies $(rs)n=r(sn)$, so it is balanced; [L1] and [L2] give ${\lambda }_N$.

The pairing $(m,r)\mapsto mr$ is balanced, so [L1] and [L2] give ${\rho }_M$. Define the additive map ${\eta }_M(m)=m\otimes 1_R$. Then ${\rho }_M{\eta }_M(m)=m$, while ${\eta }_M{\rho }_M(m\otimes r)=mr\otimes 1_R=m\otimes r$ by balance; uniqueness in [L1] makes the second composite the identity. Thus ${\rho }_M$ and ${\eta }_M$ are inverse.

The additive map ${\eta }_N:N\to R{\otimes }_{R}N$, $n\mapsto 1_R\otimes n$, satisfies ${\lambda }_N{\eta }_N(n)=n$, while ${\eta }_N{\lambda }_N(r\otimes n)=1_R\otimes rn=r\otimes n$ by balance. Uniqueness in [L1] makes the second composite the identity, so ${\lambda }_N$ and ${\eta }_N$ are inverse.

Each map commutes with any outer scalar action by associativity of that action, checked on elementary tensors. The calculations remain valid for the zero ring and for zero modules, where all displayed maps are the unique maps between zero groups.

Therefore the regular module is a left and right tensor unit with the stated natural formulas.

For $k=0$, an $R$-linear map $R\to P$ is uniquely determined by the image of $1_R$, and every $p\in P$ defines such a map by $r\mapsto rp$; this is the required representation of maps from the one-point empty product.

For $k=1$, the identity $M_1\to M_1$ represents linear maps from $M_1$ by composition.

Assume a parenthesized product $T_k$ represents $k$-linear maps. A $(k+1)$-linear map is equivalently a bilinear map $T_k\times M_{k+1}\to P$: first use the induction bijection with the last variable fixed, and then use multilinearity to see that the resulting dependence on the last variable is linear.

By [L1], the bilinear maps in step 1.3 correspond uniquely to linear maps $T_k{\otimes }_R{M}_{k+1}\to P$, proving the representing property for $k+1$.

By [L2], any two parenthesizations are joined by composites of elementary associativity isomorphisms preserving pure tensors. Any two such comparison maps agree on every pure tensor, so the representing uniqueness proved in step 2.1 makes them equal.

The base cases and induction step establish the representation for every finite $k$, including the empty and singleton cases, and step 3.1 proves independence of parenthesization.

For each $i$, the pairing $(m,n)\mapsto {\mathrm{ȷ}}_i(m)\otimes n$ is bilinear, so [L1] gives ${\varphi }_i:M_i{\otimes }_{R}N\to ({⨁}_i{M}_i){\otimes }_{R}N$. By [L4], the family $({\varphi }_i)$ induces the displayed map $\mathrm{\Phi }$.

Define $b:({⨁}_i{M}_i)\times N\to {⨁}_i(M_i{\otimes }_{R}N)$ by $b((m_i),n)=(m_i\otimes n{)}_i$. This tuple has finite support by [L3], and coordinatewise calculation shows that $b$ is bilinear.

By [L1], the pairing in step 1.2 induces $\mathrm{\Psi }:({⨁}_i{M}_i){\otimes }_{R}N\to {⨁}_i(M_i{\otimes }_{R}N)$ with $\mathrm{\Psi }((m_i)\otimes n)=(m_i\otimes n{)}_i$.

The composite $\mathrm{\Psi }\mathrm{\Phi }$ fixes every coordinate generator ${\mathrm{ȷ}}_i(m\otimes n)$, so it is the identity by [L4]; the composite $\mathrm{\Phi }\mathrm{\Psi }$ fixes every elementary tensor $(m_i)\otimes n$, so it is the identity by [L1].

Thus $\mathrm{\Phi }$ and $\mathrm{\Psi }$ are inverse isomorphisms. They are natural because maps $u_i:M_i\to M_i^{\mathrm{\prime }}$ and $v:N\to N^{\mathrm{\prime }}$ make the two composites send each coordinate generator ${\mathrm{ȷ}}_i(m\otimes n)$ to ${\mathrm{ȷ}}_i(u_i(m)\otimes v(n))$. When $I=\mathrm{\varnothing }$, [L3] makes both sides zero and the same construction yields the unique isomorphism.

For the second variable, put ${\mathrm{\Phi }}^{\mathrm{\prime }}:={\sigma }_{{⨁}_i{M}_i,N}\circ \mathrm{\Phi }\circ \left({⨁}_i{\sigma }_{N,M_i}\right)$, where the middle direct sum of the symmetries is the map induced by [L4] from the family ${\mathrm{ȷ}}_i\circ {\sigma }_{N,M_i}$. Each $\sigma$ is an isomorphism by [L5] and $\mathrm{\Phi }$ is one by step 4.1, so ${\mathrm{\Phi }}^{\mathrm{\prime }}$ is an isomorphism; it is natural as a composite of natural isomorphisms. Tracing a coordinate generator gives ${\mathrm{\Phi }}^{\mathrm{\prime }}({\mathrm{ȷ}}_i(n\otimes m))=\sigma (\mathrm{\Phi }({\mathrm{ȷ}}_i(m\otimes n)))=\sigma ({\mathrm{ȷ}}_i(m)\otimes n)=n\otimes {\mathrm{ȷ}}_i(m)$, which is the displayed formula. At $I=\mathrm{\varnothing }$ both sides are again zero.

The chosen bases identify $M$ with ${⨁}_{i\in I}R$ and $N$ with ${⨁}_{j\in J}R$, carrying $e_i,f_j$ to the corresponding standard basis vectors.

Apply [L1] in each variable and then [L2] to obtain canonical isomorphisms $M{\otimes }_{R}N\cong {⨁}_{i\in I}{⨁}_{j\in J}(R{\otimes }_{R}R)\cong {⨁}_{(i,j)\in I\times J}R=R^{(I\times J)}$.

Tracing a coordinate generator through step 2.1 sends it to $e_i\otimes f_j$; by [L3], those images therefore form a basis and every tensor has a unique finite expansion in them.

If $I=\mathrm{\varnothing }$ or $J=\mathrm{\varnothing }$, then $I\times J=\mathrm{\varnothing }$, the corresponding factor and the target free module are zero by [L3], and step 2.1 is the unique isomorphism between zero modules.

This proves the product-basis assertion in all cases.