Postmodern Algebra

Proof. The universal property of the tensor product says that to give an \(R\)-module homomorphism \(M \otimes_{R} N \longrightarrow P\) is the same as giving an \(R\)-bilinear map \(M \times N \longrightarrow P\text{.}\) Given such a bilinear map \(f\text{,}\) the map \(n \mapsto f(m \otimes n)\) is \(R\)-linear for each \(m \in M\text{,}\) so it defines an \(R\)-module homomorphism \(N \longrightarrow P\text{.}\) Now the assignment

is \(R\)-linear, \(f\) is an \(R\)-module homomorphism, and \(m \mapsto m \otimes n\) is \(R\)-linear on \(m\text{.}\)

Conversely, given an \(R\)-module homomorphism \(f \in \operatorname{Hom}_{R}\left(M, \operatorname{Hom}_{R}(N, P)\right)\text{,}\) one can check (exercise!) that \((m, n) \mapsto f(m)(n)\) is an \(R\)-bilinear map, so it induces an \(R\)-module homomorphism \(M \otimes_{R} N \longrightarrow P\text{.}\) Moreover, the two constructions are inverse to each other.

So we have constructed a bijection of Hom-sets

It’s routine to check that both of these bijections are indeed homomorphisms of \(R\)-modules, so we leave it as an exercise.

Finally, naturality means we have the following commutative diagrams:

and

We leave checking these do indeed commute as an exercise.

Proof. The adjointness translates into the fact that for all \(R\)-modules \(N\) and \(P\) there is a bijection

which is natural on \(N\) and \(P\text{,}\) which is a corollary of Theorem 3.60.

Proof. Consider the map

Fix \(f\text{.}\) For each \(m \in M\text{,}\) let \(\tau_{m}\) be the map \(N \longrightarrow P\) defined by \(\tau_{m}(n):=f(m \otimes n)\text{.}\) Note that \(\tau_{m}\) is indeed a homomorphism of \(S\)-modules, since it is the composition of two \(S\)-module maps, \(f\) and \(m \otimes_{R} \operatorname{id}_{N}\text{,}\) where \(m\) is the constant map \(M \longrightarrow M\) equal to \(m\text{.}\)

We should check that our proposed map \(\tau\) is indeed a map of abelian groups. It is immediate from the definition that \(\tau\) sends the 0-map to the 0-map. Moreover, given \(S\) module homomorphisms \(f, g: M \otimes N \longrightarrow P\text{,}\) and any \(n \in N\text{,}\) we have

so \(\tau_{m}(f+g)=\tau_{m}(f)+\tau_{m}(g)\) for all \(m \in M\text{,}\) and thus \(\tau(f+g)=\tau(f)+\tau(g)\text{.}\)

Suppose that \(\tau(f)=0\text{.}\) Then for every \(m \in M\) and every \(n \in N\text{,}\)

so \(f\) vanishes at every simple tensor, and we must have \(f=0\text{.}\) On the other hand, if we are given \(g \in \operatorname{Hom}_{R}\left(M, \operatorname{Hom}_{S}(N, P)\right)\text{,}\) consider the map \(M \times N \longrightarrow P\) defined by \(\tilde{f}(m, n)=g(m)(n)\text{.}\) Since \(g\) is a homomorphism of \(R\)-modules, it is \(R\)-linear on \(m\text{.}\) Moreover, for each fixed \(m, g(m)\) is a homomorphism of \(S\)-modules, so in particular \(g(m)\) is \(R\)-linear. Together, these say that \(\tilde{f}\) is an \(R\)-bilinear map. Let \(f\) be the homomorphism of \(R\)-modules \(M \otimes_{R} N \longrightarrow P\) induced by \(\tilde{f}\text{.}\) By definition, \(f(m \otimes n)=\tilde{f}(m, n)=g(m)(n)\text{,}\) so \(\tau(f)=g\text{.}\) We conclude that \(\tau\) is a bijection.

We leave the statements about naturality as exercises.

Proof. We need to show that for every \(R\)-module \(M\) and every \(S\)-module \(N\) there are bijections

which are natural on both \(M\) and \(N\text{.}\) By Theorem 3.62, we have natural bijections

The module \(M \otimes_{R} S\) is precisely \(f_{*}(M)\text{.}\) By Exercise \(38, \operatorname{Hom}_{S}(S, N) \cong N\) as an \(S\)-module. An isomorphism of \(S\)-modules \(\operatorname{Hom}_{S}(S, N) \longrightarrow N\) is in particular an \(R\)-linear map, and thus also an isomorphism of \(R\)-modules. So \(\operatorname{Hom}_{S}(S, N) \cong f^{*}(N)\) as \(R\)-modules. Therefore, the Hom-tensor adjuntion gives us the natural bijections we were looking for.