Rings, Modules, Fields

6 Tensor products

Given a homomorphism \(f: R \to S\), via composition \(R \to S \to End(O)\), every \(S\)-module structure on \(O\) induces an \(R\)-module structure. In particular, \(f\) gives a functor \(f^* :S\)-Mod\(\to R\)-Mod, and the map \(f\mapsto f^*\) is a functor from \(\Ring \to \Cat \). This is analogous to the functor \(\Grp \to \Cat \) for group actions.

Given an \(S\)-module \(N\), the tensor product with respect to \(N\) is the left adjoint of the functor \(f^*\Hom _S(N,-)\). In fact, we have an isomorphism \(\Hom _S(M\otimes _fN,P) \cong \Hom _R(M,f^*\Hom _S(N,P))\) that is natural in \(M,N,P\). If the map \(f\) is clear, it is also sometimes denoted \(M \otimes _SN\).

The tensor product also behaves nicely with respect to \(f\) in that is associative. More precisely, if we have \(R \xrightarrow {f} S \xrightarrow {g} T\), then we have the isomorphism \((M \otimes _f N)\otimes _gP \cong M \otimes _{g\circ f}(N\otimes _g P)\), which is coherent and natural in \(M,N,P\). Moreover, the tensor product preserves biproducts, meaning that there are natural isomorphisms \(M \otimes (N\oplus N') \cong (M \otimes N) \oplus (M \otimes N')\) and \((M\oplus M') \otimes N \cong (M \otimes N) \oplus (M' \otimes N)\).

Let’s construct the tensor product \(M \otimes _f N\) now. It will be a quotient of the module freely generated by elements denoted \(m\otimes n\), with \(m\in M,n \in N\). The relations will be \(rm\otimes n = m\otimes f(r)n\) for \(r \in R\), \(m\otimes n+m\otimes n' = m\otimes (n+n')\), \(m\otimes n+m'\otimes n = (m+m')\otimes n\). The action of \(S\) will be given by \(s(m\otimes n) = m\otimes sn\). Now given a map \(g:M\otimes _f N\to P\), the corresponding map \(g':M \to f^*Hom_S(N,P)\) is given by \(g'(m)(n) = g(m\otimes n)\).

It is worth mentioning a few special cases. We call \(M \otimes _f S\) the extension of scalars of \(M\) from \(R\) to \(S\) (via \(f\)). For example, if \(A\) is a monoid and \(I\) an ideal in \(R\), then there are natural isomorphisms \(M \otimes _{A^{-1}R} A^{-1}R \cong A^{-1}M\) and \(M \otimes _{R/I}R/I \cong M/IM\). Note also that extending a free module gives a free module. Finally another special case is when \(f\) is the identity. Then the tensor product, often written here \(\otimes _R\) or \(\otimes \), along with \(R\) as the identity, gives a symmetric monoidal structure to \(R\)-Mod (as mentioned before, \(R\)-algebras are monoids under this monoidal structure).

We would like to generalize the tensor product construction for two \(R\)-modules to a collection of modules over an arbitrary index set \(I\). To do this, we must realize that this special case is genuinely different from the general tensor product since it is symmetric. In particular, there is another way to phrase the universal property. Given \(R\)-modules, \(M,N\), we can say that a function \(M\times N \to P\) is bilinear if it is linear when only one of \(M\) or \(N\) varies. Then if Bilin\((M,N)\) is the functor taking \(P\) to the set of bilinear maps \(M \times N \to P\), Bilin\((M,N)\) is represented by \(M\otimes N\).

Rephrasing it this way makes it easy to generalize to arbitrary index sets. If \(I\) is an index set, let \(M_I\) be a set of modules indexed by \(I\), and say that a function \(\prod _I M_I \to P\) is I-multilinear if it is linear when only one index varies. If Multi\(_I(M_I)\) is the corresponding functor in this situation, then \(\bigotimes _IM_I\) is the representing object.

The construction of \(\bigotimes _IM_I\) is more or less the same as that of \(M_I\). Namely, consider the free module generated by elements of the form \(\bigotimes m_I\), and mod out by the same relations. Now given a multilinear map \(g\). send \(\bigotimes m_I\) to wherever \(g\) sends \((m_I)\).

There are other nice properties of infinite tensor products. For example, if \(A_I\) are sets indexed by \(I\), and \(M_{A_I}\) are indexed modules, then there is a natural isomorphism \(\bigotimes _I(\bigotimes _{A_I}M_{A_I}) \cong \bigotimes _{\coprod _I A_I} M_{A_I}\) (this is a general associativity). The reason natural makes sense is because tensor product is also functorial, since given \(M_I,N_I\), and maps \(g_i:M_i\to N_i\), then we get a map \(\bigotimes _IM_I \to \bigotimes _IN_I\), that is constructed by considering the multilinear map \((m_i) \mapsto \bigotimes g_i(m_i)\). Now we can try to construct the coproduct of \(R\)-algebras \(R_i, i \in I\).

The underlying \(R\)-module of \(\coprod _I R_I\) will be the submodule of \(\bigotimes _I R_I\) generated by simple tensors where all but finitely many of the entries are \(1\). The identity will be \(\bigotimes _I 1_I\), and multiplication will be the tensor product of multiplication on each \(R_I\). Concretely, that means that \((\bigotimes r_I)(\bigotimes s_I) = (\bigotimes r_Is_I)\). Associativity follows from the general associativity of the tensor product and the fact that multiplication in each \(R_i\) is associative. The inclusions \(R_i \to \coprod _I R_I\) are given by mapping every other coordinate to \(1\). Now given homomorphisms \(g_i:R_i \to S\), it is clear that if the homomorphisms factor through \(\coprod _I R_I\), the factorization is unique. Existence follows from seeing that the only possible homomorphism is indeed a homomorphism.