Tensor product of a right R-module with an (R,S)-bimodule

Question

Wikipedia says: if $M$ is a right $R$-module and $N$ is an $(R,S)$-bimodule, then $M\otimes_RN$ is a right $S$-module. What I want to ask are the following two question:

1. How is this tensor product defined? Is it true that we first forget the right $S$-module structure on $N$, then construct $M\otimes_RN$ as an abelian group, and finally endow it with a right $S$-module structure?

2. Does this tensor product satisfy any universal property, similar to the one for vector spaces?

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I propose the following "universal property."

Let $V$ be a right $S$-module, and $f:M\times N\to V$ be an $R$-middle-balanced map that is right $S$-linear in the second variable, there is a unique morphism of right $S$-modules $\bar f:M\otimes_RN\to V$ such that $$ \begin{array}{ccc} M\times N & \stackrel{\otimes}{\longrightarrow} & M\otimes_RN \end{array} $$ $$ \begin{array}{ccc} & f{\Large\searrow} & ~~~{\Large\downarrow}~ \bar f \end{array} $$ $$ \begin{array}{ccc} & & V \end{array} $$ commutes.

Any help is appreciated, thank you!

Answer 1

If you have a right $R$-module $M$ and a left $R$-module $N$, then you can define an abelian group $M \otimes_R N$ in exactly the way you expect. It's generated by pairs $m \otimes n$ subject to the constraints

• $(m_1 + m_2) \otimes n = m_1 \otimes n + m_2 \otimes n$
• $m \otimes (n_1 + n_2) = m \otimes n_1 + m \otimes n_2$
• $mr \otimes n = m \otimes rn$

where in the third bullet we remember that $M$ is a right $R$-module and $N$ is a left $R$-module.

Now if $N$ is an $(R,S)$-bimodule (that is, both a left $R$-module and a right $S$-module), then of course we can forget about the $S$-module structure and take the tensor product $M \otimes_R N$, as you've guessed. Also as you've guessed, we can use the (right) $S$-module structure on $N$ to put a (right) $S$-module structure on $M \otimes_R N$! We define

$$(m \otimes n) s = m \otimes (ns)$$

One can check that everything here is well defined, but I'll leave those details to you.

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Yes, this tensor product satisfies a universal property. First, as is mentioned on the wikipedia page for tensor products, given a right $R$-module $M$ and a left $R$-module $N$, the tensor product $M \otimes_R N$ is the universal module with an "$R$-balanced map" from $M$ and $N$. Here an "$R$-balanced map" $f : M \times N \to A$ is a bilinear map to an abelian group $A$ with the ~bonus property~ that $f(mr,n) = f(m,rn)$.

In case $N$ is an $(R,S)$-bimodule, we have the same universal property. It just happens to be that one can show the representing object is also an $S$-module. This does have some interesting categorical content behind it, though! See here for more.

However in the special case that $N = S$, there is a nice universal property! In this case $S$ is a (left) $R$-algebra and there's a forgetful functor from $\mathsf{Mod}_S \to \mathsf{Mod}_R$ taking an $S$-module $M$ and viewing it as an $R$-module by defining $mr = m(r 1_S)$. This uses first the $R$-action on $S$ to get an element $r 1_S$, and then the $S$-action on $M$ to multiply $m$ by $(r 1_S)$!

In the special case $R = \mathbb{R}$ and $S = \mathbb{C}$, then this is saying that we can view a $\mathbb{C}$-vector space as an $\mathbb{R}$-vector space by defining $vr$ to be $v(r1_\mathbb{C})$, which is of course $vr$ where we view $r$ as a complex number, since $\mathbb{R} \subseteq \mathbb{C}$!

You might ask if this forgetful functor $\mathsf{Mod}_S \to \mathsf{Mod}_R$ has a left adjoint. That is, if there's a way to freely "extend" an $R$-module $M$ to an $S$-module. So in our example, if there's a way to freely extend a real vector space to a complex vector space. It turns out the answer is yes, and we do it by tensoring with $S$! You already know that $S$ is a $(R,S)$-bimodule, so $-\otimes_R S$ is a functor $\mathsf{Mod}_R \to \mathsf{Mod}_S$. The incredibly useful fact is that it's left adjoint to the forgetful functor we described earlier! If you want to read more about this, the key word to google is extension of scalars. You can also find a worked out proof of this adjunction here.

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I hope this helps ^_^