# Tensor Products of R-Algebras from Atiyah and Macdonald

I know this part of Atiyah and Macdonald has a typo, but that is not what this question is about.

Let $$R$$ be a commutative ring and $$S$$ and $$T$$ be $$R$$ algebras. I am trying to show that $$S\otimes_R T$$ has the structure of an $$R$$ algebra. To do this I first want to show that $$S\otimes_R T$$ is a ring, so I need a multiplication map. I start with defining the map:

\begin{align*} \phi:(S\times T)\times (S\times T)&\longrightarrow (S\otimes _R T)\\ ((s_1,t_1),(s_2,t_2))&\longmapsto (s_1s_2)\otimes (t_1t_2) \end{align*} which is clearly $$R$$ linear in each entry, hence by the universal property of tensor products (or I guess more precisely a consequence of this property) factors through a map: \begin{align*} \psi:(S\otimes_R T)\otimes_R (S\otimes_R T)&\longrightarrow (S\otimes _R T) \end{align*} which satisfies: \begin{align*} \phi((s_1,t_1),(s_2,t_2))=\psi\circ \otimes^4((s_1,t_1),(s_2,t_2)) \end{align*} where $$\otimes^4:(S\times T)\times (S\times T)\rightarrow (S\otimes_R T)\otimes_R (S\otimes_R T)$$ is the map such that: \begin{align*} \otimes^4((s_1,t_1),(s_2,t_2))=s_1\otimes t_1\otimes s_2\otimes t_2 \end{align*} The above all makes sense to me, but then Atiyah and Macdonald state that by proposition $$(2.11)$$ $$\psi$$ corresponds to a bilinear map: $$\mu:(S\otimes_R T)\times(S\otimes_R T)\longrightarrow (S\otimes_R T)$$ I do not understand this step at all. Namely, proposition $$(2.11)$$ is:

Let $$0\rightarrow M_0\rightarrow M_1\rightarrow\cdots\rightarrow M_n\rightarrow 0$$ be an exact sequence of $$R$$ modules in which all the modules $$M_i$$ and the kernels of all the homomorphisms are belong to a class of modules $$C$$. Then for any additive function $$\lambda$$ on $$C$$ we have: $$\sum_{i=0}^n(-1)^i\lambda(M_i)=0$$

I believe I understand the proof of this statement, but I just do not see how it applies in this case at all. Is there a different way to see that there is a bilinear map $$\mu$$ which corresponds to the map $$\psi$$? Any help would be appreciated.

Edit:

So Atiyah and Macdonald don't actually define a class of modules from what I can see, but the way it is used seems to imply it is simply a collection of $$R$$ modules.

An additive function on $$C$$ is then a function is then a function on $$C$$ with values in an abelian group such that for every short exact sequence: $$0\rightarrow M_0\rightarrow M_1\rightarrow M_2\rightarrow 0$$ where each $$M_i\in C$$ we have: $$\lambda(M_0)-\lambda(M_1)+\lambda(M_2)=0$$

• What is the meaning of "class of modules"? Commented Feb 26, 2023 at 22:52
• @FShrike I will edit the post with the definitions. Commented Feb 26, 2023 at 22:56

Anyway, setting $$M = S \otimes_R T$$, you are asking how a linear map $$\psi \colon M \otimes_R M \to M$$ corresponds to a bilinear map $$\mu \colon M \times M \to M$$. Do you see that this is simply the universal mapping property of tensor products? That is (2.12).
For another approach to showing the tensor product of two $$R$$-algebras has a unique $$R$$-algebra structure such that elementary tensors multiply in the natural way, see Theorem 7.1 here.
• That would make sense. Perhaps I am misunderstanding something though, I thought I first needed a bilinear map $\mu$ to get a unique linear map $\psi$. Can I go the other way? Commented Feb 26, 2023 at 22:57
• Of course you can go the other way: just compose $\psi$ with the standard bilinear map $M\times M\to M\otimes_R M$ to get a bilinear map on $M\times M$. If $N$ is a normal subgroup of $G$, do you see that group homomorphisms $G/N\to H$ are in natural bijection with group homomorphisms $G\to H$ that are trivial on $N$? If $I$ is an ideal in a (commutative) ring $R$, do you see that ring homomorphisms $R/I\to S$ are in natural bijection with ring homomorphisms $R\to S$ that vanish on $I$? Similarly, linear maps $M\otimes_R N\to P$ are in natural bijection with bilinear maps $M\times M\to P$.