Tensor Products of R-Algebras from Atiyah and Macdonald

Question

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: $$\lambda(M_0)-\lambda(M_1)+\lambda(M_2)=0$$

Answer 1

That is an incorrect internal reference. It should say (2.12) instead of (2.11).

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.