Let $f: R \to S$ and $g: R \to T$ be two $R$-algebras. To show that $S \otimes_R T$ is an $R$-algebra I need to define a ring structure (multiplication) on it and a ring homomorphism $h : R \to S \otimes_R T$.
Using the universal property of the (multi-)tensor product, defining multiplication is clear to me. What I'm confused about is the map $h : R \to S \otimes_R T$. According to this answer here, $r \mapsto 1 \otimes g(r) = f(r) \otimes 1$.
How are they the same?
Best Answer
The basic fact is that for two $R$-modules $M,N$ you have:
$$r\cdot m\otimes _R n=m\otimes_R r\cdot n \quad (\text {for all} \quad r\in R,\; m\in M,\; n\in N) \quad (*)$$
In particular, in your case, you have: $$r\cdot 1_S\otimes _R 1_T=1_S \otimes_R r\cdot 1_T \quad (**)$$
In order to conclude, you just have to remember that built into the notion of algebra is the equality $r\cdot s=f(r)s$ where on the right hand side of the equality $f(r)s$ means the product of the elements $f(r),s$ in the ring $S$.
In particular $r\cdot 1_S=f(r)1_S=f(r)$ and similarly $r\cdot 1_T=g(r)$.
Transporting these last equalities into $(**)$, you get the required equation $$ f(r) \otimes_R 1_T= 1_S \otimes_R g(r) \quad (***) $$