I am trying to understand and prove the following identity, where $I$ is an ideal of a (unital) $R$-algebra $S$ and $T$ another $R$ algebra; here $R$ is a commutative ring: $$ (S/I) \otimes T = (S\otimes T)/[(I \otimes 1)( S\otimes T)]. \space\space(\ast) $$
(Exercise I-46 of The Geometry of Schemes by David Eisenbud, Joe Harris.)
The first issue is the meaning of the square bracket $[]$. If $I$ is an ideal of $S,$ then it is closed under multiplying anything in $S,$ so is it the case that $(I \otimes 1)( S\otimes T)=I \otimes T?$
If so, why are we writing things in the "more complicated" form? Is there anything special to $(I \otimes 1)?$
Now, I can prove the result $(\ast)$ by isomorphism theorem of course. But how to prove it using the universal property of tensor products? (This is the intention of the exercise.) I tried the triangular diagram of some obvious maps between the sets $(S/I)\times T, (S/I)\otimes T, (S\otimes T)/[(I \otimes 1)( S\otimes T)],$ but unable to make sure the map I get is an isomorphism, without going into details of the structure. What's the most easy way to prove this using universal property?
You have maps $$S \rightarrow S/I \rightarrow (S/I)\otimes_R T$$ $$T\rightarrow (S/I)\otimes_R T$$ composed from the quotient $S\rightarrow S/I$ and the maps of $S/I$ and $T$ to $(S/I)\otimes_R T$. Since tensor is a coproduct in the category of $R$-algebras this yields a map $$\varphi: S\otimes_R T \rightarrow (S/I)\otimes_R T$$ Now show this map is surjective and characterize the kernel to get $S\otimes_R T/\operatorname{Ker}\varphi\simeq (S/I)\otimes_R T$.
Note: the map $S/I\rightarrow (S/I)\otimes_R T$ is given by $x\rightarrow (x\otimes 1_T)$ is part of what makes $\otimes_R$ a coproduct in the category of $R$-algebras. Simlilary the map $T\rightarrow (S/I)\otimes_R T$ is given by $y \rightarrow 1_{S/I}\otimes y$. So that ultimately $\varphi(x\otimes y)=(\bar{x}\otimes 1_T)(1_{S/I}\otimes y)=\bar{x}\otimes y$ where $\bar{x}\in S/I$ for $x\in S$.