Sorry if this question isn't quite precise. Anyways, I'm reviewing for my algebra final right now and two things I don't think I quite grasp as well as I'd like are tensor products and $R$-algebras. I have a great professor but he didn't spend all that much time using these two objects (especially $R$-algebras, he just kind of defined them and hardly used them aside from a few examples).
As an example of something I'm not understanding well, he gave us this example:
Let $R$ be a ring, $S$ an $R$-algebra and $I\subset R$ an ideal. The sequence $$0\to I\to R\to R/I\to 0$$ is exact (which I understand fine). If $S$ is flat as an $R$-module, then $$0\to I\otimes S\to R\otimes S\to (R/I)\otimes S\to 0$$ is exact, and $R\otimes S\cong S$, $(R/I)\otimes S\cong S/(IS)$ so $$0\to I\otimes S\to S\to S/(IS)\to 0$$
is an exact sequence. This implies $I\otimes S\cong IS$.
So, as far as things I don't understand here; I don't see why we get the two isomorphisms $R\otimes S\cong S$ and $(R/I)\otimes S\cong S/(IS)$. I'm fairly certain the first has to do with the fact we can "pull out" terms from the $R$ part of the tensor product and put them in the $S$ side with the action of $R$ on $S$, so you can basically just get $1\otimes s$ for any pure tensor in $R\otimes S$. The other I don't really understand at all, although I'm sure it's along the same lines.
I also don't really see why $I\otimes S\cong IS$. I think it probably has to do with the first isomorphism theorem giving us something like $$S/(IS)\cong S/(I\otimes S)$$ (although clearly $S/(I\otimes S)$ isn't actually a well-defined object, that's just the idea behind it), because the kernel of the second map equals the image of the first, which is really just $I\otimes S$.
Does anybody know of any good references going over some things related to these types of things? And if anybody is feeling particularly generous, letting me know if my thoughts are in the right direction/what I'm missing? I didn't mean for this post to end up so long, sorry about that.