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.

  • $\begingroup$ If I recall corectly you have $R/I\otimes S=R\otimes S/I=\cong (R\otimes S)/I$ $\endgroup$ – Zelos Malum Dec 15 '15 at 8:06

An $R$-algebra can be thought of in a few ways, but it is just a ring $A$ such that there in a ring homomorphism $f : R \to A$. It's similar to an $R$-module in that we can "scalar multiply" elements of $A$ by elements of $R$, but $A$ also has an internal multiplicative structure. I found Prof Keith Conrad's notes on tensor products to be quite helpful when I was learning about them for the first time. Two of the big ideas are to think of tensor products (over a ring $R$) as algebraic objects satisfying a certain universal property having to do with bilinear maps (over $R$) and that we can build them by taking a free $R$-module and modding out by the relations we want to impose. The link is: http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod.pdf

To say some stuff on your questions, I think you have the right idea for why $R \otimes_{R} S \cong S$. To make your argument more precise, you can use the common technique of first taking an $R$-bilinear map $\phi: R \times S \to S$, using the universal property of the tensor product to get that it factors through $$ R \times S \overset{\rho}{\to} R \otimes_R S \overset{\psi}{\to} S $$ (i.e. $\psi \circ \rho = \phi$), and then defining a map $S \to R \otimes_R S$ and showing that it is the inverse of our map $\psi$. In this case, we can just take the map $R \times S \to S, (r,s) \mapsto rs$ (and we extend linearly). You can check this map is $R$-bilinear, and so we get a map $\psi : R \otimes_R S \to S$ that satisfies $\psi(r \otimes s) = rs$ on basic tensors. The inverse map is simply $s \mapsto 1 \otimes s$ (you can check this). This shows $R \otimes_R S \cong S$.

To show $R/I \otimes_R S \cong S/IS$, you can use the same general technique I exhibited above, but you need to pick the correct ($R$-bilinear) morphism. If you are stuck, you can look at Theorem 4.5 in the linked notes (this also gives another proof of the above as mentioned in the notes).

For your last question about why $I \otimes_R S \cong IS$, I think it follows from the fact that your second sequence is exact, so if $f, g$ are your first and second maps we have $im(f) = ker(g)$, and so $I\otimes_R S \cong im(f) \cong ker(g) \cong IS$ where the last isomorphism is given by the isomorphism $R/I \otimes_R S \cong S/IS$. You are right that $S/(I \otimes_R S)$ is not well defined, but the object is really $S$ modulo the image of $I \otimes_R S$, and the intuition is reasonable.


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