# $f:A\to B$ ring homomorphism, $S$ multiplicatively closed subset of A, what does $S^{-1}B$ look like as an $S^{-1}A$ module?

So, I'm working on Atiyah-McDonald, problem #3.4. I'll present the problem and then my confusion, I'm not looking for a solution to the problem itself, just to my confusion at the structure of the question itself.

So, the statement of the question: we have $f:A\to B$ a ring homomorphism, $S\subset A$ multiplicatively closed, $T=f(S)$. Show that $S^{-1}B$ and $T^{-1}B$ are isomorphic as $S^{-1}A$ modules.

Okay, so I get that somehow these two structures are supposed to be $S^{-1}A$ modules, but first, I'm not even sure what $S^{-1}B$ looks like.

To recall the definition I was given, $S^{-1}B$ would only be defined for $S$ being a multiplicatively closed subset of B, which it's not...its not even a subset of B! Then it would be a the set of equivalence classes: $(b_1,s_1)\equiv (b_2,s_2)$ iff $\exists u\in S$ such that $(s_2b_1 -s_1b_2)u=0$ I don't even see those products as being well defined, unless I'm supposed to map $s_1$ and $s_2$ to their images under $f$?

Then of course I'd have to be able to treat this ring as a $S^-1A$ module, which now means I need to be able to multiply fractions, one with numerators in $A$ and one with numerators in $B$....again, am I supposed to send the image of the thing in $A$ through $f$, or is something else going on here?

• It is useful to think of $S^{-1}B$ as $S^{-1}A \otimes_A B$. (they are isomorphic) Commented Oct 1, 2015 at 18:43
• @FredrikMeyer Hmm, will have to ponder that tomorrow. Still struggling with wrapping my head around tensor products
– Alan
Commented Oct 2, 2015 at 1:18

If $M$ is an $A$-module, and $S$ a multiplicatively closed set of $A$, one defines $S^{-1}M$, $M$ localized at $S$, or the module of $S$ fractions of $M$, as equivalence classes $m/s$ of pairs $(m,s)$ as you have above. The resulting module is an $S^{-1}A$ module. In my book, this is done on the bottom of page 38, after example 5ii, where the authors helpfully say that the $S^{-1}A$ action is obvious - but one should check that $a/s \cdot m/t = (am)/(st)$ on equivalence classes makes sense - the analogue of the exercise at the top of my page 37, ahead of proposition 3.1.

The ring $B$ is an $A$-module, with, as you say, $A$-action through $f$: by definition $a\cdot b = f(a)b$, where the product on the right is the $B$-product. Write the resulting $S^{-1}A$ module $S^{-1}B$.

On the other hand, writing $T = f(S)$, one defines $T^{-1}B$, where the construction is done with everything viewed "in $B$." Since $T$ is multiplicatively closed in $B$, one can define the ring $T^{-1}B$. The point is, I think, after the fact, that $T^{-1}B$ is an algebra over $S^{-1}A$ (Prop 3.1)- so a $S^{-1}A$ module.

The point is, I think, to compare the two constructions: the first with equivalence classes using pairs $(b,s)$, and the second using pairs $(b,f(s))$.

• Okay, so I do just factor everything through $f$ to make the danged things make sense? Got it, thanks.
– Alan
Commented Oct 1, 2015 at 12:19
• And, just to be pc, at least in the first version, you should write $u(s_2b_1 -s_1b_2)=0$ i.e., the $u$ on the left, if we are viewing $B$ as an left $A$-module - in your version, you wrote $u$ on the right... Commented Oct 1, 2015 at 12:24
• FYI - On page 36, AM do as you did, and write the $u$ on the right, but for modules on page 38 they write the $u$ on the left. Commented Oct 1, 2015 at 12:31
• Ehh, in this course we're dealing with commutative algebra, so there's no need to distinguish between left and right modules :)
– Alan
Commented Oct 2, 2015 at 1:18