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I am currently working through a problem in Rotman:

Let $R$ be a domain and let $Q=$Frac$(R)$. If $r\in R$ is nonzero and $A$ is an $R$-module for which $rA=0$, prove that for all $n\geq 0$, $\mathrm{Ext}_R^n(Q,A)=0=\mathrm{Tor}_n^R(Q,A)$.

I think I have the Tor part:

Since $Q$ is flat, Tor$_n^R(Q,A)=0$ for all $n\geq 1$. And Tor$_0^R(Q,A)=Q\otimes_R A=0$ because: for each $\frac{t}{s}\otimes a\in Q\otimes A$, $\frac{t}{s}\otimes a=\frac{tr}{sr}\otimes a=\frac{t}{sr}\otimes ra =0$.

However, I am having trouble with the Ext part. To do it directly I either need a projective resolution for $Q$ or an injective resolution for $A$, but I am not sure how I would resolve either of these. Or I considered using a long exact sequence for the short exact sequence $0\rightarrow R\rightarrow Q\rightarrow Q/R\rightarrow 0$, but that didn't seem to get me anywhere either.

I have been stuck on this problem for far too long so any help or hints are greatly appreciated.

user26857
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1234
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1 Answers1

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Hint: choose a projective resolution of $Q$, and consider the map $Q\to Q$ given by $q\mapsto rq$. It's an $R$-module homomorphism, so it extends to the projective resolution. It gives a map $\mathrm{Ext}^i_R(Q,A)\to\mathrm{Ext}^i_R(Q,A)$, which is also multiplication by $r$. What can you say about this map? Now, $q\mapsto \tfrac{1}{r}q$ is also an $R$-module homomorphism $Q\to Q$, so it must also extend in some way to a chain map of the projective resolution. What happens when you compose this with the chain map obtained from $q\mapsto rq$?. What does this tell you about the multiplication by $r$ map on $\mathrm{Ext}^i_R(Q,A)\to\mathrm{Ext}^i_R(Q,A)$?

  • Oh, I think I have it! I recently proved the following: If $R$ is a commutative ring and $M$,$N$ are $R$-modules then Ann$_R(M)\cup$ Ann$_R(N)\subset$ Ann$_R($Ext$_R^i(M,N))$. It seems that your hint is pushing me to prove this result. Once, I have this I can then show the map given by multiplication by $r$ on Ext$_R^i(Q,A)$ is actually the zero map, but it is also an isomorphism (I get this from a little work with a long exact sequence). Hence Ext$_R^i(Q,A)=0$. Does this sound like I am on the right track? – 1234 Jun 28 '15 at 19:55
  • @1234: Yes, that sounds good! However, for the isomorphism part, you can still use the fact that any map $Q\to Q$ extends uniquely up to chain homotopy to the projective resolution. –  Jun 28 '15 at 21:54