Exact sequences and derived functors

Question

Let $R$ be a commutative ring and $ r\in R$. Denote $R_r = \{ s\in R \mid s r =0 \}$. If $R_r $ is nonzero $0 \rightarrow R_r \rightarrow R\stackrel{r\cdot} \rightarrow R \rightarrow R/ rR \rightarrow 0$ is exact. Deduce that $$ 0 \rightarrow \mathrm{Tor}_2^R( R/ rR, M) \rightarrow R_r \otimes_RM \rightarrow M \rightarrow \mathrm{Tor}_1^R( R/ rR, M) $$ is exact as well and the most right arrow is surjective.

I try to apply the long exact sequence of $\mathrm{Tor}$. I deduce just that $\mathrm{Tor}_2^R( R/ rR, M)$ is isomorphic to $\mathrm{Tor}_1^R( R_r, M)$ but I think this is not useful. Can you give me a hint?

Maybe try to deduce from your exact sequence two short exact sequences, and consider the two homology long exact sequences. — Bernard, Jun 09 '17 at 13:43
@Bernard I proved the exactness excepting the last arrow. Can you give me concrete idea because I tried all the possibilities...... — Problemsolving, Jun 09 '17 at 20:12

score 1 · Accepted Answer · edited Jun 14 '17 at 08:39

1

The sequence you give is not exact. To see this you can take $M=R$. So the last term must be $rR\otimes_RM$.

I think if $M$ is good enough satisfying some property you can get this sequence.

edited Jun 14 '17 at 08:39

user26857

52,094

answered Jun 10 '17 at 03:16

Jian

2,470

Exact sequences and derived functors

1 Answers1