How can I induce an exact sequence?

Question

Let $$0 \rightarrow M'\rightarrow M \rightarrow M'' \rightarrow 0\quad (*)$$ be an exact sequence, and $I$ an ideal. How can I induce the following sequence: $$0\rightarrow \frac{I^nM\bigcap M'}{I^nM'}\rightarrow \frac{M'}{I^nM'}\rightarrow\frac{M}{I^nM}\rightarrow\frac{M''}{I^nM''}\quad(**)$$ and why is it exact?

Here is my work:

$M''=M/M'$ since the sequence is exact. So we have:

$M''/I^nM''=\frac{M/M'}{I^nM/M'}=\frac{M}{I^nM+M'}$, but then I'm lost. I can only get a short sequence which is:

$0\rightarrow\frac{I^nM+M'}{I^nM} \rightarrow\frac{M}{I^nM}\rightarrow\frac{M''}{I^nM''}\rightarrow 0$.

Can you please show me how to get sequence $(**)$?

Answer 1

Hint:

Tensor the exact sequence with $R/I^n$ (you didn't mention the name of the ring). You obtain the exact sequence: $$M'\otimes_RR/I^n\longrightarrow M\otimes_RR/I^n\longrightarrow M''\otimes_RR/I^n\longrightarrow 0,$$ and ask yourself what is the kernel of $$M'\otimes_RR/I^n\simeq M'/I^nM'\longrightarrow M/I^n M.$$

Some details:

As the mapping $\varphi:M'/I^nM'\longrightarrow M/I^nM$ is induced by the canonical injection $M'\hookrightarrow M$, an element $m'+I^nM'\in M'/I^nM'$ maps to $0$ in $M/I^nM$ if and only if $$\varphi(m'+I^nM')\overset{\mathrm{def}}{=}m'+I^nM =I^nM,$$ which means $m'\in I^nM$; as it also is in $M'$, this means $\;m'\in I^nM\cap M'$, so that $$\ker\varphi=(I^nM\cap M')/I^nM'.$$