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Let $0 \rightarrow M' \xrightarrow{f} M \xrightarrow{g} M'' \rightarrow 0$ be an exact sequence of modules on a commutative unitary ring.

We say that the exact sequence is split when M can be written as the direct sum of $M_1$ and $M_2$ such that $f$'s corestriction to $M_1\oplus \lbrace 0 \rbrace$ is an isomorphism and $g$'s restriction to $\lbrace 0 \rbrace \oplus M_2$ is also one.

Is it correct to say that it is split if and only if $M \simeq \text{Im}f \oplus M/\text{Im}f$ ? I think this property is always true of vector spaces and probably free modules, and I know that this kind of exact sequence is always split when it involves vector spaces.

I ask this because $f$'s corestriction to $\text{Im}f$ is obviously an isomorphism since $f$ is injective, and because $\text{Im}f=\text{ker}g$, $g$ is actually a function on the supplementary submodule alone.

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  • $\begingroup$ What is your definition of a split exact sequence? $\endgroup$ – Bernard May 25 '17 at 21:50
  • $\begingroup$ Thanks, I've edited the post. $\endgroup$ – James Well May 25 '17 at 23:01
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No, this is not correct. For instance, let the ring be $\mathbb{Z}$ and let $M=\mathbb{Z}\oplus V$ where $V$ is an infinite-dimensional vector space over $\mathbb{Z}/2\mathbb{Z}$. Let $M'=2\mathbb{Z}\oplus 0\subset M$ with $f$ the inclusion map, and let $M''=M/M'$ with $g$ the quotient map. Note that $M'\cong\mathbb{Z}$ and $M''\cong \mathbb{Z}/2\mathbb{Z}\oplus V\cong V$ since $V$ is infinite-dimensional, so $M\cong M'\oplus M''$.

However, the short exact sequence $0\to M' \to M\to M''\to 0$ does not split. Indeed, notice that if $x\in M$, then $2x\in M'$, so if $M_2\subset M$ is any submodule such that $M_2\cap M'=0$, then $2x=0$ for all $x\in M_2$. This implies every element of $M_2$ has the form $(0,v)$ for $v\in V$. But then $M'+M_2$ cannot be all of $M$, since the first coordinate of every element of $M'+M_2$ is even. So there does not exist any $M_2\subset M$ such that $M$ is the (internal) direct sum of $M'$ and $M_2$.

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  • $\begingroup$ Thanks, what is $\mathbb{Z}/2$ ? I read it as $\mathbb{Z}/2\mathbb{Z}$ is that correct ? $\endgroup$ – James Well May 25 '17 at 23:10
  • $\begingroup$ That's correct. (I have also simplified the example: you need only one copy of $\mathbb{Z}$.) $\endgroup$ – Eric Wofsey May 25 '17 at 23:10
  • $\begingroup$ It seems to me like you have split the sequence, I'll give it another read in a short while to see if I can find where I'm going wrong $\endgroup$ – James Well May 25 '17 at 23:23
  • $\begingroup$ I've added more explanation for why it doesn't split. $\endgroup$ – Eric Wofsey May 26 '17 at 1:25
  • $\begingroup$ Why is $V\oplus \mathbb Z/2\mathbb Z$ isomorphic to $V$ ? The former has torsion, e.g. $(0,1)+(0,1)=0$ whereas the latter does not, as it is a vector space. $\endgroup$ – James Well May 26 '17 at 13:58

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