2
$\begingroup$

Let $M$ be a module and $\phi: M\longrightarrow M$ be an endomorphism. Let $\ker \phi$ be a direct summand of $M.$ Does it imply that $\mbox{im}\,\phi$ is a direct summand of $M$ too?

I have not seen an explicit statement in any text have read but it seems quite clear from what I have seen that there is no such implication. The authors write as if it weren't true. (They would surely refer to it if it were.)

It confuses me because I cannot find any reason why the following should be incorrect.

Let $M=\ker \phi\oplus N.$ Let $\psi:(\ker \phi\oplus N)/\ker\phi\longrightarrow N$ be given by the formula $\psi([(k,n)])=n$ for $k\in \ker\phi, n\in N,$ and $[(k,n)]$ denoting the coset of $(k,n)$ in $(\ker \phi\oplus N)/\ker\phi.$

I would like to omit the proof that $\psi$ is an isomorphism. I have checked it several times and I don't see any mistake. If $\psi$ isn't necessarily an isomorphism, then please tell me and I will post my, apparently incorrect, proof.

Now we have $N\cong (\ker \phi\oplus N)/\ker\phi\cong M/\ker\phi\cong \mbox{im}\,\phi$ by the first isomorphism theorem. Therefore $M=\ker\phi\oplus \mbox{im}\,\phi.$ In particular, $\mbox{im}\,\phi$ is a direct summand of $M.$

Where am I making my mistake(s)?

  • 0
    Eh, of course a) a$n$d b) are equivalent... I should have thought before writing that. Thanks, I see now!2012-01-19

1 Answers 1

9

This is false, even for modules over the integers. Let M be the regular module Z, and let φ be multiplication by 2. The ker(φ) = 0 is a direct summand, but im(φ) = 2 Z is not a direct summand of Z. Of course, Z is isomorphic to 0 ⊕ 2 Z, but the direct sum is not internal, and this is what is meant by "is a direct summand" versus "is isomorphic to a direct summand."