# if the kernel of an endomorphism is a direct summand, doesn't the image have to be too?

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)?

• Thank you. I don't understand exact sequences so I'm can't know what reasoning you have in mind, but it's enough for me to know that the implication holds. However, I'm not sure why the fact that there may be many copies of $\operatorname{im}\phi$ in $M$ should be a problem. There is one concrete and clearly distinguished copy: the actual $\operatorname{im}\phi.$ – user23211 Jan 19 '12 at 1:21
• I think I misread your question at first, but I'd like to make a similar point. It's true that you have an isomorphism from $M$ to $\ker\phi \oplus \operatorname{im}\phi$. But it need not be the case that $\ker\phi \cap \operatorname{im}\phi = 0$. In other words, the direct sum is not internal. And how do you write down the isomorphism without making some choice of $N$? – Dylan Moreland Jan 19 '12 at 1:27
• Oh, I see! I'd never have noticed this. So when an author writes that a submodule $N$ of $M$ is a direct summand do they mean that there exists a) a module $X$ such that $M\cong N\oplus X,$ b) a submodule $X\subseteq M$ such that $M\cong N\oplus X,$ or c) a submodule $X\subseteq M$ such that $M\cong N\oplus X$ and the direct sum is internal? – user23211 Jan 19 '12 at 1:42
• @ymar: always c. I believe (a) and (b) are equivalent to each other, but not to (c). 2Z is not a direct summand of Z, but it is isomorphic to one. – Jack Schmidt Jan 19 '12 at 1:54
• Eh, of course a) and b) are equivalent... I should have thought before writing that. Thanks, I see now! – user23211 Jan 19 '12 at 2:06