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I want to prove that if $f:A \rightarrow A$ is an R-module homomorphism such that $f\circ f=f $ then $A= Ker f\oplus Im f $

If we think about a split exact sequence like : $ 0\rightarrow Ker f \rightarrow A \rightarrow Im f \rightarrow 0 $ then we will be done.

We need then an injection $i:Ker f \rightarrow A$ and a surjection $s:A \rightarrow Im f$ such that $Im\ i= Ker\ s$, and there exists an R-module homomorphism $g:Im f\rightarrow A$ with $gs= {I_d}_{|Im f}$ ,
or there exists an R-module homomorphism $h:A\rightarrow Ker f$ with $hi= {I_d}_{|Ker f}$ ,
the two conditions being equivalent by the characterisation of split sequences.

But $i= I_d$ and $s= f$ work ! since $Im\ I_d= Ker f$ and there exists an R-module homomorphism $h: A\rightarrow Ker f$ described by $h(a)= a- f(a)$ for all $a\in A\ \ \ $ which verify $hi= h_{|Ker f}= {I_d}_{|Ker f}$ as needed.
( $f$ being idempotent gives $f(h(a))= f(a)- f(f(a))= 0$ )

To see why $h$ is an R-module homomorphism, we notice that $h(a+b)= a+b-f(a+b)= a-f(a)+ b-f(b)= h(a)+h(b)$ and
$h(ra)= ra- f(ra)= ra- rf(a)= rh(a)$ for all $a, b$ in $A$ and $r\in R$ , so we are done.

Can you please confirm if this proof is correct ? And what if we want to prove this without using split exact sequences ? Is it possible ?

NotaChoice
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  • Why do you find it strange? and how would you make your sequence split? I initially tried it before inverting them. – NotaChoice Dec 12 '21 at 05:42
  • No, you're right. It isn't strange. I was just not thinking straight. – Arthur Dec 12 '21 at 06:44
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    Your proof is indeed correct, except that I have no idea what $I_{d}$ is supposed to mean (just take $i$ = inclusion). I don't know why you would want to avoid using the language of split exact sequences, but you can make your proof direct if you like - this amounts to unwinding your proof, more or less. Set $K = ker(f), I = Im(f)$, and consider the $R$-module morphisms $g \colon A \to K \oplus I, h \colon K \oplus I \to A$ given respectively by $g(a) = (a-f(a), f(a))$ and $h(x, y) = x+y$. Now observe that $g \circ h$ and $h \circ g$ compose to the respective identities, and you're done. – Alex Wertheim Dec 12 '21 at 07:56

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