$Ra$ is a direct summand of M $\Leftrightarrow \exists f:M\rightarrow R$ an $R$-linear map such that $f(a)=1_R$

22 Views Asked by Bumbble Comm At 27 Mar 2026 - 3:05

Let $R$ be a ring and $M$ an $R$-module and $a$ a free element of $M$

Show that: $Ra$ is a direct summand of M $\Leftrightarrow \exists f:M\rightarrow R$ an $R$-linear map such that $f(a)=1_R$

My work:

$\Rightarrow $? As $a$ is a free element then $\left \{ a \right \}$ is free, then the $R$-linear map $g:R\rightarrow M$ defined by $g(x)=xa$ is injective

Consider the following exact sequence,

$0\rightarrow R\overset{g}{\rightarrow}M\overset{p}{\rightarrow}M/Ra\rightarrow 0$ ...$(*)$

($p$ is the canonical surjection)

So, as $Im(g)=Ra$ is direct summand (by hypothesis), then $(*)$ is split exact, then $\exists f:M\rightarrow R$ such that $f\circ g=Id_R$

($f$ is a retract)

Then $f\circ g(1_R)=Id_R(1_R)=1_R\Rightarrow f(g(1_R))=1_R\Rightarrow f(a)=1_R$

Is what I did correct till now ?

I think I should also construct a new sequence to solve it. But can't I use the sequence $(*)$ ?