Mathematics Asked by Luiz Guilherme De Carvalho Lop on December 12, 2020

i dont know how show that theorem is true

Definition: A morphism $F: M longrightarrow N$ of $R$-modules is called regular if exist $G: N longrightarrow M$ such that $F circ G circ F = F$.

Theorem: $F$ a morphism of $R$-modules $M$ and $N$, is regular if and only if $ker(F)$ is a direct summand of $M$ and $im(F)$ is a direct summand of $N$.

the only "tip" is: choice good exact sequences and see when it split.

I shall work in $R-Mod$, the category of left $R$-modules. Suppose $exists G:Nrightarrow M$ such that $Fcirc Gcirc F=F $.

Consider the short exact sequence$$0 rightarrow operatorname{Ker}Fhookrightarrow Mxrightarrow{F} operatorname {Im}Frightarrow 0$$ Look at $G|_{operatorname{Im}F}:operatorname{Im}Frightarrow M$. We have $Fcirc G|_{operatorname{Im}F}=id_{operatorname{Im}F}$ by the condition. As such the above sequence is a split short exact sequence (it right splits and hence left splits as well). $therefore operatorname{Ker } F$ is a direct summand of $M$.

Consider the short exact sequence $$0rightarrow operatorname{Im}Fxrightarrow{j} N rightarrow operatorname{CoKer}Frightarrow 0$$

Look a $Fcirc G:Nrightarrow operatorname{Im}F$. Once again, we have $Gcirc Fcirc j=id_{operatorname{Im}F}$. So the above sequence left splits and hence is a split short exact sequence. $therefore operatorname{Im}F$ is a direct summand of $N$.

Now assume $operatorname{Ker}F$ is a direct summand of $M$. Then $$0 rightarrow operatorname{Ker}Fhookrightarrow Mxrightarrow{F} operatorname {Im}Frightarrow 0$$ is a split short exact sequence. So we get $g:operatorname {Im}Frightarrow M$ such that $Fcirc g=id_{operatorname{Im}F}$. We also have a sub-module $N'$ of $N$ such that $N=operatorname{Im}Foplus N'$ Define $$G:Nrightarrow M$$ $$(x,y)mapsto g(x)$$

Then $Fcirc Gcirc F(x)=Fcirc G((F(x),0))=Fcirc g(F(x))=F(x) $

Answered by Soumik on December 12, 2020

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