Let $R$ be a commutative ring. $I$,$J$ are two ideals of $R$ such that $I\cap J=0$. We have a short exact sequence,
$$0 \longrightarrow R\stackrel{f} \longrightarrow R/I\oplus R/J\stackrel{g}{\longrightarrow}R/(I+J) \longrightarrow 0$$
Where $f(1)=(1,-1)$ and $g(a,b)=a+b$. I want to prove that if $\operatorname{Ext}_R^1(R/(I+J),R)=0$, then the above sequence splits.
I know that if $\operatorname{Ext}_R^1(R/(I+J),M)$ vanishes for any module $M$, then $R/(I+J)$ is a projective module and the sequence splits. Why does the vanishing of $\operatorname{Ext}_R^1(R/(I+J),R)$ imply we can say the sequence splits?
Best Answer
By applying functor $\text{Hom}（\_,R)$ to the short exact sequence, and the condition $\text{Ext}_R^1(R/(I+J),R)=0$, we get a short exact sequence by writing out the long exact sequence:
$$0\longrightarrow \text{Hom}(R/(I+J),R)\stackrel{g^*}{\longrightarrow}\text{Hom}(R/I\oplus R/J,R)\stackrel{f^*}{\longrightarrow}\text{Hom}(R,R)\longrightarrow 0.$$
Where $g^*h=hg$, $f^*h'=h'f$. Since $f^*$ is surjective and $\text{id}_R\in\text{Hom}(R,R)$, there exists some $F\in \text{Hom}(R/I\oplus R/J,R)$ such that $f^*F=Ff=\text{id}_R$. Thus the sequence splits.