Suppose $J$ is an $R$-module, and for every cyclic $R$-module $C$ this short exact sequence splits $$0\rightarrow J\overset{f}{\rightarrow}B\overset{g}{\rightarrow}C\rightarrow 0.$$ Prove that $J$ is an injective $R$-module.
because $C$ is cyclic,so $C=Rm$ ($m\in C$) and $Rm\simeq \frac{R}{I}$ ($I$ is Ideal of $R$).
I tried to make homomorphism from $I$ to $J$ and show that $I$ can be expanded to $R$. so $J$ will be injective.but I couldn't apply my plan.I really don't know what should I do,please help me.
Consider an arbitrary (say left) ideal $I$ of $R$ and an arbitrary morphism $f:I\to J$. Let $i:I\to R$ be the inclusion. Pushing out $f$ along $i$ gives an inclusion $j:J\to B$ and a map $g:R\to X$ such that $gi=jf$. Here we can give an explicit presentation of $B$: $$B=R\oplus J/<f(x)-i(x)>$$ (so the map $g$ (resp. $j$) are induced by the natural inclusions of $R$ (resp. $J$) into $R\oplus B$). There is also an induced map $h:R/I \to X/J$ which one can prove is an isomorphism. This is to say $X/J$ is cyclic. By hypothesis then the exact sequence $0\to J\to B\to B/J\to 0$ splits which gives the requisite extension of $f$ to all of $R$. And then as you wanted, apply Baer's criterion.