If R/I is a free R-module, then I = 0. Let I be an ideal of a commutative ring R. 

Prove or disprove: If R/I is a free R-module, then I = 0.

Need some hints. I know that a free R-module has a basis and is isomorphic to R$^n$ for some finite n. 
If R/I is free R-module, why must it be isomorphic to R? 

Hint: If a module is isomorphic to $R/I$ and $i\in I$, then multiplication by $i$ is the same as multiplication by $0$.  What acts as zero on a free module?
