I is a two sided ideal in a ring R AndM is left R-module.M/IM becomes a module over the quotient ring R/I.
How can I show that if M/IM is a projective R/I-module then M is a projective R-module.
See my question is.
Let I be a two-sided ideal in a ring R and let M be a left R-module. IM contained in M is a submodule generated by all products xm where x is in I and m is in M. The quotient M/IM becomes a module over the quotient ring R/I.
Suppose that I is nilpotent, i.e., taht I^N =0 for some N>=1 and suppose that M is finitely generated flat R-module. Show that if M/IM is a projective R/I-module then M is a projective R-module.
Sorry I didnot wrote the whole question.
it's obviously false! a simple counter-example is this: let $\displaystyle p$ be a prime number. let $\displaystyle R=\mathbb{Z}, \ I=p\mathbb{Z},$ and $\displaystyle M=\mathbb{Q}.$ we know that $\displaystyle \mathbb{Q}$ is not projective $\displaystyle \mathbb{Z}-$module. but $\displaystyle M/IM$ is a projective $\displaystyle R/I$
module, because $\displaystyle R/I=\mathbb{Z}/p$ is a field and we know that every module over a field is projective.