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.

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- Mar 20th 2009, 09:36 AMpeteryellownon-Commutative algebra
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. - Mar 21st 2009, 01:48 AMNonCommAlg
- Mar 21st 2009, 01:51 AMpeteryellow
No I have no idea if this is false. This is my question.

- Mar 21st 2009, 01:55 AMpeteryellow
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. - Mar 21st 2009, 02:00 AMNonCommAlg
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. - Mar 21st 2009, 02:07 AMpeteryellow
But what about my full question do you also think that this is also wrong?

- Mar 21st 2009, 02:09 AMNonCommAlg