# non-Commutative algebra

• Mar 20th 2009, 09:36 AM
peteryellow
non-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 AM
NonCommAlg
Quote:

Originally Posted by peteryellow
I is a two sided ideal in a ring R and M 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.

you can't show that because it's false! can you find a counter-example? maybe the question is this: if M is a projective R-module, then M/IM is a projective R/I-module.
• Mar 21st 2009, 01:51 AM
peteryellow
No I have no idea if this is false. This is my question.
• Mar 21st 2009, 01:55 AM
peteryellow
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 AM
NonCommAlg
Quote:

Originally Posted by peteryellow
No I have no idea if this is false. This is my 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.
• Mar 21st 2009, 02:07 AM
peteryellow
But what about my full question do you also think that this is also wrong?
• Mar 21st 2009, 02:09 AM
NonCommAlg
Quote:

Originally Posted by peteryellow
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.

ok, well, i don't know how you forgot to mention strong conditions like these!! (Surprised) i'll think about this "new" problem and i'll get back to you when i have a solution.