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Math Help - non-Commutative algebra

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by peteryellow View Post
    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.
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    No I have no idea if this is false. This is my question.
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    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.
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  5. #5
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    Quote Originally Posted by peteryellow View Post
    No I have no idea if this is false. This is my question.
    it's obviously false! a simple counter-example is this: let p be a prime number. let R=\mathbb{Z}, \ I=p\mathbb{Z}, and M=\mathbb{Q}. we know that \mathbb{Q} is not projective \mathbb{Z}-module. but M/IM is a projective R/I

    module, because R/I=\mathbb{Z}/p is a field and we know that every module over a field is projective.
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    But what about my full question do you also think that this is also wrong?
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  7. #7
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    Quote Originally Posted by peteryellow View Post
    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!! i'll think about this "new" problem and i'll get back to you when i have a solution.
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