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Math Help - Free Modules

  1. #1
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    Free Modules

    Let R=C[x,y] the ring of polynomials in two variables with complex coefficients. Let M=(x,y) be the ideal generated by x and y. M is an R-module.

    Is M a free module?

    I am still in the habit of associating free with finitely generated, but I think the notion is more subtle than that. My instinct says yes: something along the lines of elements in M looking like

    m=r_{1}x+r_{2}y
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  2. #2
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    Quote Originally Posted by robeuler View Post
    Let R=C[x,y] the ring of polynomials in two variables with complex coefficients. Let M=(x,y) be the ideal generated by x and y. M is an R-module.

    Is M a free module?

    I am still in the habit of associating free with finitely generated, but I think the notion is more subtle than that. My instinct says yes: something along the lines of elements in M looking like

    m=r_{1}x+r_{2}y
    in general suppose M \neq (0) is an ideal of a commutative domain R. then, as an R module, M is free if and only if it's cyclic. the reason is that if M has a basis with more than one elements, then

    we can choose two elements of the basis, say a,b. but ba - ab = 0, which is impossible because a,b belong to a basis of M. conversely if M=Ra, then \{a \} would be a basis for M because R is

    a domain. so, in your question, M is not a free R module because it cannot be generated by one element. (prove it!)
    Last edited by NonCommAlg; September 21st 2009 at 05:45 PM.
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