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Thread: Free Modules

  1. #1
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    Free Modules

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

    Is $\displaystyle M$ 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 $\displaystyle M$ looking like

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

    Is $\displaystyle M$ 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 $\displaystyle M$ looking like

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

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

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