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Math Help - Prime ideal

  1. #1
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    Prime ideal

    Let l_1,...,l_k be independent linear forms in \mathbb{C}[x_1,...,x_n]. Let a_1,...,a_k \in \mathbb{C}.
    Prove that <l_1-a_1,...,l_k-a_k> is a prime ideal.
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  2. #2
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    Quote Originally Posted by KaKa View Post

    Let \ell_1,...,\ell_k be independent linear forms in \mathbb{C}[x_1,...,x_n]. Let a_1,...,a_k \in \mathbb{C}. Prove that <\ell_1-a_1,...,\ell_k -a_k> is a prime ideal.
    let I=<\ell_1 - a_1, \cdots , \ell_k - a_k> and R=\frac{\mathbb{C}[x_1, \cdots , x_n]}{I}. consider the system of equations \ell_j - a_j = 0, \ \ 1 \leq j \leq k. from elementary linear algebra we know that this system has either no

    solution (which implies that R = (0)) or a unique solution (which implies that R \cong \mathbb{C}) or infinitely many solutions (which implies that R \cong \mathbb{C}[x_{i_1}, \cdots , x_{i_m}], for some 1 \leq m < n). in either case

    R is clrearly an integral domain and thus I will always be a prime ideal.
    Last edited by NonCommAlg; March 24th 2009 at 04:19 AM. Reason: typo!
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    infinitely many solutions (which implies that R \cong \mathbb{C}[x_{i_1}, \cdots , x_{i_m}], for some 1 \leq m < n)

    Sorry, but I don't understood the last part. Thanks.
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  4. #4
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    Quote Originally Posted by Biscaim View Post

    Sorry, but I don't understood the last part. Thanks.
    the infinitely many solutions happens when the number of equations is less than the number of variables, i.e. k < n. in this case we can find some variables in terms of others. so wehen you mod

    out \mathbb{C}[x_1, \cdots , x_n] by I, you'll again get a polynomial ring over \mathbb{C} but with the number of variables less than n.
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