Let $\displaystyle l_1,...,l_k$ be independent linear forms in $\displaystyle \mathbb{C}[x_1,...,x_n]$. Let $\displaystyle a_1,...,a_k \in \mathbb{C}$.

Prove that $\displaystyle <l_1-a_1,...,l_k-a_k>$ is a prime ideal.

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- Mar 23rd 2009, 07:11 AMKaKaPrime ideal
Let $\displaystyle l_1,...,l_k$ be independent linear forms in $\displaystyle \mathbb{C}[x_1,...,x_n]$. Let $\displaystyle a_1,...,a_k \in \mathbb{C}$.

Prove that $\displaystyle <l_1-a_1,...,l_k-a_k>$ is a prime ideal. - Mar 24th 2009, 12:43 AMNonCommAlg
let $\displaystyle I=<\ell_1 - a_1, \cdots , \ell_k - a_k>$ and $\displaystyle R=\frac{\mathbb{C}[x_1, \cdots , x_n]}{I}.$ consider the system of equations $\displaystyle \ell_j - a_j = 0, \ \ 1 \leq j \leq k.$ from elementary linear algebra we know that this system has either no

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

$\displaystyle R$ is clrearly an integral domain and thus $\displaystyle I$ will always be a prime ideal. - Mar 24th 2009, 02:43 PMBiscaim
- Mar 24th 2009, 03:30 PMNonCommAlg
the infinitely many solutions happens when the number of equations is less than the number of variables, i.e. $\displaystyle k < n.$ in this case we can find some variables in terms of others. so wehen you mod

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