Prime ideal

**KaKa** · March 23rd 2009, 07:11 AM

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.

**NonCommAlg** · March 24th 2009, 12:43 AM

Originally Posted by KaKa

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.

**Biscaim** · March 24th 2009, 02:43 PM

Originally Posted by NonCommAlg

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.

**NonCommAlg** · March 24th 2009, 03:30 PM

Originally Posted by Biscaim

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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