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Suppose $k$ is an algebraically closed field. One of the formulations of Hilbert's weak Nullstellensatz is that each maximal ideal in $k[x_1, \dots , x_n]$ has the form $\langle x_1 -a_1 , \dots , x_n - a_n\rangle $, where $a = (a_1 , \dots , a_n)$ is a point in affine space $k^n$. Is the point $a$ unique?

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Yes, assume $b = (b_1,\dots,b_n)$ is another point like $a$ then: $$\langle x_1-a_1,\dots,x_n-a_n\rangle=\langle x_1-b_1,\dots,x_n-b_n\rangle .$$ So $x_1 -b_1 = \sum f_i (x_i-a_i) \ \ , \ \exists f_i \in k[x] .$ Can you explain why this is not possible if $b_1 \not= a_1$?

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  • $\begingroup$ The $f_i \in k[x]$ or $f_i \in k[x_1, \cdots x_n]$ $\endgroup$ May 5, 2020 at 2:18
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    $\begingroup$ im sorry 𝑓𝑖∈𝑘[𝑥1,⋯𝑥𝑛] $\endgroup$
    – Danny Ofek
    May 5, 2020 at 2:19
  • $\begingroup$ Thanks a lot!!! $\endgroup$ May 5, 2020 at 2:21

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