prove that in $k[x_1,...,x_n]$ Ideal of polynomials vanishing on some $(a_1,a_2,...,a_n)$ will be $\langle x_1-a_1,...,x_2-a_2\rangle$.

Question

Given $k$ is an algebraically closed set.I was thinking to do some sort of Taylor series expansion, but it is only valid when $k$ has some metric.what should I do?

This is true for any commutative ring. See link – BenediktK Aug 28 '19 at 20:36 — BenediktK, Aug 28 '19 at 20:36

score 1 · Accepted Answer · answered Aug 28 '19 at 20:43

Fix $a:=(a_1,\ldots,a_n)\in \mathbb A^n_k$ and set $I:=\langle x_1-a_1,\ldots,x_n-a_n\rangle\subseteq k[x_1,\ldots,x_n]$. Now define $J\subseteq k[x_1,\ldots,x_n]$ to be the ideal of polynomials vanishing at $a$ (one can show that this set is an ideal).

It is obvious that $I\subseteq J$ since every element of $I$ vanishes at $a$. Now use maximality of $I$ and the fact that $J\neq k[x_1,\ldots,x_n]$ to conclude that $J=I$.

By the way, proving that $I$ is maximal can be done in a few ways, but it may be easiest to see that $k[x_1,\ldots,x_n]/I\cong k$.

Ahh, that was easy! Thank you :-) – ANJAN SAMANTA Aug 28 '19 at 20:52 — ANJAN SAMANTA, Aug 28 '19 at 20:52

prove that in $k[x_1,...,x_n]$ Ideal of polynomials vanishing on some $(a_1,a_2,...,a_n)$ will be $\langle x_1-a_1,...,x_2-a_2\rangle$.

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