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Suppose $k$ is an algebraically closed field, and let $I$ be a proper ideal of $k[x_1, \dots, x_n]$. Does there exist an ideal $J \subseteq (x_1, \dots, x_n)$ such that $k[x_1, \dots, x_n]/I \cong k[x_1, \dots, x_n]/J$ as $k$-algebras?

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    $\begingroup$ Can you do this if I is maximal? $\endgroup$ – Mariano Suárez-Álvarez Jun 15 '13 at 22:03
  • $\begingroup$ @Mariano Yes, by the Nullstellensatz; take $J=(x_1, \dots, x_n)$. $\endgroup$ – user55407 Jun 15 '13 at 22:11
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    $\begingroup$ so if $I \subset m$ for some $m$, and you can move $m$ to $(x_1,\cdots,x_n)$... $\endgroup$ – user27126 Jun 15 '13 at 22:12
  • $\begingroup$ @Sanchez I see. Thanks. $\endgroup$ – user55407 Jun 15 '13 at 22:15
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This is very clear from the corresponding geometric statement: Every non-empty algebraic set $\subseteq \mathbb{A}^n$ can be moved to some that contains the zero. Actually a translation from some point to the zero suffices.

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Say $I \subseteq m:=(x-a_1, \dots, x-a_n)$ (we are using the Nullstellensatz, here). Then define an isomorphism of $k$-algebras $\phi: k[x_1, \dots, x_n] \to k[x_1, \dots, x_n]$ by $x_i \mapsto x_i + a_i$.

Clearly $\phi(m)=(x_1, \dots, x_n)$. Take $J:=\phi(I)$, and notice that one has a $k$-algebra isomorphism $$k[x_1, \dots, x_n]/I \to k[x_1, \dots, x_n]/J.$$

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