Consider a polynomial algebra $A=\mathbb{K}[x_1,\ldots,x_n]$ and its ideal $I$, such that $A/I=\mathbb{K}[y_1,\ldots, y_k]$. Is it true that there exist new polynomial generators $z_1,\ldots,z_n$ (in a sense that $A=\mathbb{K}[z_1,\ldots,z_n]$) such that $I=(z_1,\ldots,z_{n-k})$?
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2$\begingroup$ It's unclear in general what "such $A/I=K[y_1,\dots,y_k]$ means: does it mean that $A/I$ is generated by $k$ elements, or that it's freely generated by $k$ elements (i.e., is isomorphic to a polynomial algebra over $k$ indeterminates). Given the context it certainly means freely generated. Also "new variables" should be "elements". $\endgroup$– YCorMar 20, 2020 at 9:46
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$\begingroup$ Variables are elements that are part of a set of polynomial generators. $\endgroup$– Bugs BunnyMar 20, 2020 at 10:13
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$\begingroup$ It means that $y_i$ are algebraically independent. By variables I mean a part of independent generators of polynomial algebra. $\endgroup$– A.SkutinMar 20, 2020 at 10:26
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$\begingroup$ No, you don't mean that: $x^2$ and $y^2$ are algebraically independent but they are not "variables"... $\endgroup$– Bugs BunnyMar 20, 2020 at 11:02
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$\begingroup$ See if my edit is OK $\endgroup$– Bugs BunnyMar 20, 2020 at 11:05
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1 Answer
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This is called the Embedding Problem. I believe that it is false in positive characteristics and unknown in characteristic zero, except a few small cases. See Kraft's review for the extent of my knowledge of the state of the problem. The counterexample in characteristic $p$ is from this paper: $$ K[x,y]/(y^{p^2}-x-x^{2p}) $$
answered Mar 20, 2020 at 8:14