# When quotients of polynomial rings are isomorphic to polynomial rings?

This might be a silly question but is there a criterion for when the quotient of $k[X_1,\ldots, X_n]$ by some ideal is isomorphic to a polynomial ring?

For instance $k[X,Y]/\langle Y \rangle \simeq k [X]$ but $k[X,Y]/\langle Y^2 \rangle$ is not a polynomial ring.

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An obvious necessary condition is that the ideal has to be prime, which is enough to rule out $\langle Y^2\rangle$. –  Emil Jeřábek Feb 8 '13 at 11:56
And the quotient ring has to be regular; check out mathoverflow.net/questions/79956/… –  Allen Knutson Feb 8 '13 at 12:16
I doubt that there are is a (useful and nontrivial) criterion. One could use it to attack the wide open Zariski cancellation problem. –  Martin Brandenburg Feb 8 '13 at 16:41
There are probably no such non-trivial criterion. One deep result is the Abhyankar-Moh theorem which says that $\mathbb{C}[x,y]/f$ is the polynomial ring in one variable if and only if there is a $\mathbb{C}$- algebra automorphism of $\mathbb{C}[x,y]$ which transforms $f$ into a variable. –  Mohan Feb 8 '13 at 17:07
Martin: I'm not sure the cancellation problem is so wide open any more. Check out arxiv.org/abs/1208.0483. –  Neil Epstein Feb 9 '13 at 8:35

Proposition: Let $R$ be a finitely generated domain over $k$. Then $R$ is a polynomial ring over $k$ iff some (equivalently, every minimal) set of $k$-algebra generators of $R$ has size $\operatorname{tr.deg}_k R$.
In fact, if any set of $k$-algebra generators has the "right size" ($= \operatorname{tr.deg}_k R$), then they form a transcendence basis over $k$, and there are no additional algebraic relations in $R$.
The proof is a standard trick in dimension theory: if $a_1, \ldots, a_n$ is a set of $k$-algebra generators with $n = \operatorname{tr.deg}_k R$, then by Noether normalization, $n = \dim R$. The map $k[x_1, \ldots, x_n] \twoheadrightarrow R$ sending $x_i \mapsto a_i$ is a surjection between domains of the same dimension, hence is an isomorphism.
To relate this to the defining ideal $I$ of $R$ (w.r.t. some given presentation $R \cong k[y_1, \ldots, y_m]/I$), one can rephrase the above by saying that $R$ is a polynomial ring iff $I$ is prime and $R$ can be generated as a $k$-algebra by $m - \operatorname{ht}(I)$ many elements.