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.

  • $\begingroup$ An obvious necessary condition is that the ideal has to be prime, which is enough to rule out $\langle Y^2\rangle$. $\endgroup$ Feb 8, 2013 at 11:56
  • $\begingroup$ And the quotient ring has to be regular; check out mathoverflow.net/questions/79956/… $\endgroup$ Feb 8, 2013 at 12:16
  • $\begingroup$ I doubt that there are is a (useful and nontrivial) criterion. One could use it to attack the wide open Zariski cancellation problem. $\endgroup$ Feb 8, 2013 at 16:41
  • 5
    $\begingroup$ 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. $\endgroup$
    – Mohan
    Feb 8, 2013 at 17:07
  • 4
    $\begingroup$ Martin: I'm not sure the cancellation problem is so wide open any more. Check out arxiv.org/abs/1208.0483. $\endgroup$ Feb 9, 2013 at 8:35

1 Answer 1


Although I agree that there is probably no useful + nontrivial criterion, here's a simple one:

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.