# If a polynomial ring is a finite flat module over some subring, is that subring itself a polynomial ring?

Let $k$ be an algebraically closed field of characteristic zero (for simplicity), and let $A$ be a $k$-subalgebra of a polynomial ring $B := k[x_1, \dotsc, x_n]$ such that $B$ is a flat $A$-module of finite type.

Then, must $A$ also be a polynomial ring? It's true if $n=2$ (see Regular subrings of a polynomial ring by Masayoshi Miyanishi for a proof).

If not, could we classify all such $A$ ?

• Have you tried taking standard counter-examples to Luroth's problem in dimension $3$? There are subfields $K$ of $k(x_1, x_2, x_3)$, with $k(x_1, x_2, x_3)/K$ of finite degree, such that $K \not \cong k(y_1, y_2,y_3)$. Perhaps you could find one such that $k[x_1, x_2,x_3]$ is flat over $K \cap k[x_1, x_2,x_3]$. The other keyword you should know besides "Luroth" is that a variety with such $K$ as a function field is called "unirational, but not rational". – David E Speyer Feb 1 at 15:28
• Thinking harder: For finitely generated modules, flat = projective. Copying the part of Neil Strickland's answer to my question then shows that A is regular. Conversely, if B is a polynomial and is finite over some regular subring A, then B will be flat over A by the miracle flatness theorem. So the question is simply whether we can get a polynomial ring to be finite over a regular, non-polynomial, subring. – David E Speyer Feb 1 at 16:43
• How much do you care that $k$ is algebraically closed? I'd rather work with subrings of $\mathbb{Q}(x,y,z)/(x^3+y^3+z^3=3)$ than think about cubic 3-folds. (Note that Miyanishi assumes $k$-algebraically closed, so there might be two dimensional counterexamples over $\mathbb{Q}$.) – David E Speyer Feb 1 at 17:05
• @DavidESpeyer Thank you for some hints! I will think about it.. – zzy Feb 4 at 15:13