# When a ring is a polynomial ring?

In the paper (2.11) the authors show that if $k^*$ is a separable algebraic extension of $k$ and $x_1,x_2, \ldots, x_n$ are indeterminates over $k^*$ and a normal one dimensional ring $A$ with $k \subset A \subset k^*[x_1,x_2,\ldots,x_n]$ then $A$ has the form $k'[t]$ where $k'$ is the algebraic closer of $k$ in $A$.

The above is a very strict sufficient condition to ensure the polynomial property. Is there a general answer to the question:

Q. When a ring $R$ is a polynomial ring over a field?

• This is a difficult problem, even when $R$ is a finitely generated subring of a ring of polynomials. You shouldn't expect a definite answer. You might like to have a look at van den Essen's "Seven Lectures on Polynomial Automorphisms" (in id. ed., Automorphisms of Affine Space (Curaçao 1994), Springer 1995), or his book Polynomial Automorphisms and the Jacobian Conjecture (Birkhäuser 2000). Apr 27, 2016 at 12:25
• In the result you cite, there is a slight generalization. To deduce the same result for every normal, one-dimensional subring $A$ of a $k$-algebra $B$, it suffices that (i) for the integral closure $k^*$ of $k$ in $B$, the fraction field $k^*(B)$ has vanishing irregularity, i.e., there is no non-constant $k^*$-rational transformation from $\text{Spec}(B)$ to an Abelian variety, and (ii) every invertible element of $B$ is integral over $k$. Apr 27, 2016 at 13:05