Polynomial subalgebras and their fields of fractions Let $A = k[x_1,\ldots,x_n]$, let $e_1,\ldots,e_n$ be $n$ $k$-algebraically independent elements of $A$, and $C = k[e_1,...,e_n]$. Then clearly $C \simeq A$. But suppose we have another $k$-subalgebra of $A$, call it $B$, with $A \supset B \supset C$. Must it be the case that $B \simeq A$? No, consider the rings $A = k[x,y], C = k[x^2,y^2], B = k[x^2,y^2,xy]$. However, what if you add the following facts:


*

*$A$ is integral over $B$ and $B$ is integral over $C$ (which automatically means that $A$ is integral over $C$). The counterexample satisfies this.

*The field of fractions of $B$ and the field of fractions of $C$ are equal. The counterexample does not satisfy this.


Is the statement now true?

Let $K_A, K_B, K_C$ be the fields of fractions of the rings involved. By assumption $K_B = K_C$. Furthermore because all of the $e_i$ are algebraically independent $C$ is a polynomial algebra thus is integrally closed in its field of fractions. Thus by definition if $x \in K_C$ and $x$ is integral over $C$, then $x \in C$. Now, $B \subset K_B = K_C$. Further $B$ is integral over $C$. Thus by definition $B \subset C$ hence $B = C$. Please correct me if I'm wrong.
