Does the integral closure of a normal local ring in a finite extension of its fraction field have finite projective dimension?

Question

Let $A$ be a normal local domain with field of fractions $K$. Let $L$ be a finite separable extension of $K$ (if relevant, I'm happy to assume all possible ramification is tame), and let $B$ be the normalization of $A$ in $L$.

Must $B$ have finite projective dimension over $A$?

Disclaimer: I know very little about projective resolutions, so this may have a trivial answer. My motivation is to use the main result (Theorem 2.1) of:

http://www.ams.org/journals/proc/1999-127-01/S0002-9939-99-04501-3/S0002-9939-99-04501-3.pdf

to prove a version of Abhyankar's lemma for normal local rings (is Abhyankar's even true for normal local rings?)

Answer 1

The result of Kantorovitz that you link proves that in many cases $B$ cannot have finite projective dimension as an $A$-module. For instance, let $B$ be $k[x,y]$ and let $A$ be the subring $k[x^2,xy,y^2] \subset k[x,y]$. This is etale in codimension $1$, yet purity does not hold.

In fact, you can write down a free resolution of $B$ as an $A$-module, and it is infinite. Observe that the following $2\times 2$ matrix of elements of $A$, $$M=\left[ \begin{array}{cc} xy & y^2 \\ -x^2 & -xy \end{array} \right],$$ satisfies $M\cdot M = 0$. Form the related matrix, $$N=\left[ \begin{array}{cc} 0 & 0 \\xy & y^2 \\ -x^2 & -xy \end{array} \right].$$ Then one free resolution of $B$ as an $A$-module is as follows, $$ \dots \xrightarrow{M} A^{\oplus 2} \xrightarrow{M} A^{\oplus 2} \xrightarrow{N} A^{\oplus 3} \rightarrow B \rightarrow 0 $$ where $A^{\oplus 3} \to B$ is given by the row vector, $$ \left[ \begin{array}{ccc} 1 & x & y \end{array} \right].$$