Move result later due to Gabber's discovery of gap

Thanks to Ofer Gabber for finding this error:

 it is only clear that the right horizontal arrow ,from K' tensored over
 R with R[x/y,y/x] to K tensored the same, is surjective, and it is not
 obvious that it is injective; the only fix that I can think of invokes
 the result of Gruson-Raynaud that the stalks  of K are finitely
 generated...

Let us explain which the part of the argument was wrong: First
you reduce to the case that R is a local normal domain. Then you pick
x, y in R and you consider the inclusions R \subset R[x/y] and
R \subset R[y/x]. As R is a normal domain we get a short exact
sequence
$$
0 \to R \xrightarrow{(-1, 1)} R[x/y] \oplus R[y/x] \xrightarrow{(1, 1)}
R[x/y, y/x] \to 0
$$
see Algebra, Lemma \ref{algebra-lemma-silly-normal}.
If $R \not = R[x/y]$ and $R \not = R[y/x]$ then we see that
$K \otimes_R R[x/y]$ and $K \otimes_R R[y/x]$ are finitely generated
as $R[x/y][x_1, \ldots, x_n]$ and $R[y/x][x_1, \ldots, x_n]$ modules.
Thus we can find $k_1, \ldots, k_t \in K$ such that the elements
$k_i \otimes 1$ generate $K \otimes_R R[x/y]$ and $K \otimes_R R[y/x]$
as $R[x/y][x_1, \ldots, x_n]$ and $R[y/x][x_1, \ldots, x_n]$ modules.
Set $K' = \sum R[x_1, \ldots, x_n]k_i \subset K$. Tensoring the
sequence above with $K' \subset K$ we get the diagram
$$
\xymatrix{
 &
 K' \ar[d] \ar[r] &
 K' \otimes_R R[x/y] \oplus K' \otimes_R R[y/x] \ar[d] \ar[r] &
 K' \otimes_R R[x/y, y/x] \ar[d] \ar[r] &
 0 \\
 0 \ar[r] &
 K \ar[r] &
 K \otimes_R R[x/y] \oplus K \otimes_R R[y/x] \ar[r] &
 K \otimes_R R[x/y, y/x] \ar[r] &
 0
}
$$
Now we know that the vertical arrows in the middle and on the right
are isomorphisms. <<<========= This is where the mistake was, because
although we see that
K' \otimes_R R[x/y, y/x -----------> K \otimes_R R[x/y, y/x]
is surjective by choice of K', it is not clear that it is injective!