Take the tour ×  Let X be a scheme. U is an open subscheme of X.
Assume f is a global section on X which is not a zero divisor,
then the restriction of f to U is still an non-zero divisor?  If
X is affine, the answer is obvious true. I don't know the answer
for a general scheme.  This is a question raised in the
definition of sheaf of total fraction rings. Some author claim
U|-> total fraction ring of sections over U is a presheaf, but I
can't see the reason.  share|improve this question The authors
who claim that are wrong, even if Grothendieck is among them.
See mathoverflow.net/questions/28553/… –  Georges Elencwajg Aug
29 '11 at 8:01 add comment  1 Answer  up vote 5 down vote
accepted  Here's a counterexample.  Let $P=\mathbb{P}^1$,
$X=\mathbb{A}^1$, and attach $X$ to $P$ along a single point
$\{x\}$. Then there is a global section $f$ which is nonzero on
$X$ except at $x$, and is identically zero on $P$. Moreover, $f$
restricts to a zero divisor on the open subvariety $X\cup
P/\lbrace y\rbrace$, where $y