The restriction of a global section which is not a zero divisor is still an non-zero divisor?

2011-08-29T03:48:04Z

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.

Answer by Kevin Ventullo for The restriction of a global section which is not a zero divisor is still an non-zero divisor?

2011-08-29T06:42:47Z

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 $y$ is any point of $P$ other than $x$.

Now suppose $fg=0$. Then $g\equiv 0$ on $X/\lbrace x\rbrace$ implies $g(x)=0$. Since $g\mid_P$ vanishes at one point, $g\mid_P\equiv 0$, so $g= 0$. $\Box$

The correct definition of this presheaf is given in Hartshorne II.6. Basically, you only consider global sections which are not zero divisors in each local ring.

The restriction of a global section which is not a zero divisor is still an non-zero divisor? - MathOverflow

The restriction of a global section which is not a zero divisor is still an non-zero divisor?

Answer by Kevin Ventullo for The restriction of a global section which is not a zero divisor is still an non-zero divisor?