The restriction of a global section which is not a zero divisor is still an non-zero divisor? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T20:42:47Z http://mathoverflow.net/feeds/question/73936 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73936/the-restriction-of-a-global-section-which-is-not-a-zero-divisor-is-still-an-non-z The restriction of a global section which is not a zero divisor is still an non-zero divisor? MZWang 2011-08-29T03:48:04Z 2011-08-29T12:13:39Z <p>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?</p> <p>If X is affine, the answer is obvious true. I don't know the answer for a general scheme.</p> <p>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.</p> http://mathoverflow.net/questions/73936/the-restriction-of-a-global-section-which-is-not-a-zero-divisor-is-still-an-non-z/73946#73946 Answer by Kevin Ventullo for The restriction of a global section which is not a zero divisor is still an non-zero divisor? Kevin Ventullo 2011-08-29T06:42:47Z 2011-08-29T06:42:47Z <p>Here's a counterexample.</p> <p>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 $x$. </p> <p>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 $g= 0$. $\Box$</p> <p>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. </p>