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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 a 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.

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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.

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  • $\begingroup$ Very nice, Kevin! $\endgroup$
    – Georges Elencwajg
    Commented Aug 29, 2011 at 8:11
  • $\begingroup$ Thanks! But excuse me， what is attach along a closed subscheme? $\endgroup$
    – MZWang
    Commented Aug 29, 2011 at 12:34
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    $\begingroup$ @unknown I do not know. But the attaching along the point $x$ in the example is clear: Locally around the point $x$ it looks like the union of two axes in the affine plane. And away from the point it just looks like it looked before the attaching. If you wish you can look at the result as a push out, say as the coequalizer of a map $x\to \mathbb P^1$ and a map $x\to \mathbb A^1$. At the ring level this is locally given as a fibered product. $\endgroup$
    – Wilberd van der Kallen
    Commented Aug 29, 2011 at 13:37
  • $\begingroup$ @unknown: What Wilberd said is correct. The subvariety $X\cup P/\lbrace y\rbrace$ will be affine with global sections isomorphic to $k[S,T]/\langle ST\rangle$. In general, it is possible to make sense of gluing along closed subschemes. See this question: mathoverflow.net/questions/64294/gluing-along-closed-subschemes. $\endgroup$
    – Kevin Ventullo
    Commented Aug 29, 2011 at 18:09
  • $\begingroup$ No gluing is necessary. Consider $\mathbb P^2_k$ (with coordinates (u:v:w)). Kevin's scheme is the locally closed subscheme obtained by first taking the cross $uv=0$ (a closed subscheme of $\mathbb P^2_k$) and then deleting the closed point $(0:1:0)$ (say) from that cross. $\endgroup$
    – Georges Elencwajg
    Commented Aug 29, 2011 at 19:40

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