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In his book Algebraic Geometry and Arithmetic Curves Qing Liu claims in Exercise 3.4, page 56, the following for a scheme $X$ and a global function $f\in \mathcal O_X(X)$:
"The map $U\mapsto f\vert _U\mathcal O_X(U)$ for every affine open subset $U$ defines a sheaf of ideals on $X$."
The presheaf thus defined (on the base of open sets given by affines) is certainly separated but I don't see why it should be a sheaf if $X$ is not integral.
The problem is that already if you have two open affines $U,V$ and functions $s\in \mathcal O(U),t\in \mathcal O(V)$ such that $fs=ft$ in $U\cap V$ there is no reason that $s$ and $t$ coincide on $U\cap V$, so that $s$ and $t$ can't a priori be glued and I don't see why the function on $U\cup V$ obtained by gluing $fs$ and $ft$ (they can be glued since $\mathcal O_X$ is a sheaf!) could be written as a product $(f\vert_{U\cup V})w$ for some $w\in \mathcal O(U\cup V)$.
Hence my question: Am I missing something or is the statement in the exercise false without some hypothesis on the scheme $X$?

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Note that we only need consider the case (of your setup) where $U \cup V$ is affine. If the function obtained by gluing $fs$ and $ft$ is not a multiple of $f$, then it is a nonzero element in $ \mathcal O (U \cup V) /f$, hence a nonzero function on $\operatorname{Spec} ( \mathcal O (U \cup V)/f)$. But $\operatorname{Spec} ( \mathcal O (U \cup V)/f)$ is covered by $\operatorname{Spec} ( \mathcal O (U)/f)$ and $\operatorname{Spec} ( \mathcal O (V)/f)$ so the function must be nonzero on one of those, contradiction.

The general principle here is that kernels of maps of sheaves can be computed one open set at a time.

But probably the simplest proof is to reduce to distinguished affine opens and do it algebraically. Suppose we have an affine open $\operatorname{Spec} R$ covered by open sets $\operatorname{Spec} R [1/a_i]$ where $a_i$ generate the unit ideal. If a global section $x$ of the structure sheaf restricts to a multiple of $f$ on each open, then for all $i$ we have $$ a_i^{e_i} (x - f s_i) =0$$ for some $e_i$ and $s_i$.

Because the $a_i$ generate the unit ideal, we have $\sum_{i=1}^n a_ib_i=1$, so $$x = \left(\sum_{i=1}^n a_ib_i\right)^{1-n + \sum_{i=1}^n e_i } x$$ and when we expand out the multionmial we can write each term as a multiple of $f$ using the appropriate identity, so $x$ is a multiple of $f$.

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    $\begingroup$ I don't think your first argument works - it's not necessarily true that $Spec(O(U\cup V)/f)$ is covered by $Spec(O(U)/f)$ and $Spec(O(V)/f)$: consider for instance the case $f=0$ (so I can omit $f$ from the notation), $X=\mathbb A^2=Spec(k[x,y]),U=X-V(x),V=X-V(y)$. $U\cup V$ is the punctured plane, and $Spec(O(U\cup V))=Spec(k[x,y])$ isn't covered by $U=Spec(O(U)),V=Spec(O(V))$. $\endgroup$
    – Wojowu
    Mar 2, 2021 at 18:21
  • $\begingroup$ @Wojowu $U \cup V $ is supposed to be affine here - the claim is that the map for every affine open subset defines a sheaf of ideals. $\endgroup$
    – Will Sawin
    Mar 2, 2021 at 18:37
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    $\begingroup$ I see, that's true. It might be worth mentioning somewhere in the answer. Do you happen to know if $U\mapsto f|_UO(U)$ for all open $U$ can fail to be a sheaf? My example only invalidates your argument in this more general setting, not the statement. $\endgroup$
    – Wojowu
    Mar 2, 2021 at 18:44
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    $\begingroup$ Sure. Take $X = k[x,y,z]/( (x-y)z)$ and let $f= z$, $U$ to be the locus where $x$ is invertible, $V$ to be the locus where $y$ is invertible, the function $z x^{-1} $ on $U$ agrees with the function $ z y^{-1}$ on $V$, since on their intersection $z x^{-1} - z y^{-1} = x^{-1} y^{-1} z (y-x)=0$, but it is not $z$ times any function on $U \cup V$, as these functions are well-defined on the punctured plane, hence on the plane, as in your argument. $\endgroup$
    – Will Sawin
    Mar 2, 2021 at 18:57
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    $\begingroup$ @lefuneste I think stacks.math.columbia.edu/tag/009N does the job when you check a slightly more general notion of the sheaf condition which can be checked the same way (and which I also guess is not really more general, since we only need the intersection of two affine opens contained within a single affine open to be affine open, and that does not need separatedness). $\endgroup$
    – Will Sawin
    Mar 2, 2021 at 21:45

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