# On the definition of a principal ideal sheaf

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 $$X$$?

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 $$f$$.

• 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 $U=Spec(O(U)),V=Spec(O(V))$. Mar 2, 2021 at 18:21
• @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. Mar 2, 2021 at 18:37
• 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. Mar 2, 2021 at 18:44
• 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. Mar 2, 2021 at 18:57
• @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). Mar 2, 2021 at 21:45