# Is the presheaf ideal of a sheaf generated by a global section again a sheaf?

Given a sheaf of rings $$\mathcal O_X$$ on a top space $$X$$ and $$\sigma\in\mathcal O_X(X)$$ a global section, is $$\mathcal O_X\cdot\sigma$$ then a sheaf, or should I sheafify?

The locality property is clearly satisfied, so I'm concerned with the gluing property. Given a covering $$\{U_i\}$$ of $$X$$ and $$\tau_i\in\mathcal O_X(U_i)$$ such that $$\tau_i\cdot\sigma\vert_{U_i\cap U_j}=\tau_j\cdot\sigma\vert_{U_i\cap U_j}$$, is it then true that the unique element $$\eta\in\mathcal O_X(X)$$ such that $$\eta\vert_{U_i}=\tau_i\cdot\sigma\vert_{U_i}$$ belongs to $$\mathcal O_X(X)\cdot\sigma$$? We cannot show that $$\tau_i\vert_{U_i\cap U_j}=\tau_j\vert_{U_i\cap U_j}$$, since $$\sigma$$ could vanish on say $$U_i$$, and hence we don't need $$\tau_i\vert_{U_i\cap U_j}=\tau_j\vert_{U_i\cap U_j}$$ for $$\tau_i\cdot\sigma\vert_{U_i\cap U_j}=\tau_j\cdot\sigma\vert_{U_i\cap U_j}$$ to hold.

I'm trying to think of a counter example with $$\operatorname{Spec}(R)$$. If $$R=Rf+Rg$$ for some $$f,g\in R$$ and $$s\in R$$, then given $$\frac{a}{f^n}\cdot s\in R_f,\frac{b}{g^m}\cdot s\in R_g$$ such that $$\frac{a}{f^n}\cdot s=\frac{b}{g^m}\cdot s\in R_{gf}$$, does the unique element $$x\in R$$ such that $$x=\frac{a}{f^n}s\in R_f$$ and $$x=\frac{a}{f^n}s\in R_g$$ satisfy $$x\in R\cdot s$$? If $$R$$ is a UFD, I think it's true, otherwise not necessarily.

Edit: In the maths chat I got a counter example from Thorgott (see link below) which uses a $$3$$-point topological space $$X=\{x,y,z\}$$ such that $$x$$ is an open point, and the points $$y$$ and $$z$$ are closed, with sheaf $$\mathcal O_X(\{x,y\})=\mathbb Z=\mathcal O_X(\{x,z\})$$, $$\mathcal O_X(\{x\})=\mathbb Z/6\mathbb Z$$ (these data fully specify a sheaf on $$X$$).

https://chat.stackexchange.com/transcript/message/63472101#63472101

• For an affine scheme the answer is positive: let $I\subseteq R$ be an ideal and $r\in R$ an element such that $r/1\in I\cdot R_{\mathfrak{p}}$ for every prime $\mathfrak{p}\subseteq R$. Consider the ideal $J=\{s\in R\ |\ sr\in I\}$. If by contradiction $J\neq R$, then it is included in some maximal ideal $\mathfrak{m}\supseteq J$. But then there exists $s\notin\mathfrak{m}$ such that $sr\in I$ (as $r/1\in I\cdot R_{\mathfrak{m}}$), contradiction. Hence $J=(1)$, i.e. $r\in I$. Applying this to $I=(\sigma)$, we obtain that being a multiple of $\sigma$ is local, i.e. it is a sheaf. Apr 28, 2023 at 8:02
• It's good to know that it's true for affine schemes! That also shows why a scheme-theoretic counter example has to be a bit intricate. Apr 28, 2023 at 19:05

It is false even for $$X$$ a scheme. I restate the example given in the comments to the answer of this post by Will Sawin.

Let $$R=k[x,y,z]/(z(y-x))$$ and let $$X=D(x)\cup D(y)\subseteq\operatorname{Spec}(R)$$. Let $$\sigma=z|_X$$. Note that the functions $$zx^{-1}\in \mathcal{O}_X(D(x))$$ and $$zy^{-1}\in \mathcal{O}_X(D(y))$$ glue to a global section, because in $$\mathcal{O}_X(D(xy))=(k[x,y,z]/(z(y-x)))_{xy}$$ we have $$zx^{-1}-zy^{-1}=z(x-y)/(xy)=0.$$ So let $$\eta\in\mathcal{O}_X(X)$$ be this global section obtained by gluing. Suppose by contradiction that $$\eta=z|_X\cdot \tau$$ for some global section $$\tau\in\mathcal{O}_X(X)$$. We want to derive a contradiction from this.

Note that there exists an $$n>0$$ such that $$x^n\cdot\tau|_{D(x)}$$ and $$y^n\cdot\tau|_{D(y)}$$ lift to global sections of $$\operatorname{Spec}(R)$$, i.e. there exist $$r,s\in R$$ such that $$r|_{D(x)}=x^n\cdot\tau|_{D(x)}$$ and $$s|_{D(y)}=y^n\cdot\tau|_{D(y)}$$. Also, note that the annihilator of $$z$$ in $$\mathcal{O}_X(D(x))$$ resp. $$\mathcal{O}_X(D(y))$$ is the ideal generated by $$y-x$$ (as no power of $$x$$ resp. $$y$$ annihilate $$z$$ in $$R$$). Hence, as $$z\tau|_{D(x)}=zx^{-1}$$ resp. $$z\tau|_{D(y)}=zy^{-1}$$, we obtain $$\tau|_{D(x)}\equiv_{(y-x)}x^{-1}$$ resp. $$\tau|_{D(y)}\equiv_{(y-x)}y^{-1}$$. Finally, note that $$(y^nr-x^ns)|_{D(xy)}=0$$, and hence $$y^nr=x^ns$$ as $$D(xy)$$ is dense.

Putting everything together, we obtain the equations $$r|_{D(x)}=x^{n-1}+p\cdot (y-x),\quad s|_{D(y)}=y^{n-1}+q\cdot (y-x)$$ for some $$p,q$$, and by increasing $$n$$ we may lift these equations to $$R$$, i.e. $$r=x^{n-1}+p\cdot (y-x)$$ and $$s=y^{n-1}+q\cdot (y-x)$$. But then as $$y^nr=x^ns$$, this gives $$(y-x)((xy)^{n-1}+y^np-x^nq)=0$$. By restricting this to $$D(y-x)$$, every $$z$$ in the equation will be eliminated, and we obtain an equation of the form $$(xy)^{n-1}+y^n\overline{p}-x^n\overline{q}=0$$ inside $$k[x,y]_{y-x}$$, which is impossible.

• Regarding the intuitive explanation: Did you mean that $z$ is a zero-divisor on $D(x)$ and $D(y)$? Since we have $z(y-x)=0$ and $y\neq x$. Also, why is $D(x)\cup D(y)=\operatorname{Spec}(k[x,y])-\{0\}$? I don't see how we lose $z$ in the description (this might be tied to my previous question). I appreciate the explicit computations a lot, but these intuitive insights seem valuable too, as I generally lack intuition in algebraic geometry Apr 28, 2023 at 19:21
• @ShaVuklia your right, my intuitive explanation was a non-sensical. I rewrote the solution in a more scheme theoretic approach to have more of a geometric understanding of what is happening. I'll try to come up with a sensible intuitive explanation and I will post it if I come up with something :) May 1, 2023 at 13:16
• I appreciate it! I hope you don't mind if I still ask some questions: 1) Why is $D(xy)$ dense in $\operatorname{Spec}R$? 2) Why does restricting this global section to zero on a dense open imply that the section in zero? I suppose since $k[x,y,z]/(z(x-y))$ is reduced and Noetherian, we can invoke the result in the following post, but maybe there is a more elementary argument: math.stackexchange.com/questions/505224/… May 2, 2023 at 19:51
• 1) The irreducible components of $\operatorname{Spec}(R)=V(z(x-y))\subseteq \mathbb{A}^3$ are $V(z)$ and $V(x-y)$. These correspond to the minimal primes $(z),(x-y)\subseteq R$. Note that $D(xy)$ contains both of them, so the closure of $D(xy)$ contains their respective closures, i.e. both irreducible components. 2) If $\mathcal{F}$ is a coherent sheaf on a scheme $X$ and $s$ a global section of $\mathcal{F}$, then $V(s):=\{P\in X\ |\ s_P\in\mathfrak{m}_P\mathcal{F}\}$ is closed. If it is zero on a dense open, then $V(s)=X$. By restricting to an affine open cover, this gives $s=0$. May 3, 2023 at 6:15
• Very valuable, thank you. I still have to read about irreducible components and coherent sheaves (which I will do soon), and I'll keep your post in mind when I go through the theory. Many, many thanks for the effort you put in your answers! May 3, 2023 at 8:34