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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

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    $\begingroup$ 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. $\endgroup$
    – imtrying46
    Apr 28, 2023 at 8:02
  • $\begingroup$ 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. $\endgroup$
    – Sha Vuklia
    Apr 28, 2023 at 19:05

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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.

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  • $\begingroup$ 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 $\endgroup$
    – Sha Vuklia
    Apr 28, 2023 at 19:21
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    $\begingroup$ @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 :) $\endgroup$
    – imtrying46
    May 1, 2023 at 13:16
  • $\begingroup$ 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/… $\endgroup$
    – Sha Vuklia
    May 2, 2023 at 19:51
  • $\begingroup$ 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 $s=0$. $\endgroup$
    – imtrying46
    May 3, 2023 at 6:15
  • $\begingroup$ 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! $\endgroup$
    – Sha Vuklia
    May 3, 2023 at 8:34

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