0
$\begingroup$

Let $X=\mathrm{Spec}(A)$ be an affine integral scheme of finite type over $\mathbb{C}$ and $\phi:\mathcal{F} \to \mathcal{G}$ be a surjective morphism of coherent sheaves on $X$. Let $f \in A$, $U:=D(f)$ the open set defined by $f$ and $s \in H^0(\mathcal{G})$. Let $s' \in \Gamma(U,\mathcal{F})$ be a section such that $\phi(s')=s|_U$. Does there exist an open set $V \subset X$ containing $\mathrm{Supp}(s)$ and a section $s_0 \in \Gamma(V,\mathcal{F})$ such that $\phi|_V(s_0)=s|_V$ and $s_0|_{V \cap U} =s'|_{V \cap U}$?

$\endgroup$

1 Answer 1

0
$\begingroup$

No, that is not true as posed. Let $A=\mathbb{C}[x]$, let $f$ be $x$, let $\mathcal{F}$ be $\widetilde{A}$, and let $\mathcal{G}$ be the zero sheaf. Let $s'$ be $1/x$. Do you want to allow modifications of $s'$ and $s$ by multiplying by powers of $f$?

$\endgroup$
2
  • $\begingroup$ @Starr: I have edited the question slightly. $\endgroup$
    – Ron
    May 4, 2015 at 18:15
  • $\begingroup$ @Ron. The answer is still no, even with the edits. Let $\mathcal{F}$ be $\widetilde{A}\oplus \widetilde{A}$, let $\mathcal{G}$ be $\widetilde{A}$, let $\phi$ be projection onto the second factor, let $s$ be $1$ and let $s'$ be $(1/x,1)$. If you hope for something like this to be true, you typically allow modifications of $s$ and $s'$ by multiplying by powers of $f$, cf. Lemma 5.3 of Chapter II of Hartshorne's "Algebraic Geometry". $\endgroup$ May 4, 2015 at 20:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.