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Consider two schemes $X,Y$ over a locally noetherian scheme $S$. Let $p \in X$ and assume that $X$ is irreducible and not affine spectrum of a semilocal ring.

We assume moreover we have a morphism $f: \operatorname{Spec}O_{X,p} \to Y$. One may ask what are the neccessary conditions to extend $f$ to an open subscheme $U \subset X$ which contains $p$, that is to extend to a morphism $h: U \to Y$ which restricts under composition with $ \operatorname{Spec}O_{X,p} \to U$ to $f$.

Certainly, a well known sufficient condition is if we assume that $Y$ is locally of finite type over $S$. Then we can choose an affine subscheme $\operatorname{Spec} \ R= S_0 \subset S$ and open subscheme $\operatorname{Spec} \ T= Y_0 \subset Y$. Since $Y$ locally of finite type $T= R[x_1,x_2,..., x_n]/I$. Since $R$ is noetherian, $R[x_1,x_2,..., x_n]$ is also noetherian and the ideal $I$ is finitely generated. Let $\operatorname{Spec} \ A\subset X$ be an affine open neighbourhood of $p \in X$.

Then we consider a morphism $\phi:R[x_1,\dots,x_n]/I\rightarrow \mathscr{O}_ {X,p}$. Now, write $\phi(x_i)=a_i/r$ for some $r \in A$ not vanishing at $p$ and let $s \in A$ not vanishing at $p$ be such that $s \cdot g(\phi(x_1),\dots,\phi(x_n))=0$ for all $g \in I$. Here it is crucial that $I$ is finitely generated ideal, otherwise such $s$ might not exist. then the open set $D(sf)$ makes the job, where $l:V→Y$ corresponds to the morphism $\psi:k[x_1,\dots,x_n]/I\rightarrow A[(sf)^{-1}]$ mapping $x_i$ to $a_i/r$.

Now I ask if to require $Y$ is locally of finite type over $S$ around $y=f(x)$ is also a neccessary condition to obtain the extension above. I guess so but I haven't found a conterexample.

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    $\begingroup$ If $X$ is a point then the extension property is vacuously satisfied without any assumptions on $Y$. Do you maybe mean something like "if $Y$ is locally of finite type over $S$ near $y$, then for every $S$-scheme $X$ with a morphism $\operatorname{Spec} \mathcal O_{X,x} \to Y$ taking $x$ to $y$ there exists an extension on an open"? $\endgroup$ – R. van Dobben de Bruyn Apr 29 at 18:47
  • $\begingroup$ No, you are right, I just overlooked this trivial case if $X$ is finite or more general discrete. But, no I not meant the "if $Y$ is locally of finite type over $S$ near $y$, then for every $S$-scheme $X$ with a morphism $Spec \ O_{X,x} \to Y$ taking $x$ to $y$ there exists an extension on an open" since we know that in this case it is always true, see my sketch above. I'm interested in the case if we assume that $X$ is irreducible of dimension $\ge 1$. Is it neccessary to require that $Y$ is locally of finite type over $S$ near $y$? I will improve my question... $\endgroup$ – user7391733 Apr 29 at 19:01
  • $\begingroup$ In other words I'm asking if the statement that the solubility of the extension problem above (with $X$ irreducible of dim $\ge 1$ is equivalent to $Y$ is locally of finite type over $S$ near $y$. One implicaition I solved. For the other I conjecture that it also true but I need a conterexample in order to argue by contraposition $\endgroup$ – user7391733 Apr 29 at 19:09
  • $\begingroup$ Oh right, I wrote the converse of what I meant; my question was really about varying $X$. But 'irreducible of dimension $\geq 1$' will not help much, because my objection still exists for (semi)local schemes. $\endgroup$ – R. van Dobben de Bruyn Apr 29 at 20:15
  • $\begingroup$ Yes, I see the issue. The also obvious semilocal case we have also to exclude. see update $\endgroup$ – user7391733 Apr 29 at 20:24

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