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Let $X,Y$ over a say locally noetherian scheme $S$. (later we have to discuss if the locally noetherian scheme condition for $S$ is really important.)

Let $p \in X$ be a point and we have a morphism $f: \operatorname{Spec}O_{X,p} \to Y$. We 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 $f$.

A 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 $p \in X$.

Consider a morphism 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 $a_i/r$. (that's an argument by Gaussian from here. The discussion there motivates this question.

Now I ask if to require $Y$ is locally of finite type over $S$ is also a neccessary condition. I guess so but I haven't found a conterexample.

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  • $\begingroup$ How do you quantify the condition? Is it "for every $X$, every $p$ and every $f$", or are some of the data fixed? $\endgroup$ – Pavel Čoupek Apr 25 '20 at 3:48
  • $\begingroup$ @PavelČoupek: Yes, my conjecture was: For every datum $(X, p \in X, f: \operatorname{Spec}O_{X,p} \to Y)=(X,p,f)$ holds: the $f$ is locally extendable (in the sense I explained above) $\Leftrightarrow $ $Y$ is locally (only around $f(p)$(!), indeed a good point) of finite type over $S$. One direction Gaussian proved, now to finish the proof I have to find a conterexample: that is a datum $(X,p,f)$ with $Y$ (indeed, $Y$ is a part of datum specified by $f$) that is locally around $f(p)$ not of finite type over $S$, such that $f$ is not extendable. $\endgroup$ – user7391733 Apr 25 '20 at 19:22

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