# Extendability criteria for Morphisms of Schemes

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.

• How do you quantify the condition? Is it "for every $X$, every $p$ and every $f$", or are some of the data fixed? – Pavel Čoupek Apr 25 '20 at 3:48
• @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. – user7391733 Apr 25 '20 at 19:22