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Let $X$ be an integral scheme over $S$ and let $Y$ be a scheme of finite type over $S$. Let $x \in X$. How can I show that a morphism from $Spec \ \mathcal{O}_{X,x}$ to $Y$ can be extended to a morphism from some neighbourhood of $x$?

Well, I am not very good even with first steps towards a solution. The only idea is that we probably should find a way to extend a homomorphism to the local ring (more precisely, algebra) to a homomorphism to some $\mathcal{O}_{X}(U)$.

Hypsoline
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  • "from some $\mathcal{O}_{X,x}(U)$" Are not you confused? Functions are going in opposite direction! – quinque May 29 '15 at 00:32
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    This answer might be helpful. – c_c_chaos May 29 '15 at 06:56
  • Indeed, the link provided by @c_c_chaos yields a perfect answer. – Georges Elencwajg May 29 '15 at 07:26
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    @c_c_chaos Thank you. The only thing I am confused about is that the original conditions of my problem are somewhat different. Firstly, in this case $S$ is not Noetherian (hence, we cannot choose a finite set of relations). And the fact that $X$ is an integral scheme remains unused. – Hypsoline May 31 '15 at 11:34
  • @quinque Oh, yes, you are right! – Hypsoline May 31 '15 at 11:36
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    Dear @Hypsoline, I think you indeed have to make some modifications. I'll freely use the notation used in the answer I linked to. Then the crucial fact is that any natural morphism $A_f \to A_\mathfrak{p}$ is injective (where $f \in A \setminus \mathfrak{p}$) because $X$ is integral. For this reason, you do not have to define a morphism $B \to A_f$; instead, it suffices to show that the image of the given morphism $B \to A_\mathfrak{p}$ is contained in the subring (via the canonical injective morphism alluded to above) $A_f \subset A_\mathfrak{p}$ for some $f$ . Does this help? – c_c_chaos May 31 '15 at 15:15
  • @c_c_chaos Thank you! Well, in fact, this case proves to be less complicated than that in the answer you linked to. – Hypsoline May 31 '15 at 17:43

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First of all it is enough to proof in case where $S$ and $Y$. Why? By definition of finite type at the point applied to image of $x$ in $Y$.

So we can think that $Y = Spec B$, $Z = Spec A$. Also we can think that $X$ is affine (we are constructing map from a neighbourhood of $x$). Denote $X = Spec A$.

Finally, we have map of $C$ algebras $f: B \rightarrow A_{\mathfrak{p}}$, where $\mathfrak{p}$ is the ideal of $x$. $B$ has finitely many generates $b_1, \dots b_k$.

$$f(b_i) = \frac{x_i}{y_i} $$

$X$ is integral scheme i.e. $A$ does not have zero divisors. Then $A_{y_1, \dots , y_k} \subset A_{\mathfrak{p}}$.

Notice that $\text{Im} f \subset A_{y_1 , \dots , y_k}$. So we have the map

$$ f: B \rightarrow A_{y_1 , \dots , y_k} $$

quinque
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  • "it is enough to proof in case where S and Y." What do you mean by that? Do you use somewhere that $X$ is an integral scheme over S? Could you please describe more precisely this map $f \colon B \to A_{y_1,…,y_k}$ (I am rather interested not in the definition itself but why it is correct)? – Hypsoline May 31 '15 at 11:42
  • Oh, I am sorry, I forgot the fact that morphisms to the local ring are injective. Now everything is clear. Thank you! – Hypsoline May 31 '15 at 17:49