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Let $X$ be a scheme, $x\in X$. Then we have a canonical morphism $\mathrm{Spec}\,\mathcal O_{X,x}\to X$ defined as follows. Let us take an affine open set $U\ni x$. The canonical homomorphism $\mathcal O_X(U)\to \mathcal O_{X,x}$ induces a morphism $\mathrm{Spec}\,\mathcal O_{X,x}\to U$ . Composing this morphism with the open immersion $U \to X$, we obtain a morphism $\mathrm{Spec}\,\mathcal O_{X,x}\to X$ . How to show this morphism does not depend on the choice of $U$?

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  • $\begingroup$ Do you know how to show it for two open sets $V\subset U$? $\endgroup$ – A.G Sep 25 '16 at 14:46
  • $\begingroup$ $\mathrm{Spec} A_g\to \mathrm{Spec} B\to \mathrm{Spec} A$ $\endgroup$ – AG2016 Sep 25 '16 at 15:43
  • $\begingroup$ $\mathrm{Spec}\,B_f\to\mathrm{Spec}\,A_g\to \mathrm{Spec}\,B\to\mathrm{Spec}\,A$? $\endgroup$ – AG2016 Sep 25 '16 at 15:59
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First step. Show it for $V \subset U$:

This is purely formal: The map $\mathcal O_X(U) \to \mathcal O_{X,x}$ coincides with the composition $\mathcal O_X(U) \to \mathcal O_X(V) \to \mathcal O_{X,x}$, i.e. by functorality the induced map $\operatorname{Spec} \mathcal O_{X,x} \to U$ coincides with $\operatorname{Spec} \mathcal O_{X,x} \to V \to U$. This is all we need. Just further compose with the inclusion $U \to X$ and note that the compositions of inclusions $V \to U \to X$ is equal to the inclusion $V \to X$.

Second step. This suffices. Let $U,V \ni x$ be two arbitrary affine open. Choose an affine open $W \subset U \cap V$. By the first step the constructions of the morphism via $U$ and $V$ both give the same result as the construction via $W$. In particular the construction does not depend on whether we pick $U$ or $V$.

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