# How to show the following morphism is independent of the choice of neighborhoods?

In Görtz and Wedhorn's AGI (3.4)

Let $$X$$ be a scheme. Let $$x ∈ X$$, and let $$U \subset X$$ be an affine open neighborhood of $$x$$, say $$U = \mathop{Spec} A$$. Denote by $$p \subset A$$ the prime ideal of $$A$$ corresponding to $$x$$. Then $$\mathcal{O}_{X,x} = \mathcal{O}_{U,x} = A_p$$, and the natural homomorphism $$A \to A_p$$ gives us a morphism $$j_x \colon \mathop{Spec} \mathcal{O}_{X,x} = \mathop{Spec} A_p \to \mathop{Spec}A = U \subset X$$ of schemes. By Proposition 3.2 (2) this morphism is independent of the choice of $$U$$.

The Proposition 3.2 (2) is the following:

Let $$X$$ be a scheme. The affine open subscheme are a basis of the topology.

My question: Why is this morphism independent of the choice of $$U$$?

## 1 Answer

Let's start with the Affine Communication Lemma

Let $$X$$ be a scheme. $$U, \ V$$ are open affine subschemes of $$X$$. Then given $$x \in U \cap V \ \ \exists W \subset U \cap V$$ such that $$x \in W$$ is distinguished open in both $$U$$ and $$V$$.

So in your case you get two morphisms $$j_1 : Spec \ \mathcal O_{X,x} \rightarrow U$$ and $$j_2 : Spec \ \mathcal O_{X,x} \rightarrow V$$

Since $$W$$ is a distinguished open affine in $$U$$ the first morphism factors through $$Spec \ \mathcal O_{X,x} \rightarrow W \to U$$ and the second factors through $$Spec \ \mathcal O_{X,x} \rightarrow W \to V$$.

The first arrow in both the factors is the same morphism by commutativity while the second is just inclusion in both cases. So the composition gives you the same morphism.

• You don't really need distinguished affine if you know $$Mor(X,Y) \simeq Hom_{Rings} (\Gamma (Y,\mathcal O(Y)) , \Gamma (X,\mathcal O(X) ) )$$ if Y is affine. – Ignorant Mathematician Apr 14 at 6:38