Two morphisms are equivalent if and only if they have the same germ

Question

Let $X, Y$ be two schemes. Assume $U,V$ are two dense open sets in $X$. Let $f:U \longrightarrow Y$ and $g: V \longrightarrow Y$ be morphisms. We say $f$ and $g$ are equivalent if they coincide on an open subset of $U \cap V$.

I see a paragraph in EGA : Let $X$ be an irreducible scheme, then any open subsets of $X$ is dense. Assume the generic point of $X$ is $x$. Then any two morphisms from some open subsets to $Y$ are equivalent if and only if they have the same germ at $x$. The paragraph comes from Volumn I, Grothendieck's EGA, (7.1.4), in the section about rational functions.

I do not how to prove the last sentence. I can only prove that if $f:U \longrightarrow Y$ and $g: V \longrightarrow Y$ are equivalent morphisms, then they have the same germ at $x$. This is easy since $f$ and $g$ coincide on an open set containing $x$.

Answer 1

From what I understand :

• Let $X$, $Y$ be two schemes with $U, V \subset X$ two nonempty dense open subsets and $f : U \to Y$, $g : V \to Y$ two morphisms. We define an equivalence relation $$f \mathcal R g \Longleftrightarrow \exists W \neq \emptyset,~ W \subset U \cap V ~;~ f\vert_W = g \vert_W$$ where $W$ is open and dense in $U \cap V$. We say that $f$ and $g$ are equivalent if and only of $f \mathcal R g$.

• A germ at $x \in X$ is an equivalence class of morphisms defined on a nonempty dense open subset of $X$ containing $x$ for the relation (notice the subscript $x$) : $$(f : U \to Y) \mathcal R_x (g : V \to Y) \Longleftrightarrow \exists W \ni x \subset U \cap V ; f\vert_W = g\vert_W$$ where $W$ is open and dense in $U \cap V$.

• As far as we know, $\forall x \in X :$ $f \mathcal R_x g \implies f \mathcal R g$, but the converse may be false. Two morphisms can be equivalent but it does not imply that there exists an open subset of $X$ containing $x$ such that $f$ and $g$ coincide on it.

If $X$ is irreducible, any nonempty open subset of $X$ is dense, in particular it contains the generic point of $X$ that I will denote by $\eta$. In particular, if $U$ and $V$ are such open subsets of $X$, any nonempty open subset of $U \cap V$ is dense in $X$ (and so, à fortiori, dense in $U \cap V$) and contains $\eta$. But now in this configuration it is clear that $f \mathcal R g \implies f \mathcal R_{\eta} g$ so that $\mathcal R = \mathcal R_{\eta}$. In other words, if $X$ is irreducible, two morphisms are equivalent if and only if they have the same germ at $\eta \in X$, the generic point of $X$.