There is a statement in Huybrechts complex geometry that says that
analytic hypersurface(analytic subvariety of codimension 1) is the
union of locally finite irreducible components.
I was trying to prove this claim, and I found it quite likely to use the fact that: an analytic germ is a finite union of irreducible components, and the decomposition is unique.
My attempt:
First pick any point $x\in A \subset X$ where $A$ is analytic subvariety, then $(A,x)$ is an analytic germ by the definition. Assume $A = \cup A_i$ with $A_i$ are irreducible analytic subvariety.
if it's not locally finite at some point $y\in A$, then my idea is $(A,y)$ will have infinite many component, however I don't know how to compute the germ of the union $(\cup A_i,y) = \cup_{j\in J} (A_j,y)$ ?
Best Answer
First, each component of the germ $(A,x)$ is contained in a unique irreducible component of the hypersurface $A$: Let $Z$ be an irreducible component of the germ $(A,x)$. Then $Z$ is an irreducible subset of $A$, and since the irreducible components $A_1, A_2, \dotsc$ of $A$ are maximal irreducible subsets, $Z$ is contained in one component, say $Z \subset A_1$. If $Z$ is contained in another component, say $Z \subset A_2$, then $Z \subset A_1 \cap A_2$. But both germs $(A_1,x)$ and $(A_2,x)$ have distinct irreducible components, which are irreducible components of $(A,x)$. So $Z$ is contained in two different irreducible components of $(A,x)$, which contradicts the maximality of $Z$ with respect to irreducible germs.
This means you get a map $$\{\, \text{components of } (A,x) \} \to \{\, \text{components of } A\,\}.$$ Clearly, every component of $A$ that contains $x$ is in the image of that map. Since the domain is a finite set, there are only finitely many components of $A$ which contain $x$.