Irreducible hypersurface germs and their components

Fix $n\ge 1$ and $p\in {\mathbb{C}}^n$, and let $X$ be a complex-analytic hypersurface germ at $p$, that is, a nonempty proper set germ of the form $X=(Z(f),p)$ for a nonzero nonunit germ $f$ (Complex-analytic hypersurface germ and its reduced equation).

A hypersurface subgerm of $X$ is a hypersurface germ $Y$ at $p$ with $Y\subseteq X$ as set germs; by the definition of a hypersurface germ, every such $Y$ has the form $(Z(g),p)$ for a nonzero nonunit $g$, and by Square-free reduction of a holomorphic equation and Reduced holomorphic germ for a hypersurface the equation may be taken reduced.

The germ $X$ is reducible when there are hypersurface subgerms $Y_1,Y_2\subseteq X$ with

$Y_1\mathrm{\ne }X,Y_2\mathrm{\ne }X,X=Y_1\cup Y_2$

as set germs; it is irreducible when no such pair exists.

An irreducible component of $X$ is an irreducible hypersurface subgerm $Y\subseteq X$ that is maximal among the irreducible hypersurface subgerms of $X$: if $Y\subseteq Y^{\mathrm{\prime }}\subseteq X$ and $Y^{\mathrm{\prime }}$ is an irreducible hypersurface subgerm, then $Y^{\mathrm{\prime }}=Y$.

The notions only involve the set germ $X$: containment and union of set germs are defined by containment and union of representatives on a common neighbourhood of $p$, and the resulting notions do not depend on the chosen representatives or on the defining equation.

The pair condition in the definition of reducibility also covers finite decompositions. If $X=Y_1\cup \cdots \cup Y_s$ is a finite union of hypersurface subgerms with $Y_i=Z(g_i)$, then the identity $Z(g_1)\cup \cdots \cup Z(g_s)=Z(g_1\cdots g_s)$ writes the union as a single hypersurface subgerm, and grouping the factors into two products writes $X$ as the union of the two corresponding subgerms $Z(g_1\cdots g_k)$ and $Z(g_{k+1}\cdots g_s)$. Thus $X$ is reducible exactly when it is a finite union of hypersurface subgerms properly contained in it.