Finite unique irreducible components of a hypersurface germ

By [F1] and reducedness of $f=(f_{\mathrm{r}\mathrm{e}\mathrm{d}})$, write $f=u{q}_1\cdots q_r$ with $u$ a unit, $r\ge 1$ and $q_1,\dots ,q_r$ pairwise nonassociate irreducibles. Since a product of complex values vanishes exactly when one factor vanishes and $u$ has no zero near $p$ by [F5], the zero sets agree: $X=Z(f)={\bigcup }_{i=1}^{r}Z(q_i)$.

Each $q_i$ is reduced: if $q_i$ were divisible by the square of an irreducible germ, say $q_i={\rho }^{2}h=\rho \cdot (\rho h)$, then both factors $\rho$ and $\rho h$ would be nonunits by [F5], contradicting irreducibility of $q_i$ in [F1]. Consequently $I_p(Z(q_i))=(q_i)$ by [F3].

Each $Z(q_i)$ is irreducible. Suppose $Z(q_i)=Z(g_1)\cup Z(g_2)$ with hypersurface subgerms $Z(g_j)⊊Z(q_i)$, the $g_j$ taken reduced by [F2]. Then $g_1{g}_2$ vanishes on $Z(q_i)$, so $g_1{g}_2\in I_p(Z(q_i))=(q_i)$ by step 2.1 and [F3], that is, $q_i\mid g_1{g}_2$. By primality of $q_i$ in [F4] we get $q_i\mid g_1$ or $q_i\mid g_2$, say $g_1=q_{i}h$; then $Z(q_i)\subseteq Z(g_1)$, so $Z(g_1)=Z(q_i)$, contradicting that $Z(g_1)$ is a proper subgerm. Hence $Z(q_i)$ admits no such cover and is irreducible.

The germs $Z(q_1),\dots ,Z(q_r)$ are pairwise incomparable. If $Z(q_i)\subseteq Z(q_j)$ with $i\mathrm{\ne }j$, then $q_j$ vanishes on $Z(q_i)$, so by [F3] and step 2.1 we have $q_j\in I_p(Z(q_i))=(q_i)$, that is, $q_i\mid q_j$. Since $q_j$ is irreducible, the other factor in $q_j=q_{i}h$ must be a unit; hence $q_i$ and $q_j$ are associates, contradicting their pairwise nonassociateness in step 1.1.

For every subset $J\subseteq \{1,\dots ,r\}$, every irreducible hypersurface subgerm $Y\subseteq {\bigcup }_{j\in J}Z(q_j)$ equals $Z(q_j)$ for some $j\in J$. Write $Y=Z(g)$ with $g$ reduced by [F2]. The product ${\prod }_{j\in J}{q}_j$ vanishes on $Y$, so it lies in $I_p(Z(g))=(g)$ by [F3], and $g$ divides that product. Factoring $g$ into irreducibles, each factor divides some $q_j$ by primality [F4], hence is associate to that irreducible $q_j$; since $g$ is reduced, $g$ is a unit multiple of ${\prod }_{j\in J_0}{q}_j$ for a nonempty subset $J_0\subseteq J$. Thus $Y={\bigcup }_{j\in J_0}Z(q_j)$. If $\mathrm{\mid }{J}_0\mathrm{\mid }\ge 2$, choose $j_0\in J_0$ and put $K=J_0\setminus \{j_0\}$, which is nonempty. Then $Y=Z(q_{j_0})\cup Z({\prod }_{j\in K}{q}_j)$. Both terms are hypersurface subgerms of $Y$ and both are proper: equality of either with $Y$ would, by [F3], make its reduced defining equation associate to $g$, although $g$ has distinct irreducible factors indexed by all of $J_0$. This contradicts irreducibility of $Y$ [F2]. Hence $\mathrm{\mid }{J}_0\mathrm{\mid }=1$ and $Y=Z(q_{j_0})$.

Taking $J=\{1,\dots ,r\}$ in step 4.1 and using step 1.1, a hypersurface subgerm $Y\subseteq X$ is irreducible if and only if $Y=Z(q_i)$ for some $i$: one direction is step 4.1 and the other is step 3.1. The germs are pairwise incomparable by step 3.2, so each is maximal among the irreducible subgerms of $X$; hence they are exactly the irreducible components in the sense of [F2].

Suppose $X={\bigcup }_{s=1}^{s_0}{Y}_s$ with the $Y_s$ pairwise distinct irreducible hypersurface germs. Applying step 4.1 to each $Y_s\subseteq X$ shows that $Y_s=Z(q_{j(s)})$ for some index $j(s)$, and distinctness makes $s\mapsto j(s)$ injective. Conversely, each $Z(q_i)\subseteq X={\bigcup }_{s}Z(q_{j(s)})$, so step 4.1 applied to this union gives $Z(q_i)=Z(q_{j(s)})$ for some $s$; pairwise incomparability in step 3.2 gives $j(s)=i$. Thus $s\mapsto j(s)$ is a bijection, $s_0=r$, and the two sets of germs agree.

Each component $Z(q_i)$ determines its reduced defining germ up to a unit (Complex-analytic hypersurface germ and its reduced equation), so the multiset of associate classes of $q_1,\dots ,q_r$ depends only on $X$. Steps 1.1, 5.1 and 5.2 prove all three assertions.