Proposition 9.3.1477256F93241

Proof. A decreasing sequence of analytic germs is eventually constant, because the corresponding ideals increase and are eventually constant by proposition 9.2.4, and $V(I_{A,x})=(A,x)$. Split a reducible germ into two analytic subgerms different from it, split each reducible piece again, and so on. If this process did not terminate, the resulting tree, in which every vertex has at most two children, would be infinite, and choosing at each level a child with infinitely many descendants would produce a strictly decreasing infinite sequence of analytic germs, which is impossible. Hence a finite decomposition into irreducible germs exists; removing members contained in others, we may assume that no member is contained in another. An irreducible germ contained in a finite union of analytic germs is contained in one of them, since it is the union of its intersections with them. Now let $\bigcup _ iA_ i=\bigcup _ jB_ j$ be two such decompositions. Each $A_ i$ is contained in some $B_ j$, which in turn is contained in some $A_ k$; then $A_ i\subset A_ k$ forces $k=i$ and $A_ i=B_ j$. By symmetry the two decompositions agree.

Let $(A,x)$ be irreducible. Its ideal is proper, since $x\in A$. If $fg$ vanishes on $(A,x)$, then $(A,x)$ is the union of its intersections with $\{ f=0\} $ and $\{ g=0\} $, so one of them is all of $(A,x)$, and $f$ or $g$ belongs to $I_{A,x}$; thus $I_{A,x}$ is prime. Conversely, suppose that $(A,x)$ is the union of two analytic subgerms $(B,x)$ and $(C,x)$ different from it. Since $V(I_{B,x})=(B,x)\neq (A,x)$, there is a germ $f\in I_{B,x}\setminus I_{A,x}$, and similarly $g\in I_{C,x}\setminus I_{A,x}$. Then $fg$ vanishes on $(A,x)$ while neither factor does, so $I_{A,x}$ is not prime. $\square$