Irreducible analytic germ

Remarks

Equivalently, the ideal $\mathcal{I}(A_a)$ of holomorphic germs vanishing on $A_a$ is a prime ideal of $\mathcal{O}_{n,a}$. Example: the germ at $0$ of the cusp $A = \set{z_1^2 = z_2^3}$ in $\C^2$ is irreducible. The germ $f = z_1^2 - z_2^3$ is irreducible in $\mathcal{O}_{2,0}$, since it is a Weierstrass polynomial of degree $2$ in $z_1$. A factorization would give a germ $h$ with $h(z_2)^2 = z_2^3$, which is impossible by comparing orders of vanishing. Since $\mathcal{O}_{2,0}$ is a unique factorization domain, the ideal $(f)$ is prime, hence radical, and the Rückert Nullstellensatz gives $\mathcal{I}(A_0) = (f)$, a prime ideal. Non-example: the germ at $0$ of $\set{z_2^2 = z_1^2(1 + z_1)}$ is reducible. With $s = \sqrt{1 + z_1}$, the convergent binomial series, it is the union of the germs of the two smooth curves $\set{z_2 = z_1 s}$ and $\set{z_2 = -z_1 s}$, and neither contains the other. Every germ of an analytic set has a decomposition $A_a = A_{1,a} \cup \cdots \cup A_{k,a}$ into irreducible germs, unique up to order when no $A_{i,a}$ is contained in another.