Irreducible holomorphic germs are prime

Assume $q\mid ab$, so that $ab=qc$ for some germ $c$, and suppose for contradiction that $q\nmid a$ and $q\nmid b$.

If $a=0$, then $a=q\cdot 0$ gives $q\mid a$; similarly $b=0$ gives $q\mid b$. Both contradict the supposition of step 1.1, so $a\mathrm{\ne }0$ and $b\mathrm{\ne }0$, and then $c\mathrm{\ne }0$ as well because ${\mathcal{O}}_{{\mathbb{C}}^n,0}$ is an integral domain by [F2].

If $a$ were a unit, then $b=q(c{a}^{-1})$ would give $q\mid b$, and if $b$ were a unit then $a=q(c{b}^{-1})$ would give $q\mid a$; both contradict step 1.1. Hence $a$ and $b$ are nonunits.

The germ $c$ is a nonunit. If $c$ were a unit, then $q=a\cdot (b{c}^{-1})$ would be a factorisation of the irreducible germ $q$ into the nonunit $a$ and the nonunit $b{c}^{-1}$ — the latter because $b$ is a nonunit and $c^{-1}$ is a unit, so [F3] applies — contradicting irreducibility of $q$ in [F1].

By [F2] factor the nonzero nonunits $a,b,c$ of steps 2.1, 2.2 and 3.1 into irreducibles, say $a=u{a}_1\cdots a_r$, $b=v{b}_1\cdots b_s$ and $c=w{c}_1\cdots c_t$ with $u,v,w$ units and all displayed factors irreducible. Then $ab=uv{a}_1\cdots a_r{b}_1\cdots b_s$ and $qc=qw{c}_1\cdots c_t$ are equal, so the products of irreducibles $a_1\cdots a_r{b}_1\cdots b_s$ and $q{c}_1\cdots c_t$ differ by the unit $(uv{)}^{-1}w$.

Setting $q^{\mathrm{\prime }}:=(uv{)}^{-1}wq$, the germ $q^{\mathrm{\prime }}$ is associate to $q$, hence irreducible, and step 4.1 gives the equality of products of irreducibles $a_1\cdots a_r{b}_1\cdots b_s=q^{\mathrm{\prime }}{c}_1\cdots c_t\mathrm{.}$

By the uniqueness clause of [F2] applied to the two products of step 5.1, the irreducible $q^{\mathrm{\prime }}$ is associate to one of the irreducibles $a_1,\dots ,a_r,b_1,\dots ,b_s$.

If $q^{\mathrm{\prime }}$ is associate to some $a_i$, then $q^{\mathrm{\prime }}\mid a_i$ and $a_i\mid a$ because $a_i$ is one of the factors of $a$, hence $q^{\mathrm{\prime }}\mid a$; since $q^{\mathrm{\prime }}$ is associate to $q$, also $q\mid a$, contradicting step 1.1. The same argument with some $b_j$ gives $q\mid b$, again contradicting step 1.1. Hence the supposition of step 1.1 is impossible, so $q\mid a$ or $q\mid b$; this proves that the irreducible germ $q$ is prime.