Prepared factorizations correspond to germ factorizations

Suppose $f=gh$. Restricting to the axis $z^{\mathrm{\prime }}=0$ gives $f(0,\zeta )=g(0,\zeta )h(0,\zeta )\mathrm{.}$ Because $f$ is regular of order $d$, [L3] makes the left-hand side a product of ${\zeta }^d$ and a nonvanishing holomorphic function. Hence neither factor on the right is identically zero, and [L3] gives integers $e$ and $d-e$ such that $g(0,\zeta )$ has exact order $e$ and $h(0,\zeta )$ has exact order $d-e$. Thus $g$ and $h$ are regular in $z_m$.

Conversely, if $W=GH$ with $G$ and $H$ Weierstrass of positive degree, then $f=uW=(uG)H$. Step [L1] makes both $G$ and $H$ nonunits, so this is a nontrivial factorization of $f$ in the germ ring.

Prepare the factors: $g=u_{g}G,h=u_{h}H\mathrm{.}$ Then $f=(u_g{u}_h)(GH)\mathrm{.}$ The product $GH$ is monic of degree $d$ in $z_m$, and its lower coefficients still vanish at $z^{\mathrm{\prime }}=0$, so $GH$ is a Weierstrass polynomial of degree $d$. By the uniqueness part of [L2], the prepared polynomial of $f$ is unique, hence $W=GH$.

Step 2.1 shows that every nontrivial factorization of $f$ yields a nontrivial factorization of $W$, and step 1.2 shows the converse. Therefore $f$ is irreducible exactly when $W$ is irreducible in ${\mathcal{O}}_{m-1,0}[z_m]$.