Reduced preparation has nonzero discriminant

By [F1] and [F2] write $f=u{\prod }_{i=1}^r{p}_i$ with $r\ge 1$, the $p_i$ irreducible and pairwise nonassociate, and no irreducible factor repeated.

For each $i$ factor $f=p_i{h}_i$ with $h_i:=u{\prod }_{j\mathrm{\ne }i}{p}_j$. By [F4] both $p_i$ and $h_i$ are regular in the last variable, the preparation $f=uW$ satisfies $W=W_i{V}_i$ where $p_i=u_i{W}_i$ and $h_i=v_i{V}_i$ are preparations, and $p_i$ is a nonunit, so its order $d_i=\mathrm{deg}{W}_i$ is at least $1$.

Applying the consequence in [F4] to the irreducible $p_i$ shows that $W_i$ is irreducible in ${\mathcal{O}}_{n-1,0}[T]$. Each $W_i$ is monic by [F3], hence primitive, so by Gauss's lemma [F5] $W_i$ is irreducible in $K[T]$.

The $W_i$ are pairwise distinct: if $W_i=W_j$ for $i\mathrm{\ne }j$, then $p_i=u_i{W}_i=u_i{u}_j^{-1}{p}_j$ makes $p_i$ and $p_j$ associates, contradicting step 1.1.

The product ${\prod }_{i=1}^r{W}_i$ is a Weierstrass polynomial: it is monic of degree ${\sum }_i{d}_i$ with coefficients in ${\mathcal{O}}_{n-1,0}$, and at $z^{\mathrm{\prime }}=0$ each factor equals $T^{d_i}$ by [F3], so the product equals $T^{{\sum }_i{d}_i}$. Since step 2.1 gives $f=\left(u{\prod }_i{u}_i\right){\prod }_i{W}_i$ and $f=uW$ is a preparation, uniqueness of the prepared polynomial [F3] yields $W={\prod }_{i=1}^r{W}_i$.

Hence $W$ is square-free in $K[T]$: an irreducible $P\in K[T]$ dividing $W$ twice would, by uniqueness of factorisation in the UFD $K[T]$ from [F6], be associate to two of the distinct monic irreducibles $W_i$; being monic it would equal both, contradicting step 4.1.

For the discriminant, let $E$ be a splitting field of $W$ over $K$ and write $W={\prod }_{k=1}^d(T-{\alpha }_k)$ as in [F7]. The roots of $W$ are the roots of the factors $W_i$. Two distinct factors $W_i,W_j$ are coprime in $K[T]$: their monic gcd divides the irreducible $W_i$, so it is $1$ or an associate of $W_i$, and in the second case it would also be an associate of $W_j$, forcing $W_i=W_j$; thus $1=A{W}_i+B{W}_j$ for some $A,B\in K[T]$ by [F9], and a common root would give $1=0$. A root of exactly one factor $W_i$ that were repeated for $W$ would be a repeated root of $W_i$, since the complementary product does not vanish there.

Each $W_i$ is separable by step 3.1 and [F8], so it has no repeated root in the extension $E$; combined with step 7.1, all roots ${\alpha }_1,\dots ,{\alpha }_d$ of $W$ in $E$ are pairwise distinct. The root formula in [F7] then gives ${\mathrm{Disc}}_T(W)={\prod }_{k<l}({\alpha }_k-{\alpha }_l{)}^2\mathrm{\ne }0$ in $K$.

Finally, ${\mathrm{Disc}}_T(W)$ is the coefficient expression $D_d(-a_1,\dots ,(-1{)}^d{a}_d)$ in the coefficients $a_j\in {\mathcal{O}}_{n-1,0}$ of $W$ by [F7], hence is a holomorphic base germ $D_W\in {\mathcal{O}}_{n-1,0}$; since it is nonzero as an element of $K=\mathrm{Frac}({\mathcal{O}}_{n-1,0})$ by step 8.1, it is a nonzero germ.