Weierstrass division theorem

By [L1] and [L2], after shrinking representatives of $W$ and $f$ if needed there are $r>0$ and a neighbourhood $V$ of $0$ such that $W(z^{\mathrm{\prime }},\zeta )\mathrm{\ne }0$ whenever $z^{\mathrm{\prime }}\in V$ and $\mathrm{\mid }\zeta \mathrm{\mid }=r$. Define $q(z^{\mathrm{\prime }},z_m):=\frac{1}{2\pi i}{\int }_{\mathrm{\mid }\zeta \mathrm{\mid }=r}\frac{f(z^{\mathrm{\prime }},\zeta )}{W(z^{\mathrm{\prime }},\zeta )(\zeta -z_m)}d\zeta ,$ $r(z^{\mathrm{\prime }},z_m):=\frac{1}{2\pi i}{\int }_{\mathrm{\mid }\zeta \mathrm{\mid }=r}f(z^{\mathrm{\prime }},\zeta )\frac{W(z^{\mathrm{\prime }},\zeta )-W(z^{\mathrm{\prime }},z_m)}{W(z^{\mathrm{\prime }},\zeta )(\zeta -z_m)}d\zeta \mathrm{.}$ As in the preparation proof, [L3] makes $q$ and $r$ holomorphic on $V\times \{\mathrm{\mid }{z}_m\mathrm{\mid }<r\}$.

For fixed $z^{\mathrm{\prime }}$ and $\zeta$, the quotient $\frac{W(z^{\mathrm{\prime }},\zeta )-W(z^{\mathrm{\prime }},z_m)}{\zeta -z_m}$ is a polynomial in $z_m$ of degree at most $d-1$: expand the monic polynomial $W$ in powers of $z_m$ and factor each difference ${\zeta }^j-z_m^j=(\zeta -z_m)({\zeta }^{j-1}+\cdots +z_m^{j-1})$. Therefore $r(z^{\mathrm{\prime }},z_m)$ is itself a polynomial in $z_m$ of degree $<d$ with coefficients holomorphic in $z^{\mathrm{\prime }}$.

Adding the two integral formulas from step 1.1 gives $q(z^{\mathrm{\prime }},z_m)W(z^{\mathrm{\prime }},z_m)+r(z^{\mathrm{\prime }},z_m)=\frac{1}{2\pi i}{\int }_{\mathrm{\mid }\zeta \mathrm{\mid }=r}\frac{f(z^{\mathrm{\prime }},\zeta )}{\zeta -z_m}d\zeta \mathrm{.}$ By [L4], the right-hand side is exactly $f(z^{\mathrm{\prime }},z_m)$ for $\mathrm{\mid }{z}_m\mathrm{\mid }<r$. Thus $f=qW+r$ with ${\mathrm{deg}}_{z_m}r<d$.

Suppose also $f=q^{\mathrm{\prime }}W+r^{\mathrm{\prime }}$ with ${\mathrm{deg}}_{z_m}{r}^{\mathrm{\prime }}<d$. Then $r-r^{\mathrm{\prime }}=(q^{\mathrm{\prime }}-q)W\mathrm{.}$ For each fixed $z^{\mathrm{\prime }}$, the left-hand side is a one-variable polynomial of degree $<d$ divisible by the monic degree-$d$ polynomial $W(z^{\mathrm{\prime }},\cdot )$. Hence $r(z^{\mathrm{\prime }},\cdot )-r^{\mathrm{\prime }}(z^{\mathrm{\prime }},\cdot )=0$ as a polynomial, so $r=r^{\mathrm{\prime }}$. Then $(q-q^{\mathrm{\prime }})W=0$, and on the nonempty open set where $W\mathrm{\ne }0$ one has $q=q^{\mathrm{\prime }}$; [L5] forces $q=q^{\mathrm{\prime }}$ everywhere. Therefore both quotient and remainder are unique.