A hyperplane-supported coherent quotient on a polydisc

On $D$, multiplication by $z_1$ gives a morphism ${\mathcal{O}}_D\to {\mathcal{O}}_D$. Restriction of holomorphic functions to $H_D$ is a morphism ${\mathcal{O}}_D\to i_{\ast }{\mathcal{O}}_{H_D}$ by [F3]–[F4], and it kills the image of multiplication by $z_1$. It therefore induces a morphism $\bar{r}:\mathcal{Q}\to i_{\ast }{\mathcal{O}}_{H_D}$. The quotient is coherent by [F1] and closure under cokernels in [F2].

At a point $a=(0,a^{\mathrm{\prime }})\in H_D$, a germ $h$ holomorphic on $H_D$ extends to $\tilde{h}(z_1,z^{\mathrm{\prime }})=h(z^{\mathrm{\prime }})$, so ${\bar{r}}_a$ is surjective by [F3]. If $f$ is a holomorphic germ at $a$ whose restriction to $z_1=0$ vanishes, write its convergent expansion on a smaller product polydisc as $f(z_1,a^{\mathrm{\prime }}+{\zeta }^{\mathrm{\prime }})={\sum }_{k,{\alpha }^{\mathrm{\prime }}}{c}_{k,{\alpha }^{\mathrm{\prime }}}{z}_1^k({\zeta }^{\mathrm{\prime }}{)}^{{\alpha }^{\mathrm{\prime }}}$. Since the restriction is the zero germ, [F6] implies $c_{0,{\alpha }^{\mathrm{\prime }}}=0$ for every ${\alpha }^{\mathrm{\prime }}$. The series $g(z_1,a^{\mathrm{\prime }}+{\zeta }^{\mathrm{\prime }})={\sum }_{k\ge 1,{\alpha }^{\mathrm{\prime }}}{c}_{k,{\alpha }^{\mathrm{\prime }}}{z}_1^{k-1}({\zeta }^{\mathrm{\prime }}{)}^{{\alpha }^{\mathrm{\prime }}}$ still converges on every smaller product polydisc, because shifting each positive $z_1$ exponent down by one preserves the positive radius of convergence. Thus $g$ is holomorphic and $f=z_{1}g$. The kernel of restriction on this stalk is exactly $(z_1)$, so ${\bar{r}}_a$ is an isomorphism.

On this arbitrary $D$, [F7] gives the exact smooth Dolbeault sheaf resolution of ${\mathcal{O}}_D$. Its terms are modules over smooth real-valued functions; [F8] makes them acyclic and identifies $H^q(D,{\mathcal{O}}_D)$ with the cohomology of their global complex. By [F9] that complex has zero cohomology for $1\le q\le N$, and its terms are zero above degree $N$. Hence $H^q(D,{\mathcal{O}}_D)=0$ for every $q>0$.

If $a\in D\setminus H_D$, then $z_1(a)\mathrm{\ne }0$, so the germ of $z_1$ is a unit and ${\mathcal{Q}}_a=0$. The stalk $(i_{\ast }{\mathcal{O}}_{H_D}{)}_a$ is zero by [F3]. Together with step 1.2 this proves that $\bar{r}$ is an isomorphism on every stalk; [F5] gives the sheaf isomorphism.

Multiplication by $z_1$ is injective on every holomorphic stalk. Away from $H_D$ it multiplies by a unit. At $a=(0,a^{\mathrm{\prime }})\in H_D$, if $z_{1}f=0$ as a germ, expand $f$ on a sufficiently small product polydisc about $a$ using [F6]. Multiplication shifts every $z_1$ exponent up by one; uniqueness of the power-series coefficients therefore forces all coefficients of $f$ to vanish. Thus $f=0$. By [F10] multiplication is a sheaf monomorphism; its cokernel is $\mathcal{Q}$ by definition, giving the stated short exact sequence on the whole $D$.

Apply [F11] to the short exact sequence of step 2.2. For $q>0$, its segment $H^q(D,{\mathcal{O}}_D)\to H^q(D,\mathcal{Q})\to H^{q+1}(D,{\mathcal{O}}_D)$ has both outer groups zero by step 1.3, so $H^q(D,\mathcal{Q})=0$. Its degree-zero segment also makes $\mathcal{O}(D)\to \mathrm{\Gamma }(D,\mathcal{Q})$ surjective with kernel $z_1\mathcal{O}(D)$, because $H^1(D,{\mathcal{O}}_D)=0$. Finally the sheaf isomorphism of step 2.1 and the direct-image formula [F3] give $\mathrm{\Gamma }(D,\mathcal{Q})=\mathrm{\Gamma }(D,i_{\ast }{\mathcal{O}}_{H_D})=\mathcal{O}(H_D)$. Thus both the global quotient computation and all positive-degree vanishings hold for every coordinate polydisc containing $0$, using only scalar acyclicity and this multiplication sequence.