A holomorphic relation kernel on a bidisc

Fix $x=(x_1,x_2)\in D$ with $x\mathrm{\ne }0$. If $z_1(x)\mathrm{\ne }0$, then $z_1$ is a unit germ and a relation $z_{1}a+z_{2}b=0$ satisfies $(a,b)=(b\mathrm{/}{z}_1)(-z_2,z_1)$. If $z_1(x)=0$, then $z_2(x)\mathrm{\ne }0$, so $z_2$ is a unit germ and $(a,b)=(-a\mathrm{/}{z}_2)(-z_2,z_1)$. Thus the stated section generates every relation stalk away from the origin.

At the origin, let $a{z}_1+b{z}_2=0$ as a holomorphic germ. Restricting to $z_1=0$ gives $z_{2}b(0,z_2)=0$ as a one-variable germ, so [F5] makes $b(0,z_2)=0$. Write the convergent series $b(z_1,z_2)={\sum }_{m,n\ge 0}{c}_{mn}{z}_1^m{z}_2^n$. Uniqueness of coefficients makes $c_{0n}=0$ for every $n$, so $b=z_{1}c$ for the holomorphic germ $c={\sum }_{m\ge 1,n\ge 0}{c}_{mn}{z}_1^{m-1}{z}_2^n$.

Substituting $b=z_{1}c$ into the relation gives $z_1(a+z_{2}c)=0$. The germ ring is a domain by [F5], so $a=-z_{2}c$ and $(a,b)=c(-z_2,z_1)$. Hence the same section generates the relation stalk at the origin.

Steps 1.1 and 2.1 show that the generated subsheaf and $\ker \mu$ have equal stalks at every point, where a stalk is the germ module from [F6]; [F4] identifies them. Therefore the relation sheaf is locally of finite type. Under the assumed Axiom of Choice [A1], [F2] makes ${\mathcal{O}}_D$ coherent; ${\mathcal{O}}_D^{\oplus 2}$ is coherent by the split short exact sequence $0\to {\mathcal{O}}_D\to {\mathcal{O}}_D^{\oplus 2}\to {\mathcal{O}}_D\to 0$ and [F3]. Applying [F3] to $\mu$ shows that its kernel is coherent.