How "many" non-reduced spaces with smooth reduction are there?

Suppose $(M,\mathcal{O}_M)$ is an analytic subspace of $\mathbb{C}^n$, such that its reduction is simply some $\mathbb{C}^k\subseteq \mathbb{C}^n$. How "many" such structure sheaves are there?

It is clear that there are at least countably many, by considering the ideals generated by monomials $z_{k+1}^{p_1},...,z_{n}^{p_{n-k}},(z_{k+1}\cdot z_{k_2})^{p_{12}},...,(z_{k+1}\cdot {\dots} \cdot z_n)^{p_{1...n-k}}$, with $p_1\neq 0,...,p_{n-k}\neq 0$. Let me call these ideals $I_{p_1,...,p_{1\dots n-k}}$. Moreover, for any defining ideal $J$ of such an $M$ it follows that there exist numbers $q_1,...,q_{1\dots n-k}$ such that

$I_{q_1,...,q_{1\dots n-k}}\subseteq J$.

This is a consequence of the Nullstellensatz. It follows that elements of $J/I_{q_1,...,q_{1\dots n-k}}$ are only polynomials in $z_{k+1},...,z_n$ with coefficients in $\mathcal{O}_{\mathbb{C}^k}$. Showing that the ideal $J$ is generated by polynomials $P_i$ in $z_{k+1},...,z_n$ with coefficients in $\mathcal{O}_{\mathbb{C}^k}$.

May one further deduce that $J$ is generated by monomials in $z_{k+1},...,z_n$ with coefficients in $\mathcal{O}_{\mathbb{C}^k}$? How "bad" can the ideal $J$ still be?

Solution 1:

There are many more than what you've listed. Specifically, for any closed analytic subspace $X\subset M$, we can find a complex analytic structure on $M$ which is non-reduced along $X$ by taking the structure sheaf $\mathcal{O}_{\Bbb C^n}/(\mathcal{I}_M\cdot\mathcal{I}_X)$. This will give you at least uncountably many. On the other hand, I think the amount of non-reduced structures is at most uncountable: every complex-analytic space is locally isomorphic to a model space of the form $V(I)\subset U\subset \Bbb C^n$ where $I$ is a finitely-generated ideal of the ring of holomorphic functions on the open subset $U$, so since $\Bbb C^n$ is second-countable, we can specify $M$ by choosing finitely many things from an uncountable set countably many times. (Apologies to anyone who's a serious set-theorist if I've butchered anything in the previous sentence.)

Here is an explicit example: let $n=2$, $m=1$, and take some finite list of values $z_1,\cdots,z_l\in\Bbb C$. Then the ideal $$(\prod (x_1-z_i),x_2))\cdot(x_2)=(x_2\prod(x_1-z_i),x_2^2)$$ gives a copy of $\Bbb C$ with nilpotence at each of the $z_i$. This also shows that your "may one further..." question is false.