Why does the variety of ideals in this quaternion type algebra have a non-reduced structure?

Question

Let $A$ be the $\mathbb{C}$-algebra generated by elements $i,j$ with relations $i^2=j^2=0$ and $ij=-ji$, i.e. we have $A=\mathbb{C}\oplus\mathbb{C}i\oplus\mathbb{C}j\oplus\mathbb{C}ij$.

Let $\mathcal{B}\mathcal{S}(A): Sch_{\mathbb{C}}^{op}\rightarrow Sets$ be the functor that sends the $\mathbb{C}$-scheme $X$ to the set of left ideals $I\subset A\otimes\mathcal{O}_X$ such that $(A\otimes\mathcal{O}_X)/I$ is a locally free sheaf of rank 2 on $X$. Then this functor is representable by a $\mathbb{C}$-scheme $BS(A)$.

Why is $BS(A)$ given by the double line $Proj(\mathbb{C}[u,v,w]/(w^2))$?

I see that we have a $\mathbb{P}^1$ lurking around: if $I$ is an ideal in $A$ such that $A/I$ is of dimension 2, then $I$ cannot contain elements from the copy $\mathbb{C}$ because $i$ and $j$ are nilpotent. Also because of the relations, every ideal needs to contain $ij$ and hence the whole copy $\mathbb{C}ij$. So it remains to find a one dimensional subspace in the two dimensional space $\mathbb{C}i\oplus\mathbb{C}j$, whch is parametrized by a $\mathbb{P}^1$.

But I don't see where the non-reduced structure comes from. Does any one see this? I found the claim, that $BS(A)$ is the double line in Lemma 7.11 in the article [1], where it just says "we know that BS(A) is the double line", without proof. Or is this obvious?

[1] https://arxiv.org/pdf/1101.1705.pdf

Answer 1

It is not immediately obvious (at least to me) but it is an easy calculation. The following is too long for a comment, so I post it here:

Here is a simpler example of the same phenomenon: Let $A=\mathbb C[t]/(t^2)$ be the ring of dual numbers. Consider the moduli scheme parametrizing dimension one ideals in $A$ (defined in the same way as in your question). On the level of points, there is only one such ideal: $(t)$. However, the scheme has a non-reduced structure that you can easily see in coordinates: all one-dimensional subspaces in $A$ are of the form $\mathbb C(x_0+x_1t)\subset A$, and so they are parametrized by $(x_0:x_1)\in\mathbb P^1$. Such a subspace is an ideal exactly when $$tx_0=t(x_0+x_1t)\in\mathbb C(x_0+x_1t),$$ which amounts to the equation $$0=\det\begin{pmatrix}x_0&0\\x_1&x_0\end{pmatrix},$$ that is, $x_0^2=0$. You now clearly see that the scheme is actually a double point, essentially because $(\epsilon+t)$ is an ideal if $\epsilon^2=0$.

P.S. Of course, in this example, there is a geometric way of explaining why we get a double point: we are studying ideals of dimension one in $A$, which are basically points of $\mathrm{Spec}(A)$, so the moduli scheme is in bijection with $\mathrm{Spec}(A)$ tautologically. However, the coordinate method generalizes to other rings, like the one you are interested in.