# Classify all schemes of degree $2$ and $3$ over $\mathbb{R}$ supported at the origin in $\mathbb{A}_\mathbb{R}^2$

Classify all schemes of degree $2$ and $3$ over $\mathbb{R}$ supported at the origin in $\mathbb{A}_\mathbb{R}^2$. In particular, show that while any such scheme $X$ whose complexification $X \times_{\text{Spec}\,\mathbb{R}} \text{Spec}\,\mathbb{C}$ is isomorphic to $\text{Spec}\,\mathbb{C}[x]/(x^3)$ is itself isomorphic to $\text{Spec}\,\mathbb{R}[x]/(x^3)$, there are exactly two nonisomorphic schemes $X$ whose complexification is isomorphic to $\text{Spec}\,\mathbb{C}[x, y]/(x^2, xy, y^2)$.

• The way you ask the question by giving an order ("Classify"!) is rude and seems to indicate that you just copied some exam question, hoping the community here will do your homework. That you accept an "answer" which doesn't solve the problem seems to show that you don't even understand that question. Jul 26, 2016 at 7:04

As in here or here, we only need to classify $\mathbb{R}$-local algebras $A$ of lengths $2$ and $3$. Let $m$ be the maximal ideal. Then we consider the filtration associated to $m$. If $m/m^2$ is one dimensional, the same reasoning as in what I linked to above implies that$$A \simeq \mathbb{R}[x]/\langle x^l\rangle.$$Otherwise, the length must be three. The trick to solve this problem is to realize that a scheme $X$ over $\mathbb{R}$ is the same as a scheme $Y$ over $\mathbb{C}$, together with an involution $\sigma$, which lifts complex conjugation,$$\require{AMScd} \begin{CD} Y @>\sigma>> Y \\ @VVV @VVV \\ \text{Spec}\,\mathbb{C} @>>> \text{Spec}\,\mathbb{C}. \end{CD}$$ So suppose we start with$$Y = \text{Spec}\,\mathbb{C}[x, y]/\langle x^2, xy, y^2\rangle.$$The diagram above, reflecting in a vertical mirror and using the equivalence of categories, becomes$$\require{AMScd} \begin{CD} \mathbb{C}[x, y]/\langle x^2, xy, y^2\rangle @>\sigma>> \mathbb{C}[x, y]/\langle x^2, xy, y^2\rangle \\ @AAA @AAA \\ \mathbb{C} @>>> \mathbb{C}. \end{CD}$$ The only ring homomorphism, up to linear change of coordinates, which is not the standard lift of complex conjugation, is given by$$f(x, y) \to g(y, x),$$where $g$ is the polynomial obtained by taking the complex conjugate of $f$. Note that this does indeed fix the maximal ideal squared. The invariants, modulo $m^2$, are then $x + y$ and $i(x - y)$.

• What is the non obvious $\mathbb R$- algebra of dimension $3$ that the OP is asking about? (And what is that strange $0$ in your first diagram?) Jul 25, 2016 at 20:30

This is an excellent question -- although it appears to have been answered in various forms elsewhere on this site. I would however like to remark that it speaks to the following interesting open problem: if $k$ is a field, how do we describe the length-$m$ closed subschemes of $\mathbb{A}_k^2$ supported at the origin? By describe,'' I mean not only give an algebraic characterization (which @ErinHagood seems to give), but also a geometric picture. More precisely, we would like to describe the Hilbert scheme of length-$m$ closed subschemes of $\mathbb{A}_k^2$ supported at the origin.

For simplicity, let us work over an algebraically closed field $k$. To begin with, notice that a closed subscheme $X \subset \mathbb{A}_k^2 = \text{Spec}\, k[x,y]$ is supported at the origin if the ideal cutting out $X$ has radical equal to $(x,y)$.

Case (1): $m = 1$. In this case, we want to find all ideals $I \subset k[x,y]$ such that $k[x,y]/I$ is $1$-dimensional over $k$. The only way for this to happen is to have $I = (x,y)$, so there is only one length-$1$ closed subscheme of $\mathbb{A}_k^2$ supported at the origin; i.e., the Hilbert scheme is a point in this case.

Case (2): $m = 2$. In this case, we want to find all ideals $I \subset k[x,y]$ such that $k[x,y]/I$ is $2$-dimensional over $k$. Intuitively, we expect that the locus of such ideals forms a $\mathbb{P}_k^1$, because there is a $\mathbb{P}_k^1$ of possible tangent directions at the origin. To see why this intuition is correct, let $I$ be such an ideal. Then $k[x,y]/I \simeq k[\varepsilon]/(f(\varepsilon))$, where $f(\varepsilon) \in k[\varepsilon]$ is a quadratic polynomial. Unless $f$ is a square, you can check that $I$ will not be supported at a single point, so in fact we find that $k[x,y]/I \simeq k[\varepsilon]/(\varepsilon^2)$. The preimage of $\varepsilon$ under this isomorphism is a linear form $ax + by$, and you can check that this linear form determines what the ideal $I$ must be. There are a $\mathbb{P}_k^1$ of linear forms, so in this case the Hilbert scheme is $\mathbb{P}_k^1$.

An alternative way of dealing with this case is by viewing length-$2$ schemes supported at the origin as the limit of two points approaching the origin. In fact, you can take one point to already be fixed at the origin, and have another point coming in to the origin along a line. There are precisely a $\mathbb{P}_k^1$'s worth of ways for the second point to approach the origin.

Case (3): $m = 3$. This case is decidedly harder than either of the previous two cases, and the problem of identifying the Hilbert scheme is not readily amenable to an algebraic approach. But the idea of using limiting points provides a nice picture. If we fix one point at the origin and then have two other points limiting toward the origin, you end up with what appears, on first glance, to be a $\mathbb{P}_k^1 \times \mathbb{P}_k^1$'s worth of possibilities. However, we need to be careful when the two points coincide; what you actually obtain looks geometrically like $\mathbb{P}_k^1 \times \mathbb{A}_k^1$, with the loose ends'' of all the $\mathbb{A}_k^1$ fibers compactified in a point (corresponding to when the two limiting points coincide). As it happens, the Hilbert scheme in this case is the cone over a curve.

• Dear all reading the above solution: I am very much interested in this problem, so if anyone would like to discuss it in greater length with me, please shoot me an email! Jul 25, 2016 at 21:21
• You write : "This is an excellent question". Indeed, but you don't answer it since the OP asks about $\mathbb R$-schemes and you assume $k$ algebraically closed. Jul 26, 2016 at 7:07