Analogy between the nodal cubic curve $y^2=x^3+x^2$ and the ring $\mathbb{Z}[\sqrt{-3}]$?

Question

I'm trying to motivate a bit of algebraic geometry in an abstract algebra course (while simultaneously trying to learn a bit of algebraic geometry), and I thought that it might be nice to present an example of the analogy between Riemann surfaces and algebraic integers (since one might say that the entire subject originates from this analogy --- see Dedekind and Weber, 1880). I'm also hoping to motivate the concept of "integral closure" and to make the case that "unique factorization" is spiritually the same as "smooth".

Anyway, I thought that perhaps the best examples from each side of the analogy would be $y^2=x^2(x+1)$ and $\mathbb{Z}[\sqrt{-3}]$. The coordinate ring of $y^2=x^2(x+1)$ is $\mathbb{C}[t^2-1,t^3-t]$, which is not integrally closed in $\mathbb{C}(t)$. The integral closure is $\mathbb{C}[t]$. Geometrically $y^2=x^2(x+1)$ is $\mathbb{C}P^1$ with two points identified and the integral closure separates the two points to obtain $\mathbb{C}P^1$.

Similarly, $\mathbb{Z}[\sqrt{-3}]$ is not a UFD (read: smooth) and since UFD implies integrally closed, one might naively hope that the integral closure of $\mathbb{Z}[\sqrt{-3}]$ is a UFD. Indeed it is. The integral closure of $\mathbb{Z}[\sqrt{-3}]$ is the ring of Eisenstein integers $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]=\mathbb{Z}[\omega]$, where $\omega=e^{2\pi i/6}$. Then one can use the very nice geometry of $\mathbb{Z}[\omega]$ to prove that it is Euclidean, hence UFD. One can view the inclusion map $\mathbb{Z}[\sqrt{-3}]\hookrightarrow \mathbb{Z}[\omega]$ as a projection of spectra $\mathrm{Spec}\,\mathbb{Z}[\omega]\twoheadrightarrow \mathrm{Spec}\,\mathbb{Z}[\sqrt{-3}]$. I'm hoping that this projection "looks like" identifying two points of $\mathbb{C}P^1$.

(I realize that in general the integral closure of $\mathbb{Z}[\alpha]$ is not UFD, but only has unique factorization of ideals. Pedagogically, though, that seems like too many complications in one example.)

Can anyone help me flesh out the analogy? Since it's an abstract algebra class (not an algebraic geometry class) I would like to keep things in naive terms. Maybe you can think of a more appropriate pair of examples? Thanks.

Answer 1

I think this will be a needlessly confusing example. In algebraic geometry over an algebraically closed field, there are two basic examples of nonnormal curves: the node and the cusp. Explicit equations are $y^2=x^2+x^3$ and $y^2 = x^3$. respectively.

If $X$ has a node, and $\tilde{X}$ is its normalization, then $X$ is obtained by identifying two different points of $\tilde{X}$. An analogous example in number theory is $\mathbb{Z}[\sqrt{-7}]$, with normalization $\mathbb{Z}[(1+\sqrt{-7})/2]$. There are two maps $\mathrm{Spec} \ \mathbb{F}_2 \to \mathrm{Spec} \ \mathbb{Z}[(1+\sqrt{-7})/2]$ and $\mathrm{Spec} \ \mathbb{Z}[\sqrt{-7}]$ is the equalizer of these maps. This is what I would compare to $y^2 = x^2 + x^3$.

If $X$ has a cusp, and $\tilde{X}$ is its normalization, then $\tilde{X} \to X$ is bijective on points, but elements in the coordinate ring of $\tilde{X}$, if they vanish at the cusp, must vanish to order $>1$. An analogous number theory example is $\mathbb{Z}[\sqrt{8}]$, with normalization $\mathbb{Z}[\sqrt{2}]$. If you squint, the ring $\mathbb{Z}[y]/(y^2 - 2^3)$ even looks like $k[x,y]/(y^2-x^3)$.

The example of $\mathbb{Z}[\sqrt{-3}]$ exhibits a phenomenon which you simply don't see over an algebraically closed field: The normalization map is bijective on the set of closed points, but there is a residue field extension. In other words, the two maps from $\mathrm{Spec} \ \mathbb{F}_4$ which Qiaochu mentions have the same image, but differ by the automorphism of $\mathbb{F}_4$. This phenomenon does happen with curves over a non-algebraically closed field: Consider $x^2+y^2=x^3$ over $\mathbb{R}$. But you probably don't want to discuss that in an intro course. I would just use one or both of the examples in the previous paragraphs, and only talk about $\mathbb{Z}[\omega]$ if the students made me.

Answer 2

I wrote down the details in this blog post. Briefly, there are exactly two morphisms $\text{Spec } \mathbb{F}_4 \to \text{Spec } \mathbb{Z}[\omega]$, and their coequalizer (gluing them together) is $\text{Spec } \mathbb{Z}[\sqrt{-3}]$, at least in the category of affine schemes.

The geometric example is probably easier to start with: gluing together two points $a, b$ on the affine line corresponds to looking at the subalgebra

$$\{ f \in k[t] : f(a) = f(b) \}$$

of a polynomial algebra $k[x]$. (For $a = -1, b = 1$ this algebra is precisely $k[t^2 - 1, t^3 - t]$.)

Answer 3

As a "public service," so-to-speak, here is a plot of $y^2 = x^3 + x^2$:
         

Answer 4

By a result that I know by the name of the "Kummer-Dedekind theorem" (see Chapter 3 of these notes), there is a unique prime ideal $\mathfrak{p}=(2,1+\sqrt{3})$ of $R = \mathbb{Z}[\sqrt{-3}]$ lying over $2$. By the same result, the rational prime $2$ is inert in $\widetilde{R} = \mathbb{Z}[\frac{1+\sqrt{-3}}{2}]$, i.e. $\mathfrak{q}=(2)$ is a prime ideal in $\widetilde{R}$. So for the corresponding map of prime spectra $\operatorname{Spec}(\widetilde{R})\rightarrow \operatorname{Spec}(R)$, the preimage of the "singular" point $\mathfrak{p}$ of $\operatorname{Spec}(R)$ consists solely of the "regular" point $\mathfrak{q}$ of $\operatorname{Spec}(\widetilde{R})$. I would therefore say that, at least in this sense, the situation is more like that of the cuspidal point $(0,0)$ on the singular curve $y^2=x^3$, which also has one point lying over it on the normalization.