MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 13 down vote accepted

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.

share|cite|improve this answer
I'm unfamiliar with the "equalizer" idea. Is there a more pedestrian way to say that $\mathrm{Spec}\,\mathbb{Z}[\sqrt{-7}]$ is obtained from $\mathrm{Spec}\,\mathbb{Z}[(1+\sqrt{-7})/2]$ by identifying two points? Which two ideals are collapsing to which ideal? Thanks. – Drew Armstrong Feb 16 '14 at 4:05
@Drew: roughly speaking, the equalizer of a pair of maps $f, g : X \to Y$ is $\{ x \in X : f(x) = g(x) \}$ (at least this is how it is calculated in many familiar concrete categories). Here the pair of maps are the two homomorphisms $\mathbb{Z}[\alpha] \to \mathbb{F}_2$ (where $\alpha = \frac{1 + \sqrt{-7}}{2}$) whose kernels are the two prime ideals over $(2)$, namely $(2, \alpha)$ and $(2, 1 - \alpha)$, and these become the single ideal $(2, 2\alpha)$ in $\mathbb{Z}[2\alpha]$. (This equalizer becomes a coequalizer in the opposite category of affine schemes. Roughly speaking, the coequalizer – Qiaochu Yuan Feb 16 '14 at 4:36
...of a pair of maps $f, g : X \to Y$ is the quotient of $Y$ by the equivalence relation generated by the relation $f(x) \sim g(x)$ for all $x \in X$. – Qiaochu Yuan Feb 16 '14 at 4:36
@David: I think this is only confusing if you've already internalized the idea that the underlying set of an affine scheme is its set of prime ideals. From a functor-of-points perspective you're still just gluing points together, you're just using a richer notion of point than prime ideals. – Qiaochu Yuan Feb 16 '14 at 4:40
@QiaochuYuan An earlier version of this post contained a sentence like: "This assumes that you are teaching from the perspective that a scheme is a locally ringed space whose underlying set is the set of prime ideals; the functor-of-points perspective makes this issue much less confusing." Then I thought "wait a minute, no one teaches a first course from the functor-of-points perspective!" and deleted it. But, of course, you are right that this is one of the things functor of points is good for. – David Speyer Feb 16 '14 at 11:18

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 $k[t^2 - 1, t^3 - t]$.)

share|cite|improve this answer

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.

share|cite|improve this answer
With a cusp, the preimage of the singular point is non-reduced. With a node, the preimage of the singular point is two points. Here it is neither, but rather an extension of the residue field. As I spell out further in my answer, I think neither one is a good comparison. (Although, if forced, I would say the node is closer.) – David Speyer Feb 16 '14 at 0:50
Yes, I agree with that (and with your parenthetical statement as well). This is the reason that I later inserted the hedge phrase "at least in this sense". :) – René Feb 16 '14 at 0:52

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.