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## Z[sqrt{-3}] is the Eisenstein integers glued together at two points

Today’s post is a record of a very small observation from my time at PROMYS this summer. Below, by $\text{Spec } R$ I mean a commutative ring $R$ regarded as an object in the opposite category $\text{CRing}^{op}$.

Once you’ve learned how to solve Diophantine equations like $y^2 + 1 = x^3$ using unique factorization in the Gaussian integers $\mathbb{Z}[i]$, it’s natural to want to generalize the method to a Diophantine equation like $y^2 + 3 = x^3$ by using unique factorization in $\mathbb{Z}[\sqrt{-3}]$. However, $\mathbb{Z}[\sqrt{-3}]$ does not have unique factorization, since for example

$\displaystyle 2 \cdot 2 = (1 + \sqrt{-3})(1 - \sqrt{-3})$.

To fix this, we can instead work with the Eisenstein integers $\mathbb{Z}[\omega]$ where $\omega = \frac{1 + \sqrt{-3}}{2}$, which do have unique factorization. More abstractly, we want to work with the ring of integers of $\mathbb{Q}(\sqrt{-3})$ because it is integrally closed and hence a Dedekind domain, so even if unique factorization of elements fails we still have unique factorization of ideals.

$\mathbb{Z}[\sqrt{-3}]$ may not be integrally closed, but it is still a perfectly good Noetherian integral domain of Krull dimension $1$, so putting our algebraic geometry hats on we may still think of $\text{Spec } \mathbb{Z}[\sqrt{-3}]$ as some kind of curve. Its failure to be integrally closed has the geometric interpretation that it is a singular curve. More precisely, since an integral domain $D$ is a Dedekind domain if and only if all localizations $D_P$ at prime ideals are discrete valuation rings, in this context we can interpret a localization that fails to be a discrete valuation ring as a point at which the curve is singular, which in the case of $\mathbb{Z}[\sqrt{-3}]$ is $P = (2, 1 + \sqrt{-3})$.

The very small observation is that this singular point may be thought of heuristically as coming from a gluing procedure, as follows.

First, recall that a coequalizer of a parallel pair of arrows $f, g : a \to b$ in a category is a morphism $\text{coeq}(f, g) : b \to \text{Coeq}(f, g)$ which coequalizes $f, g$ in the sense that $\text{coeq}(f, g) \circ f = \text{coeq}(f, g) \circ g$ and which is universal with respect to this property. In $\text{Top}$, the coequalizer may be thought of as obtained by gluing $b$ to itself along two subspaces which look like $a$. For example, the coequalizer of the two inclusions of an endpoint into $[0, 1]$ is $S^1$.

Proposition: $\text{Spec } \mathbb{Z}[\sqrt{-3}]$ is the coequalizer of the unique pair of distinct morphisms $\text{Spec } \mathbb{F}_4 \to \text{Spec } \mathbb{Z}[\omega]$.

In other words, $\text{Spec } \mathbb{Z}[\sqrt{-3}]$ is obtained from $\text{Spec } \mathbb{Z}[\omega]$ by gluing it together at two points (where “point” here means generalized point and not prime ideal). One heuristically therefore expects the curve to be singular at the glued point.

Proof. Writing $\mathbb{Z}[\omega] = \mathbb{Z}[x]/(x^2 + x + 1)$, any morphism $\mathbb{Z}[x]/(x^2 + x + 1) \to \mathbb{F}_4$ factors through the quotient $\mathbb{F}_2[x]/(x^2 + x + 1)$, which is itself $\mathbb{F}_4$. Hence there are exactly two morphisms $\mathbb{Z}[\omega] \to \mathbb{F}_2[x]/(x^2 + x + 1)$, one of which sends $\omega$ to $x$ and one of which sends $\omega$ to $x + 1$. The equalizer of these two morphisms in $\text{CRing}$ (hence the coequalizer in $\text{CRing}^{op}$) is the subring of $\mathbb{Z}[\omega]$ on which they agree, which is $\mathbb{Z}[2\omega] = \mathbb{Z}[\sqrt{-3}]$ by inspection. $\Box$

An example of the above phenomenon involving an actual algebraic curve can be obtained by considering the nodal cubic $\text{Spec } \mathbb{C}[x, y]/(y^2 - x^3 - x^2)$. The ring of functions on the nodal cubic, which like $\mathbb{Z}[\sqrt{-3}]$ is not integrally closed, embeds into $\mathbb{C}[t]$ as the subalgebra $\mathbb{C}[t^2 - 1, t^3 - t]$ (to see this, let $t = \frac{y}{x}$), which is precisely the subalgebra of polynomials which have the same value at $1$ and $-1$. In other words, it is the equalizer of two morphisms

$e_1, e_{-1} : \mathbb{C}[t] \to \mathbb{C}$

where $e_p$ is the evaluation map at $t = p$. In other other words, the nodal cubic is obtained from the affine line by gluing together two points. And, indeed, a picture of the real points of the nodal cubic confirms that it certainly looks like it was made by gluing a line together at two points, and the gluing point certainly looks pretty singular:

### 3 Responses

1. I know it came up independently at PROMYS, but I just happened to notice that you mentioned this as something you’d blog about almost three years ago, at the end of this post:
https://qchu.wordpress.com/2010/01/04/the-arithmetic-plane/

• Ah, yes, that’s before I learned my lesson about not promising material in future blog posts.

2. Nice! If $\mathbb{Z} [\omega]$ is faithfully flat over $\mathbb{Z} [\sqrt{-3}]$ then this should also be a coequaliser in the full category of schemes.