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Drew Armstrong
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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}\hookrightarrow \mathbb{Z}[\omega]$$\mathbb{Z}[\sqrt{-3}]\hookrightarrow \mathbb{Z}[\omega]$ as a projection of spectra $\mathrm{Spec}\mathbb{Z}[\omega]\twoheadrightarrow \mathrm{Spec}\mathbb{Z}[\sqrt{-3}]$$\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.

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 $\mathbb{Z}\hookrightarrow \mathbb{Z}[\omega]$ 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.

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.

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Drew Armstrong
  • 3.8k
  • 19
  • 41

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

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 $\mathbb{Z}\hookrightarrow \mathbb{Z}[\omega]$ 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.