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 $\mathbb{Z}[(1+\sqrt{-7})/2]$. Writing $\alpha = (1+\sqrt{-7})/2$, there are two maps $\mathrm{Spec} \ \mathbb{F}_2 \to \mathrm{Spec} \ \mathbb{Z}[\alpha]$ 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.

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.

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 $\mathbb{Z}[(1+\sqrt{-7})/2]$. Writing $\alpha = (1+\sqrt{-7})/2$, there are two maps $\mathrm{Spec} \ \mathbb{F}_2 \to \mathrm{Spec} \ \mathbb{Z}[\alpha]$ 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 paragraph, and only talk about $\mathbb{Z}[\omega]$ if the students made me.

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 $\mathbb{Z}[(1+\sqrt{-7})/2]$. Writing $\alpha = (1+\sqrt{-7})/2$, there are two maps $\mathrm{Spec} \ \mathbb{F}_2 \to \mathrm{Spec} \ \mathbb{Z}[\alpha]$ 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.