This might seem like a really silly question, but what are those weird curves connecting $(x^2 + 1)$ and $(5, x+2)$ in Mumford's picture of $\text{Spec}\,\mathbb{Z}[x]$?
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It's the curve $V (x^2 + 1) \subset \operatorname{Spec} \mathbb{Z} [x]$. Since $x^2 + 1$ is irreducible in $\mathbb{Z} [x]$, the curve is also irreducible, so the only non-closed point on the curve is the prime ideal $(x^2 + 1)$. The closed points on the curve are the maximal ideals $(p, f)$ where $p$ is a prime and $f$ is an irreducible factor of $x^2 + 1$ in $\mathbb{F}_p [x]$. There are three possibilities:
- $x^2 + 1$ is irreducible in $\mathbb{F}_p [x]$ (i.e. $-1$ is not a quadratic residue mod $p$) in which case we have the maximal ideal $(p, x^2 + 1)$ on the curve.
- $x^2 + 1$ factors as $(x + a)(x - a)$ in $\mathbb{F}_p [x]$ (i.e. $-1$ is a quadratic residue mod $p$), in which case we have the maximal ideals $(p, x + a)$ and $(p, x - a)$ on the curve.
- $x^2 + 1$ factors as $(x + a)^2$ in $\mathbb{F}_p [x]$ (i.e. $p = 2$), in which case we have the maximal ideal $(p, x + a)$ on the curve.
Zhen Lin
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