Here is what seems to be a fun little exercise in algebraic geometry. Take a DVR $R$ and automorphism of $Frac(R)$; glue $Spec(R)$ to itself via this automorphism. Can the glued scheme be separated non-affine?

For separatedness, the automorphism must fail to preserve $R$ (https://mathoverflow.net/a/335962). For some DVRs, no such automorphism exists, e.g. $k[[x]]$ (https://mathoverflow.net/a/336022) and $\mathbb{Z}_p$ (https://math.stackexchange.com/a/449465).

  • $\begingroup$ I'm voting to close this question as off-topic because the OP is trolling. $\endgroup$ – Steven Landsburg Jul 15 '19 at 4:52
  • 4
    $\begingroup$ @StevenLandsburg The Wikipedia page on Internet trolls says that they post "inflammatory and digressive, extraneous, or off-topic messages." Does this question satisfy any of these descriptions? IMHO this is a reasonable on-topic question. $\endgroup$ – user143116 Jul 15 '19 at 11:37

Answer: no, if the glued scheme is separated, it is affine.

We have a DVR $R$ and an automorphism $\phi:Frac(R)\rightarrow Frac(R)$ such that $\phi(R)$ is not a subset of $R\subset Frac(R)$. We glue $\mathrm{Spec}\:R$ to itself along the generic point by $\phi$. Denote the resulting separated scheme by $X$.

First, let us compute $\Gamma(X, \mathcal{O}_X)$. If we take an open cover, giving a global section of $\mathcal{O}_X$ is the same as giving a section of $\mathcal{O}_X$ on each member of the cover so that the sections agree on the overlaps. So $\Gamma(X, \mathcal{O}_X)$ is identified with the ring $R\cap \phi(R)\subset Frac(R)$.

Note that $R$ is not a subset of $\phi(R)$ because if it were then $\phi^{-1}$ would preserve the order induced by the valuation; then $\phi$ would preserve the order but that is impossible because $\phi(R)$ is not a subset of $R$ by assumption. Thus Theorem 12.2 in "Commutative ring theory" (Matsumura) gives us that $R\cap \phi(R)$ is a PID that has exactly two distinct maximal ideals: $\mathfrak{m}\cap \phi(R)$ and $R\cap \phi(\mathfrak{m})$.

Consider the canonical map $X\rightarrow \mathrm{Spec}\:\Gamma(X, \mathcal{O}_X)$. A point $x\in X$ is mapped to (the point corresponding to) the kernel of the canonical homomorphism $\Gamma(X, \mathcal{O}_X)\rightarrow k(x)$. So the generic point is mapped to $\{0\}$, and two closed points are mapped $\mathfrak{m}\cap \phi(R)$ and $R\cap \phi(\mathfrak{m})$ respectively. As remarked above, the ideals are distinct so the map $X\rightarrow \mathrm{Spec}\:\Gamma(X, \mathcal{O}_X)$ is injective. Because $X$ and $\mathrm{Spec}\:\Gamma(X, \mathcal{O}_X)$ both have 3 points, it is bijective and one directly verifies that it is a homeomorphism so $X$ is affine by Stacks 04DE. Admittedly, one could show that the map $X\rightarrow \mathrm{Spec}\:\Gamma(X, \mathcal{O}_X)$ is affine more directly by looking at the inverse images of the affine opens but I like this mode of argumentation more.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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