# When does glueing affine schemes produce affine/separated schemes?

Let $$X$$ be an affine scheme with an open affine subscheme $$U\subset X$$. Given an automorphism of $$U$$, we can glue $$X$$ with itself along $$U$$ to get a new scheme. Is there a description in terms of commutative algebra of automorphisms such that the resulting scheme is affine/separated?

If $$U=\mathrm{Spec}\:B$$, $$X=\mathrm{Spec}\:A$$, then $$B$$ is an $$A$$-algebra of finite presentation so there's a chance to be explicit.

For example, if $$X=\mathrm{Spec}\:k[x]$$ and $$U=\mathrm{Spec}\:k[x, \frac{1}{x}]$$ if we take the identity on $$U$$ the result is non-separated and if we take $$x\rightarrow \frac{1}{x}$$ the result is separated.

I'm especially interested in what happens for $$X$$ the spectrum of a discrete valuation ring or a PID.

Here is a commutative algebra description of open immersions between affine schemes (not a very convenient one I'd guess).

Here are some thoughts in the case of gluing a DVR along an automorphism of its fraction field:

Setup: Let $$A$$ be a DVR with uniformizer $$\pi$$ and fraction field $$K$$, and let $$\varphi : K \to K$$ be a ring automorphism. Let $$S$$ be the gluing of two copies of $$\operatorname{Spec} A$$ along the automorphism of $$\operatorname{Spec} K$$ corresponding to $$\varphi$$.

On separatedness: I claim that $$S$$ is separated if and only if $$\varphi(A) \not\subseteq A$$. By Tag 01KP we have that $$S$$ is separated if and only if the ring map \begin{align} \mu_{\varphi} : A \otimes_{\mathbb{Z}} A \to K \end{align} sending $$a_{1} \otimes a_{2} \mapsto a_{1} \cdot \varphi(a_{2})$$ is surjective. If $$\varphi(A) \not\subseteq A$$, then $$\mu_{\varphi}$$ is surjective since the image of $$\mu_{\varphi}$$ is a subring of $$K$$ strictly larger than $$A$$. On the other hand, if $$\varphi(A) \subseteq A$$ then the image of $$\mu_{\varphi}$$ is $$A$$.

On affineness: EDIT: user "m.mor" proves in this answer that if $$S$$ is separated then it is affine. I don't have a complete answer. We may consider using Serre's criterion (e.g. Tag 01XF). Let $$\mathcal{I} \subseteq \mathcal{O}_{S}$$ be an ideal sheaf of $$S$$. By the Mayer-Vietoris sequence, we have that $$\mathrm{H}^{1}(S,\mathcal{I}) = 0$$ if and only if for all $$e_{1},e_{2} \in \mathbb{Z}_{\ge 0}$$ the addition map \begin{align} \alpha_{e_{1},e_{2},\varphi} : \pi^{e_{1}}A \oplus \pi^{e_{2}}A \to K \end{align} sending $$(a_{1},a_{2}) \mapsto a_{1} + \varphi(a_{2})$$ is surjective. Thus certainly if $$\varphi(A) \subseteq A$$ then $$S$$ is not affine. Here I tried some examples (e.g. $$A = k[x]_{(x)}$$ and the automorphism of $$k(x)$$ sends $$x \mapsto \frac{1}{x}$$, or $$A = k[x,y]_{(x)}$$ and the automorphism of $$k(x,y)$$ switches $$x,y$$) and the resulting glued schemes were affine (these examples generalize to Lemma 1 below). Another strategy would be to view everything with the $$\pi$$-adic topology and try to use Lemma 2 below, but I don't know whether $$\varphi(A)$$ is a dense subgroup of $$K$$.

Lemma 1: Let $$A$$ be a Dedekind domain with fraction field $$K$$. Suppose that $$A$$ has exactly two maximal ideals $$\mathfrak{p}_{1},\mathfrak{p}_{2}$$. Then for any $$s_{1},s_{2} \in \mathbb{Z}_{\ge 0}$$ the addition map \begin{align} \mathfrak{p}_{1}^{s_{1}}A_{\mathfrak{p}_{1}} \oplus \mathfrak{p}_{2}^{s_{2}}A_{\mathfrak{p}_{2}} \to K \end{align} sending $$(a_{1},a_{2}) \mapsto a_{1}+a_{2}$$ is surjective.

Proof: Since $$A$$ has finitely many maximal ideals, it is a PID. Let $$\pi_{i}$$ be a generator of $$\mathfrak{p}_{i}$$; then $$K := A[(\pi_{1}\pi_{2})^{-1}]$$ as $$A$$-algebras. Moreover $$\pi_{1}^{e_{1}}-\pi_{2}^{e_{2}}$$ is a unit for all $$e_{1},e_{2} \ge 1$$. As an $$A$$-module, $$K$$ is generated by the elements $$1/\pi_{1}^{e_{1}}\pi_{2}^{e_{2}}$$ for all $$e_{1},e_{2} \ge 0$$. We may assume that $$e_{1},e_{2} \ge 1$$. Then \begin{align} \textstyle \frac{1}{\pi_{1}^{e_{1}}\pi_{2}^{e_{2}}} = \pi_{1}^{s_{1}} \cdot (\frac{1}{\pi_{1}^{s_{1}+e_{1}}-\pi_{2}^{s_{2}+e_{2}}}) \frac{1}{\pi_{2}^{e_{2}}} + \pi_{2}^{s_{2}} \cdot (-\frac{1}{\pi_{1}^{s_{1}+e_{1}}-\pi_{2}^{s_{2}+e_{2}}}) \frac{1}{\pi_{1}^{e_{1}}} \end{align} so we have the desired result.

Lemma 2: Let $$G$$ be a topological group. Let $$U$$ be an open subgroup of $$G$$ and let $$H$$ be a dense subgroup of $$G$$. Then $$G = UH$$.

• what is $k(x)$ in your answer? I believe formal Laurent series are infinite only in one direction so there is no automorphism $x\rightarrow \frac{1}{x}$? – user142965 Jul 12 '19 at 5:51
• For me $k(x) := \operatorname{Frac}(k[x])$, namely the field of rational functions in the variable $x$ over $k$. – Minseon Shin Jul 12 '19 at 6:30
• Hi @rbarc, I don't think that's a correct description of the ring of global functions; for example it's not closed under addition (e.g. consider $\frac{x+1}{1}-\frac{1}{1}$) – Minseon Shin Aug 1 '19 at 6:24
Proposition: Let $$X_1, X_2$$ be a separated $$S$$-schemes, $$U_i$$ open subschemes in $$X_i$$ (for $$i=1, 2$$), and $$f:U_1 \to U_2$$ an $$S$$-isomorphism. Then the $$S$$-scheme $$X$$ obtained as a gluing of $$X_1$$ and $$X_2$$ along the isomorphism $$f$$ is separated if and only if the diagonal'' morphism $$U_1 \to X_1\times_S X_2$$ is a closed immersion.
In the situation $$U_1=\operatorname{Spec} A$$ and $$X_1=X_2=\operatorname{Spec} B$$ are affine the criterion says that separtedness of $$X$$ is equivalent to surjectivity of the map $$\phi:B\otimes_{\mathbf Z} B \to A$$ defined by $$\phi(a\otimes a')=af^*(b)$$.