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Let $A$ be a commutative ring with unity. Consider its associated affine scheme $(\operatorname{Spec}(A),\mathcal{O}_A)$. I was wondering if the restriction morphism, $$A \xrightarrow{r|^X_U} \mathcal{O}_A(U)$$ would induce an open immersion of $\operatorname{Spec}(\mathcal{O}_A(U))$ into $\operatorname{Spec}(A)$. I know this for the case of distinguished opens. What happens in the general case?

Abellan
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  • It is true if $A$ is a UFD. My first source to look for a counterexample would be a Dedekind domain with a non-prinicipal ideal. I am not sure though. – MooS Nov 29 '16 at 16:24
  • I am sorry my previous example was wrong .. but you can instead use the examples given in this post http://math.stackexchange.com/questions/143954/isolated-non-normal-surface-singularity – skeptic Nov 29 '16 at 20:00
  • Just a comment--not fleshed out. Let's assume that $A$ is a domain. Then if $U=X-Z$ for $\text{codim}(Z)\geqslant 2$ then 'algebraic Hartog's lemma' implies that $r\mid_U^X$ is an isomorphism, and so no question there. So, the case when $A$ is normal one only has to worry about complements of codimension $1$ primes. One might then, assuming that $A$ is locally factorial (e.g. if it's regular) reduce to the local case since every codimension $1$ prime (i.e. every divisor) is locally principal--i.e. reduce it to the distinguished affine case. – Alex Youcis Nov 30 '16 at 02:03
  • @AlexYoucis These were precisely the thoughts I had in mind, when I said that its true for UFDs. If the closed complement has codimension $2$, we are done by Hartog's lemma and in the codimension $1$ case, the ideal will be principal, hence it is a distinguished affine open.But how do you reduce to the local case, if the codimension $1$ ideals are only locally principal, i.e. for Dedekind domains with non-trivial ideal class group? Are you saying that it is always an open immersion if $A$ is regular? – MooS Nov 30 '16 at 06:35

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Let me work out the counter example proposed in the comments.

Let $X = \operatorname{Spec}A$ be a non-normal isolated surface singularity, as constructed here: isolated non-normal surface singularity. Let $p$ be the singular point and $U = X \setminus \{p\}$ be the open complement. It turns out, that this will always be a counter example, and luckily we dont have to do any concrete computations.

Let $A \to \bar A$ be the integral closure (which is a finite extension, since $A$ is of finite type over a field), and $Y = \operatorname{Spec} \bar A \xrightarrow f X$ be the induced finite morphism of affine schemes. This induces an isomorphism over the normal locus, i.e. $Y \setminus f^{-1}(p) \xrightarrow{\sim} U$ .

Since the map is finite, we have that $f^{-1}(p)$ has codimension two in the normal scheme $Y$, hence by Hartog's lemma:

$$\mathcal O_X(U) = \mathcal O_U(U) = \mathcal O_{Y \setminus f^{-1}(p)}(Y \setminus f^{-1}(p)) = \mathcal O_Y(Y \setminus f^{-1}(p)) = \mathcal O_Y(Y) = \overline A$$

Hence you are asking whether the map $$A \to \mathcal O_X(U) = \overline A$$ induces an open immersion. Of course we expect the map to be the inclusion into the integral closure, thus the induced map is the map $f$. To show this rigorously, note that the sheaf map $\mathcal O_X \to f_* \mathcal O_Y$ gives rise to a commutative diagram

$$\require{AMScd} \begin{CD} A @>>> \overline A\\ @VVV @VV\operatorname{id}V \\ \mathcal O_X(U) @>\cong>> \mathcal O_Y(Y \setminus f^{-1}(p))=\overline A \end{CD}$$

By functoriality of $\operatorname{Spec}$, we get a commutative diagram

$$\require{AMScd} \begin{CD} Y @>\cong>> \operatorname{Spec} \mathcal O_X(U)\\ @V\operatorname{id}VV @VVV \\ Y @>f>> X \end{CD}$$

Thus the map $\operatorname{Spec} \mathcal O_X(U) \to X$ is an open immersion if and only if $f$ is one. But $f$ is surjective and not an isomorphism, hence it can not be an open immersion.

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MooS
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    i believe it is unfair that you forgot to remark that your answer is just an exapnded version of my comment above. – skeptic Nov 30 '16 at 15:32
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    @skeptic you are right. I added that remark. If you had shown any signs of filling in the details, I would not have posted that answer. – MooS Nov 30 '16 at 15:48