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I am interested to know to what extent the notion of normality makes sense on a non-noetherian scheme.
Specifically, I can ask the following question: let $\pi:X\to Y$ be a formally smooth morphism of schemes. Assume that $Y$ is noetherian and normal. Let $U\subset Y$ be an open subset such that the complement has codimension $\geq 2$. Let $f$ be a regular function on $\pi^{-1}(U)$.

$\mathbf{Question:}$ Is it true that $f$ extends to all of $X$?

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1 Answer 1

No, it is not true. I'm going to describe everything in terms of commutative algebra, so take $X=Y=\operatorname{Spec} R$.

Let $G$ be the group $\mathbb Z \times \mathbb Z$ with the lexicographic ordering. Let $R$ be any valuation ring with value group $G$ and then $R$ has dimension $2$ since it has exactly three prime ideals, forming the chain: $$\mathfrak m = \{ r \in R \mid v(r) \geq (0, 1) \} \supset \mathfrak p = \{ r \in R \mid v(r) = (a, b) \mbox{ where } a \geq 1\} \supset 0,$$ where $v$ denotes the valuation. Let $t$ be any element of valuation $(0,1)$ so that the vanishing set of $t$ is the single closed point $\mathfrak m$. Thus, $1/t$ is a regular function on $U = X \setminus \mathfrak m$, but $1/t$ is not in $R$ even though $\mathfrak m$ has codimension 2.

I think about the result for Noetherian normal rings in terms of two separate results: First, if $R$ is a Noetherian domain, then $R$ is the intersection of $R_{\mathfrak p}$ as $\mathfrak p$ ranges over all associated primes of principal ideals. Second, in a normal Noetherian domain, all associated primes of principal ideals have codimension 1 (this is condition S2 of Serre). I suspect that the first might generalize to non-Noetherian rings, provided that you have the right definition of associated prime. The second fails pretty much as badly as possible, since a single element can generate a prime ideal of arbitrary codimension, or even infinite codimension.

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Is your $Y=\operatorname{Spec} R$ noetherian, as Alexander requires in his hypotheses? –  Charles Staats Jan 30 '12 at 21:16
    
@Charles, no, I missed the requirement that $Y$ be Noetherian. This is an example of a normal, but non-Noetherian scheme where a function being regular in codimension 1 does not imply regular. It might be possible to give a variation on this where $Y$ is Noetherian and $X$ is formally smooth over $Y$, but I'm not sure at the moment. –  Dustin Cartwright Jan 30 '12 at 22:44
    
Thanks, but I am really interested in the case when $X$ is equivalent to something normal and noetherian in the formally smooth topology. I actually suspect that in this case nothing "pathological" can happen, but I don't know how to prove this. –  Alexander Braverman Jan 31 '12 at 10:27
    
I still think that codimension is not well-behaved enough on non-Noetherian schemes for the answer to be yes. Here's an extension of the idea in my example: Take $Y = \operatorname{Spec} k$, where $k$ is a field. Put the $\mathbb Z \times \mathbb Z$ valuation on $k(t,u)$ as in the example and let $R$ be the valuation ring, and $X = \operatorname{Spec} R$. If $X \rightarrow Y$ is formally smooth, then this is would be a counter example. I can't prove that it is formally smooth, but I don't see that it isn't either. –  Dustin Cartwright Jan 31 '12 at 11:38
    
Well, that still wouldn't be a counterexample to what I am asking - in my case the open subset on which the function is defined a priori is assumed to be pulled back from $Y$. –  Alexander Braverman Jan 31 '12 at 14:37

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