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This is of course true in the affine case, so it seems like it should be true in general, because $\mathcal{O}_X(X)$ should be "smaller" for a non-affine scheme than for a similar affine scheme (e.g. all global sections over a projective scheme are constants). Just to be clear on notation:

$X$ is a scheme, not necessarily affine.

$f_1, \ldots, f_n \in \mathcal{O}_X(X)$.

$X_f := \{ x \in X \; | \; f_x \not \in \mathfrak{m}_{X, x} \}$, where $\mathfrak{m}_{X, x}$ is the maximal ideal of the stalk $\mathcal{O}_{X, x}$

$X = X_{f_1} \cup \cdots \cup X_{f_n}$

Does it follow that $(f_1, \ldots, f_n) = (1)$ in $\mathcal{O}_X(X)$?

We have $(f_1|_U, \ldots, f_n|_U) = (1)$ in $\mathcal{O}_X(U)$ where $U \subset X$ is open affine, but I don't see how to extend this to all of $X$.

Daniel McLaury
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  • What happens if you look at the morphism $\alpha: X \rightarrow Y :=\operatorname{Spec} \mathcal{O}X(X)$? Since the latter scheme is affine, if your conclusion was false we would have to have $Y \neq \cup Y{f_i}$. But I don't think that can happen; isn't it true that $X_{f_i} = \alpha^{-1} (Y_{f_i})$? (I haven't worked this out, so I might be completely off-base here.) –  May 15 '13 at 21:57
  • It is true that $X_{f_i} = \alpha^{-1}(Y_{f_i})$, but just because the $f^{-1}(Y_{f_i})$ cover $X$ it doesn't seem to follow that the $Y_{f_i}$ necessarily cover $Y$. – Daniel McLaury May 15 '13 at 23:12
  • If we had a map in the opposite direction something like this might work but I don't think such a map exists in general. – Daniel McLaury May 15 '13 at 23:13
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    The ideal sheaf generated by $f_1,\dotsc,f_n$ is $\mathcal{O}_X$ (because this holds locally). Globally this won't be the case, there are cohomological obstructions. I will try to find a simple example ... Note that for infinitely many sections it is easily seen to be wrong (take an infinite disjoint union of non-empty schemes and the obvious idempotent sections, the ideal consists of those sections which are supported only on finitely many schemes), but for $n=1$ it is true. – Martin Brandenburg May 15 '13 at 23:19

2 Answers2

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No. Let $X = \mathbb{A}^2 \setminus (0,0)$. Then $\mathcal{O}(X) = k[x,y]$ and $X = X_x \cup X_y$. However, the ideal $\langle x,y \rangle$ in $k[x,y]$ is not $1$.

The plane with a point deleted should be in your standard toolkit to test things against.

David E Speyer
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    Ah, so the idea is quite simple and more general: Find some $Y$ with global sections $f_1,\dotsc,f_n$ which do not generate the unit ideal and define $X = \cup_i Y_{f_i}$. If $Y$ is normal, noetherian and $X \setminus Y = V(f_1,\dotsc,f_n)$ has codimension at least $2$, then the global sections of $X$ and $Y$ coincide. Thus $X$ is a counterexample. – Martin Brandenburg May 16 '13 at 23:10
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This is a long comment, which might lead to a counterexample.

The question is equivalent to: Let $f : \mathcal{O}_X^n \to \mathcal{O}_X$ be a surjective homomorphism of quasi-coherent modules on $X$. Is it also surjective on global sections? If $K$ denotes the kernel of $f$, the long exact sequence of cohomology groups associated to (*) $0 \to K \to \mathcal{O}_X^n \to \mathcal{O}_X \to 0$ tells us that a sufficient condition is that $H^1(X,K)=0$. Note that $K$ is a vector bundle of rank $n-1$, since (*) splits locally.

This shows on the one hand that some schemes satisfy the property (affine schemes and $\mathbb{P}^1$, perhaps someone can add more examples), but on the other hand it shows how to construct counterexamples: Find $X$ which admits some vector bundle $K$ which fits into some sequence (*) such that $H^1(X,\mathcal{O}_X)=0$, but $H^1(X,K) \neq 0$.

Probably there are already examples when $X$ is a variety which is covered by two open affines.

Martin Brandenburg
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