I have been thinking in the following problem

Let $X$ be a scheme. Is it true that $\dim \mathcal{O}_X(U) \leq \dim X$ for every open set $U\subset X$?

I think it's true but I can't prove it. Here are my ideas in particular cases:

  • If $U$ is an affine open subset then $\dim \mathcal{O}_X(U)=\dim U\leq \dim X$.

  • If $X$ is an integral $k$-scheme of finite type and $\mathcal{O}_X(U)$ is a finitely generated $k$-algebra (this is not true in general) we can take $V$ an open affine subset contained in $U$. Then we have $k\subseteq \mathcal{O}_X(U)\subseteq \mathcal{O}_X(V)$ so $$\dim \mathcal{O}_X(U)= \text{tr.deg}_k\ \text{Frac}(\mathcal{O}_X(U))\leq \text{tr.deg}_k\ \text{Frac}(\mathcal{O}_X(V))=\dim V \leq \dim X$$

It would be great if someone can comment about the general case or give a proof in other cases.

Edit 04/03/2018:

  • If $X=\text{Spec}(A)$ is affine we have the inequality above for any $U$. This follows directly from the fact that $$\mathcal{O}_X(U)=S^{-1}A \text{ where } S=A\setminus \bigcup_{\mathfrak{p}\in U} \mathfrak{p}$$ So $\dim \mathcal{O}_X(U) =\dim \text{Spec}(S^{-1}A)\leq \dim X$ because $\text{Spec}(S^{-1}A)$ is homeomorphic to a subset of $X$.

This is not true for arbitrary schemes (though I would not be surprised if it is true with some sort of finiteness hypotheses). For a simple example, let $X$ be a disjoint union of infinitely many copies of $\operatorname{Spec}\mathbb{Z}$. Then $\dim X=1$, but $\mathcal{O}_X(X)$ is an infinite product of copies of $\mathbb{Z}$, which has infinite Krull dimension (see Spectrum of $\mathbb{Z}^\mathbb{N}$).


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.