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 $X$.
Edit: 06/02/2020:
- I saw this result in mathoverflow and it made me remember this problem. Using it we have that for any integral $k$-scheme $X$ of finite type and any open set $U\subseteq X$ we have $k\subseteq \mathcal{O}_X(U)\subseteq K(X)$, and hence $$\dim U \leq \mathrm{tr.deg}_k(\mathrm{Frac}\,\mathcal{O}_X(U))\leq \mathrm{tr.deg}_k\, K(X)=X$$ so the result is true for any open set in an integral variety.
