Let $X$ be an irreducible scheme. Can the Krull dimension of $\mathcal{O}_X(X)$ exceed that of $X$?
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1$\begingroup$ See meta.mathoverflow.net/questions/4200/flood-of-new-users $\endgroup$– Yemon ChoiCommented May 6, 2019 at 22:55
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Yes, this happens if $X$ is the punctured spectrum of a two dimensional regular local ring.
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$\begingroup$ what is the Krull dimension of such $X$? I believe $X$ has closed points, so puncturing should not really change anything, right? $\endgroup$– user138661Commented May 4, 2019 at 20:19
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1$\begingroup$ @schematic_boi The Krull dimension of $X$ is $1$. $\endgroup$– ssxCommented May 4, 2019 at 23:50
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1$\begingroup$ can this happen if $X$ is a scheme locally of finite type over $\mathbb{Z}$? $X$ of finite type over $\mathbb{Z}$? $\endgroup$– user138661Commented May 5, 2019 at 5:49