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Let $X$ be an irreducible scheme. Can the Krull dimension of $\mathcal{O}_X(X)$ exceed that of $X$?

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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? $\endgroup$ – user138661 May 4 at 20:19
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    $\begingroup$ @schematic_boi The Krull dimension of $X$ is $1$. $\endgroup$ – Mere Scribe May 4 at 23:50
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    $\begingroup$ can this happen if $X$ is a scheme locally of finite type over $\mathbb{Z}$? $X$ of finite type over $\mathbb{Z}$? $\endgroup$ – user138661 May 5 at 5:49

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