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Let $X'\rightarrow X$ be a blow-up of a finitely dimensional scheme $X$ in a center $D$.

Under which assumptions one has $\dim X'=\dim X$? Do you know a proof or a reference for a proof? Do you know a example for $\dim X'\neq\dim X$?

In my research in resolution of singularities I need $\dim X'\leq\dim X$ if $X$ is an excellent scheme, normally flat along $D$ and if $D$ is regular and contains no generic points of $X$. It seems to be well known but I did not find a proof yet.

Edit: Thanks to the comments there are the following positive/negative answers:

If $X$ is locally Noetherian and $D$ contains no generic points of $X$, then $\dim X'=\dim X$. To see this one may assume that $X$ is reduced, since $(X')_\mathrm{red}=(X_\mathrm{red})'$. Further one may assume that $X$ is irreducible because the irreducible components of $X'$ are the strict transforms of the irreducible components of $X$. Then $X'\rightarrow X$ is proper and birational, by U. Görtz, T. Wedhorn, Algebraic Geometry I, Corollary 13.97. As diverietti commented we get $\dim X'=\dim X$ (cf. Q. Liu, Algebraic Geometry and Arithmetic Curves, Chap.8, Corollary 2.7).

If $X$ is locally Noetherian and $D$ contains all irreducible components $Z$ of $X$ with $\dim Z=\dim X$, then $\dim X'<\dim X$.

For $X$ not Noetherian one may have $\dim X'\neq\dim X$. For example let $k$ be a field, $X=\mathrm{Spec}(R)$ for the valuation ring $R=k[[y]]+x\cdot k((y))[[x]]$ and $D=V(\langle xy^{-n}\,|\,n\in\mathbb N\rangle)\subseteq X$. Then $X'$ is covered by open affine subschemes isomorphic to $\mathrm{Spec}(R_y)$.

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    $\begingroup$ The dimension of an algebraic variety is a birational invariant (it equals the trascendental degree of the function field) so it is invariant under birational modifications, in particular invariant under blow-up. $\endgroup$
    – Francesco Polizzi
    Commented Apr 22, 2015 at 11:18
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    $\begingroup$ This is certainly true if $X$ is a locally Noetherian integral scheme and you blow-up a non-zero quasi-coherent sheaf of ideals. In this case, the blow-up morphism is proper and birational, and $X'$ integral. Finally, if $f\colon Z\to Y$ is any proper birational morphism, where $Y$ is a locally Noetherian integral scheme, then $\dim Z=\dim Y$. $\endgroup$
    – diverietti
    Commented Apr 22, 2015 at 11:56
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    $\begingroup$ If you blow up the zero ideal, you get the empty scheme. If you blowup an irreducible component, then that component goes away. But yes, outside of that situation everything is fine as the other comments said. $\endgroup$
    – Karl Schwede
    Commented Apr 22, 2015 at 14:47
  • $\begingroup$ Thank you for your comments! I had a completely false perception of blow-ups in centers which contain a irreducible component. For $X$ locally Noetherian I should be able to to reduce my problem to the case that $X$ is a locally Noetherian integral scheme. $\endgroup$
    – Alexander Voitovitch
    Commented Apr 23, 2015 at 6:53

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