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Let $X$ be an irreducible separated scheme of finite type over an algebraically closed field $k$. In the proof of Theorem 8.15 Hartshorne claims that for any closed point $x\in X$ we have $dim\mathcal{O}_{X,x}=dim X$. This is an exercise in Hartshorne in the case that X is integral, i.e. also reduced. But the proof i know does not work in this more general situation and i was not able to find this statement anywhere else. Does anyone know how this works in this case ?

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It is false in general. – only Oct 22 '12 at 19:09
Could you give an example ? – Dieter Meint Oct 22 '12 at 19:49
Sorry, I misread your question; I thought it was asking about removing the "of finite type over an algebraically closed field" condition. – only Oct 22 '12 at 23:14

As the dimension can be computed locally, you can suppose $X$ is affine defined by a finitely generated $k$-algebra $A$. Let $N$ be the ideal of nilpotent elements of $A$. Let $m$ be the maximal ideal of $A$ corresponding to $x$.

Then $A/N$ defined an integral algebraic variety, and $m/N$ is a maximal ideal. Let $Y$ be the affine integral algebraic variety associated to $A/N$ and let $y$ be the point of $Y$ corresponding to $m/N$. Then $\dim Y=\dim X$ because the closed immersion $Y\to X$ is actually a homeomorphism (it is surjective). On the other hand, $O_{Y,y}=O_{X,x}/(NO_{X,x})$. Similarly to $Y$, we have $\dim Y=\dim O_{Y,y}=\dim O_{X,x}$. So $\dim X=\dim O_{X,x}$.

It is not necessary to suppose $k$ algebraically closed, and $X$ needs not separated.

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