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 ?
Dimension of local rings at closed points in an irreducible scheme
3
$\begingroup$
algebraic-geometry
-
0It is false in general. – 2012-10-22
-
0Could you give an example ? – 2012-10-22
-
0Sorry, I misread your question; I thought it was asking about removing the "of finite type over an algebraically closed field" condition. – 2012-10-22