I'm reading the Gortz's Algebraic Geometry, Lemma 14.109 and stuck at some point :
Why the underlined statement is true? By 'the correspondence of $x$ to a minimal ideal of $\mathcal{O}_{X,x}\otimes \kappa(y)$', I think that it means the Theorem 3.3 in next linked pdf : http://virtualmath1.stanford.edu/~conrad/216APage/handouts/irreddim.pdf
And note that $\mathcal{O}_{f^{-1}(y),x} \cong R:= \mathcal{O}_{X,x} \otimes_{\mathcal{O}_{Y,y}} \kappa(y)$ (c.f. Showing that $\mathcal{O}_{f^{-1}(y),x} \cong \mathcal{O}_{X,x} \otimes_{\mathcal{O}_{Y,y}} \kappa(y)$) So, my question is, $\operatorname{dim}\mathcal{O}_{f^{-1}(y),x} =0$?
In our case, the local ring $\mathcal{O}_{f^{-1}(y),x}$ satisfies that its maximal ideal is minimal prime? Then may be $\operatorname{dim}\mathcal{O}_{f^{-1}(y),x} =0$.
What will be good to notice?
Can anyone help?
EDIT : Below SiSi answered but I think that it is not complete : 1) the statement "The fibres of $f$ are equidimensional because $f$ is dominant and of finite type" can be false? c.f. : For a morphism $f: X\to Y$ dominant, locally of finite type ($Y$ is some scheme), each fiber is equidimensional?
2) And the reductibility to the case that "$f:\operatorname{Spec}(A)→\operatorname{Spec}(B)$ is dominant where $B$ is a discrete valuation ring with residue field $κ$ and field of fraction $K$ and $A⊂B$ is a finite type algebra over $B$." , is somewhat ambiguous : c.f. : For $y \in Y$, does there exists an open neighborhood $\operatorname{Spec}B$ where $B$ is DVR and containing the $y$ as the special point?. I don't know how we can reduce to that case.
Can anyone provide a more exact proof? Anyway I will struggling with the underlined statement.
Furthur progress : I found an associated proof : $\operatorname{dim}\mathcal{O}_{f^{-1}(y),x} =0$ is true ; c.f. Lemma 3.2 of http://virtualmath1.stanford.edu/~conrad/216APage/handouts/irreddim.pdf
I'm now trying to understand the underlined statement.
Can anyone explain?
Best Answer
use $\mathcal{O}_{X,x}\otimes \kappa(y)=\mathcal{O}_{f^{-1}y, \:x}$ . Then $dim(f^{-1}y)=dim(\overline{\{x\}})=dim(Z)$ as all components of the fibre $f^{-1}y$ have the same dimension, hence (the prime corresponding to) $x$ has height $0$ in the ring $\mathcal{O}_{f^{-1}y}$. So localising at $x$ gives $\mathcal{O}_{f^{-1}y,x}$ a zero dimensional (=Artinian) ring.
Edit The fibres of f are equidimensional because $f$ is dominant and of finite type see also here. Why? You may reduce to the case $f: \mathbf{Spec}(A) \rightarrow \mathbf{Spec}(B)$ is dominant where $B$ is a discrete valuation ring with residue field $\kappa$ and field of fraction $K$ and $A\subset B$ is a finite type algebra over $B$. Then $dim(A\otimes \kappa)=dim(A\otimes K)$, so the dimension of the special fibre is the dimension of the generic fibre and that IS equidimension by definition.