I want to prove the conclusion below:
If $f:X\to Y$ is a quasi-finite flat morphism, then $\dim \mathcal{O}_{X,x}=\dim \mathcal{O}_{Y,f(x)}$.
For the part $\dim \mathcal{O}_{X,x}\leq\dim \mathcal{O}_{Y,f(x)}$, we can deduce from Zariski Main Theorem and I don't illustrate the details here.
My main question is the part $\dim \mathcal{O}_{X,x}\geq\dim \mathcal{O}_{Y,f(x)}$. We can see that the induced map $Spec\mathcal{O}_{X,x}\to Spec\mathcal{O}_{Y,f(x)}$ is faithfully flat, hence is surjective. Then for any prime ideal in $\mathcal{O}_{Y,f(x)}$, there is a prime ideal of $\mathcal{O}_{X,x}$ corresponding to it. But for $\mathfrak{p}\subset\mathfrak{p'}$ prime ideals of $\mathcal{O}_{Y,f(x)}$, if $\mathfrak{q}$ and $\mathfrak{q'}$ are the corresponding prime ideals to them respectively, how can we know the inclusion is kept?