Let X be a k-scheme (k a field) locally of finite type, Z an irreducible component. Then if K is some field extension of k, let $p:Y= X \times_k K \rightarrow X$ be the projection, And Z' an irreducible component of Y, such that $\overline{p(Z')}=Z$. Show that in this case, $dim Z = dim Z'$. What I've been trying so far is to argue along the following steps, and I would like some pointers as to whether this is correct.
$1.$ We reduce to the case where $X=Z$ and further to where X is an integral scheme.
$2.$ We note that the dimension of X can be computed on any non-empty open subset of X, so we can assume that X is affine, $X= Spec A$ say, and now, we're in the case that $X = Spec A$ and $Y =Spec A \otimes_k K$.
Are these okay reductions? I'm very unsure about the second step, but I can't seem to point my finger as to where the fault lies. So, if it's wrong please tell me, and further, I'd like useful heuristics for dealing with questions regarding dimension of schemes locally finite type over a field.
