Suppose $f:R\to S$ is a homomorphism of rings of finite type. Then why is that the fibers of $f^*:\mathrm{Spec}~S\to\mathrm{Spec}~R$ a Noetherian subspace of $\mathrm{Spec}~S$? Does this have any geometric meaning?
The fiber of $f^*$ over a point $\mathfrak{p}$ is $\mathrm{Spec}~S_{\mathfrak{p}}/\mathfrak{p}S_{\mathfrak{p}}$.
The fiber over the point $p\subset R$ is given by: $$ \operatorname{Spec}S\times_{\operatorname{Spec}R}\operatorname{Spec}k(p) = \operatorname{Spec}(S\otimes_R k(p)), $$ where the finite type hypothesis means that $S\cong R[x_1,\dots,x_n]/I$ and by definition $k(p)=R_p/pR_p$. Then, the fiber is: $$ (f^*)^{-1}(p)=\operatorname{Spec}\left(k(p)[x_1,\dots,x_n]/I\right), $$ which is noetherian by Hilbert's basis theorem and the fact that the quotient of a noetherian ring is again noetherian.