I have some questions on scheme morphisms. I ask pardon for posting them in one thread as they are most likely not worth to be distributed into several threads.
Let $X=Spec R$ be a noetherian scheme.
For a maximal ideal $m$ of $R$ one has a morphism $Spec (k(m))\to X$ induced by the ring homomorphism $R\to R_m\to R_m/R_mm=R/m$ which is a closed immersion. It maps the only point of $Spec (k(m))$ to $m$. If one takes a prime ideal $p$ instead of a maximal ideal, there is a scheme morphism induced by $R\to R_p\to R_p/R_p p=Quot(R/p)$. Is the image of the only point $(0)$ of $Spec (Quot(R/p))$ the generic point of the irreducible subscheme of $X$ correspoinding to $p$?
For a prime ideal $p$ one has, as above, scheme morphisms $ Spec(Quot(R/p))\xrightarrow{g_p} Spec(R_p)\xrightarrow{f_p} X $ and if $p$ is maximal $Spec(Quot(R/p))$ can be thought of as a point. What is $Spec(R_p)$ geometrically? Is $f_p$ or $g_p$ respectively an open/closed immersion? How can the image of $f_p$ or $g_p$ respectively be thought of?
If one has a morphism $f:Y\to X$, the base change of $f$ by $Spec (k(m))\to X$ is the fiber of $f$ at the point $m$. Is the base change of $f$ by $Spec(R_m)\to X$ a kind of 'thickened fiber'? In how far can I reconstruct the morphism $f$ if I know what it does on the topological spaces and if I know all the fibers or 'thickened fibers' respectively?