I have two simple questions:
1) Given a ring $R$ we can consider $R_{f}$ for $f \in R$, the localization at $f$. What elements go to zero in the localization?
2) If given two distinguished open sets $D(f) \subseteq D(h)$ for $f, h \in R$, how do you explicitly define the morphism $R_{h} \rightarrow R_{f}$?
Thanks.
This is pretty lazy of you, since your first question is easily answered by reading the definition of localization, which I found pretty quickly on wikipedia.
Anyway...
If $S\subset R$ is the multiplicative set that you're inverting, then $S^{-1}R$ is precisely the symbols $\{r/s\}$ with the equivalence relation that $r_1/s_1 = r_2/s_2$ iff there exists some $t\in S$ such that $$t(r_1s_2 − r_2s_1) = 0$$ It's clear that $0/1 = 0$ in your localization, so setting $r_2/s_2 = 0/1$, we see that this relation gives that $r_1/s_1 = 0$ iff there is some $t\in S$ such that $tr_1 = 0$ in $R$. I.e., the elements of $R$ going to zero are exactly those which are killed by some element of $S$.
For your second question, first consider the case of the inclusion $D(f)\subseteq \text{Spec }R$. The corresponding morphism is clearly just the localization map $R\mapsto R_f$. Now apply this to the case $R = R_h$.