$\text{Spec}$ is a topological space on the prime ideals of a ring. What fails if we try to make the ideals into a topological space?
We might try something like this: the points of this space are ideals. For $f_1, ..., f_n \in A$, put $D(f_1, ..., f_n)$ to be the set of ideals not containing any of $f_1, ..., f_n$. Closing under unions (including the empty union) gives a topological space.
We could try to make this into a sheaf $\mathcal{F}$ by declaring $\mathcal{F}(D(f_1, ..., f_n))$ to be the localization of $A$ at the multiplicative set generated by $f_1, ..., f_n$.
Something seems wrong here, but I don't quite see it.