# Topological Space on all Ideals

$$\text{Spec}$$ 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.

• I don't think anything fails, the result just isn't particularly useful. The motivation for Spec is not "let's make a topological space out of prime ideals" but "let's generalize the relationship between $k^n$ and $k[x_1,\dots,x_n]$". – Eric Wofsey Mar 8 at 23:01
• I think there is more to say than that Spec is a generalization of the relationship between $k^n$ and $k[x_1, ..., x_n]$ while the other isn't. One specific line of thought: Spec is unique in that it is right adjoint to the global sections functor from locally ringed spaces to rings. Supposing one would like to represent a ring as a locally ringed space whose global sections are that ring, Spec is the universal way of doing so. I went through the proof of this and don't quite understand why the above construction wouldn't also be universal as such. But adjoints are unique, so something is off. – Dean Young Mar 8 at 23:10
• Secondly, many other candidates for representing a ring as a locally ringed space have been developed, historically and currently, and these do not generalize the relationship between $k^n$ and $k[x_1, ..., x_n]$. E.g. Huber's work, $\text{Spv}$, and the Zariski Riemann space. This shows that motivating $\text{Spec}$ that way is sufficient but not necessary. – Dean Young Mar 8 at 23:13
• I think it is but that's just a guess (I haven't thought about the details). Certainly it is at least a presheaf that you can sheafify. – Eric Wofsey Mar 8 at 23:15
• The topological space you suggest is well-defined, and you're free to study it! But it's not clear that the "sheaf" you suggest is well-defined. For this we would need that if $D(f_1,\dots, f_n) = D(g_1,\dots,g_m)$ (in your notation), then the localizations inverting the $f_i$ and inverting the $g_i$ agree. This isn't obvious to me. @EricWofsey – Alex Kruckman Mar 9 at 2:12