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$\text{Spec}$ 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.

Eric Wofsey
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Ronald J. Zallman
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    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 08 '19 at 23:01
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    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. – Ronald J. Zallman Mar 08 '19 at 23:10
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    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. – Ronald J. Zallman Mar 08 '19 at 23:13
  • Well, I wouldn't expect to even get a locally ringed space at all. Why would the stalk at a non-prime ideal be local? – Eric Wofsey Mar 08 '19 at 23:13
  • I hadn't thought of that, thanks. But do you think it would still be a sheaf? – Ronald J. Zallman Mar 08 '19 at 23:14
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    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 08 '19 at 23:15
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    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 09 '19 at 02:12
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    @AlexKruckman: If $D(f_1,\dots,f_n)=D(g_1,\dots,g_m)$, then in particular the prime ideals not containing any $f_i$ are the same as the prime ideals not containing any $g_j$. That implies they give the same localization by the same proof as for the usual Spec. – Eric Wofsey Mar 09 '19 at 02:20
  • @EricWofsey Thanks! – Alex Kruckman Mar 09 '19 at 02:29
  • Also note that the set of ideals of a ring may be viewed as a top space via the Hilbert scheme construction. – Ben Mar 09 '19 at 03:29
  • Oh, thanks so much @Alex. That must be the error then. You see, my goal is not to study this space, but to understand what fails if we don't make the prime assumption. And it seems we don't get a sheaf. If you leave this as an answer, I would of course accept it. – Ronald J. Zallman Mar 09 '19 at 04:14

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