Assume I know the topology of the spectrum $\operatorname{Spec}(R)=(X,\mathcal{O}_{X})$ of a reduced ring $R$, and I know what the local rings $\mathcal{O}_{X,x}$ look like for every $x\in X$. How much information does that give me about $R$? Does it define $R$ uniquely?

More generally: Assume I know the topology of a scheme $X=(X,\mathcal{O}_{X})$, and I know some of the rings $\mathcal{O}_{X,x}$. How much does that tell me about $\mathcal{O}_{X}$?

A slight variation: Assume I know the topology of the spectrum $\operatorname{Spec}(R)=(X,\mathcal{O}_{X})$ of a reduced normal ring $R$, and I know what the Zariski tangent spaces $T_{X,x}$ look like for every $x\in X$. How much information does that give me about $R$? What happens if I only know what some of the $T_{X,x}$ look like?

What happens if I let go of the reducedness/normality conditions?

  • $\begingroup$ When you say, I know what the Zariski tangent spaces look like, what exactly do you mean? You know, say its vector space dimension (assuming you have a variety over a field) or something more? $\endgroup$
    – Mohan
    May 4, 2020 at 21:19
  • $\begingroup$ @Mohan I mean I know what the tangent spaces are isomorphic to. And I am not talking about varieties, but about schemes. I am however particularly, though not exclusively, interested in schemes over fields, yes. $\endgroup$ May 4, 2020 at 21:25
  • $\begingroup$ Isomorphic as what? Vector spaces? The point I am trying to make is, even among varieties you probably can say very little knowing the dimension of the tangent spaces, so for schemes it seems even more implausible. $\endgroup$
    – Mohan
    May 4, 2020 at 21:27

1 Answer 1


Even the case where you know the isomorphism type of the underlying topological space of $X=\operatorname{Spec} R$ and the isomorphism type of every local ring $\mathcal{O}_{X,x}$, this is not enough information to determine $X$ uniquely. Consider $\operatorname{Spec} \Bbb C[t]$ and $\operatorname{Spec} \Bbb C[t,t^{-1}]$. As topological spaces, these are homeomorphic (each have one generic point, $|\Bbb C|$ closed points, and the open sets consist of the empty set and sets who's complement is finitely many closed points) and every local ring is isomorphic to either $\Bbb C(t)$ (the generic point), or $\Bbb C[t]_{(t)}$ (the closed points). But $\Bbb C[t]\not\cong\Bbb C[t,t^{-1}]$.

In the more general situation where you only know some of the local rings, things are even worse: you could have nilpotents supported at the points where you don't know the local rings and you will never be able to detect these. The basic issue here is that while local rings tell you a lot (any single local ring of an irreducible, generically reduced scheme determines the function field and thus the birational class of such a variety), you still have a wide latitude to alter your scheme on closed subsets. In some sense, this is the foundational question of birational geometry: if we know two varieties are birational, how much more can we say?

Your question about tangent spaces suffers from the same issues because knowing dimensions of tangent spaces is a strictly weaker condition that knowing isomorphism types of local rings. For instance, any two smooth curves have all tangent spaces of dimension one at each closed point, but there's a smooth curve of every non-negative integer genus and they are all non-isomorphic.


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