In the proof of Chapter II, Proposition 2.6 of Hartshorne's Algebraic Geometry book, the author states that if $V$ is a variety over $k$, then $(t(V),\alpha_*(\mathcal{O}_V))$ is a scheme over $k$. In the book, $\alpha:V\to t(V)$ is defined by $\alpha(P)=\overline{\{P\}}$. $t$ is defined to be the functor that sends a topological space $X$ to a the set of (nonempty) irreducible closed subsets of $X$, with the topology of $t(X)$ defined by having sets of the form $t(Y)\subseteq t(X)$ be closed, where $Y\subseteq X$ is closed.

The author then states that since any variety can be covered by open affine subvarieties, it will be sufficient to prove the above statement in the case that $V$ is affine. Since a scheme is a locally ringed space $(X,\mathcal{O}_X)$ in which every point has an open neighborhood $U$ such that the restricted sheaf $\mathcal(O)_X|_U$ is affine, I am assuming that the author was thinking of covering $t(V)$ with open sets of the form $t(U_i)$, where $U_i$ are isomorphic to affine or quasi-affine varieties and $\{U_i\}$ cover V.

My confusions are that (1) sets of the form $t(U_i)$ don't necessarily cover $t(V)$ and (2) sets of the form $t(U_i)$ aren't necessarily open in $t(V)$. Was the author thinking of a different covering of $t(V)$ in his proof?

  • $\begingroup$ You want to use $U_i$ which are open affines in $V$ (saying that they're affine means that they're isomorphic to affine varieties, not just homeomorphic) and then (1) and (2) are actually true - why do you think they're false? $\endgroup$
    – KReiser
    Commented Jan 26, 2022 at 20:35
  • $\begingroup$ @KReiser Thanks, I should've said isomorphic. For (1), V is an irreducible subset of V, but V is not in $t(U_i)$ for any $i$ in the case that each $U_i$ is a proper subset of $V$. For (2), Suppose that $U\subsetneq V$ is a nonempty open subset. Suppose now that $t(U)$ is open in $t(V)$. Then $t(U)=t(V)\backslash t(Z)$ for some closed subset $Z\subseteq V$. In the case that that $Z$ is not $V$, we have that $V\in t(V)\backslash t(Z)$, but $V\not\in t(U)$, leading to a contradiction. Therefore, $Z=V$, implying $t(U)=\emptyset$. Therefore, $\{t(U_i)\}$ couldn't cover $t(V)$. $\endgroup$
    – Charuvinda
    Commented Jan 26, 2022 at 20:55
  • 1
    $\begingroup$ I think that the problem is notation. For a non closed $U$, you do not have $t(U) \subset t(V)$ since elements of $t(U)$ are not closed in $V$. For open $U$, define the map $f:t(U) \to t(V)$ sending each irreducible set to its closure in $X$. I think it is in this context where you have to prove that $f$ is a homeomorphism onto its image, which is the open set $t(V) \setminus t(V \setminus U)$, and conditions (1) and (2) are clearly satisfied since, for example, for the case you wrote to KReiser, $V$ irreducible implies $V$ is the closure of $U$ so it is in the image of $f$. $\endgroup$ Commented Jan 26, 2022 at 21:47
  • $\begingroup$ My apologies, I was incorrect in my first comment. I've explained in an answer below. $\endgroup$
    – KReiser
    Commented Jan 26, 2022 at 21:55

1 Answer 1


You're right, this isn't quite what Hartshorne had in mind here. One should not take $t(U)$ to be an open subset of $t(V)$- it's not even clear what that would mean, since the closed subsets of $U$ need not be closed subsets of $V$ in general. Instead, the bijection between open subsets of $V$ and open subsets of $t(V)$ proceeds in as follows: given an open subset $U$ with closed complement $W$, the open subset of $t(V)$ corresponding to $U$ is the open subset $t(V)\setminus t(W)$.

This fixes your covering problem: if $\{U_i\}$ cover $V$, then $\bigcap U_i^c=\emptyset$ and since $t$ commutes with intersections of closed sets, we see that $\bigcap t(U_i^c)=\emptyset$, so the sets $t(V)\setminus t(U_i^c)$ have union all of $t(V)$.

  • $\begingroup$ Still, one has to prove that $t(V) \smallsetminus t(W)$ is homeomorphic to $t(U)$ in order to show that one can reduce to the case where $V$ is affine $\endgroup$ Commented Jan 26, 2022 at 22:16
  • $\begingroup$ @AitorIribarLopez I feel like that's pretty clear. $\endgroup$
    – KReiser
    Commented Jan 26, 2022 at 22:19

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