Reducing the statement that a variety can be interpreted as a scheme to the case of an affine variety (Harshorne II. 2.6)

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?

• 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? Commented Jan 26, 2022 at 20:35
• @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 $t(V)$. Commented Jan 26, 2022 at 20:55
• 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 $f$. Commented Jan 26, 2022 at 21:47
• My apologies, I was incorrect in my first comment. I've explained in an answer below. Commented Jan 26, 2022 at 21:55

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 $$t(V)$$.
• 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 Commented Jan 26, 2022 at 22:16