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This question comes from Proposition 2.6 in Chapter 2 of Hartshorne's Algebraic Geometry. In my edition, that's on page 78.

For a variety $V$, Hartshorne defines the topological space $t(V)$ to consist of the nonempty closed irreducible subsets of $V$, where the closed sets of $t(V)$ are of the form $t(Y)$ for $Y$ closed in $V$. He then defines a map $\alpha: V \rightarrow t(V)$ where P gets sent to {P} in $t(V)$. The claim is that $(t(V), \alpha_*(\mathcal{O}_V))$ is a scheme. I understand why this is true if $V$ is affine, but I have been unable to show $(t(V), \alpha_*(\mathcal{O}_V))$ is a scheme for an arbitrary variety $V$.

I had hoped to show that if $U$ is an affine open subset of $V$, then $t(U)$ is isomorphic to an open subset of $t(V)$. I used the map from $t(U)$ into $t(V)$ where we send an irreducible subset $W$ in $U$ to the smallest irreducible subset of $V$ containing $W$. However, although the image of of $t(U)$ is contained in $[t(U^c)]^c$, I don't believe these are equal.

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To show that $(t(V),\alpha_*\mathcal{O}(V))$ is a scheme, you must show that $t(V)$ has an open cover on which this ringed space is isomorphic to an affine scheme.

Take an affine open cover $\{U_i\}$ of $V$. Since you believe the affine case, it suffices to show that $\{t(U_i)\}$ is an open cover of $t(V)$, and

$(t(V),\alpha_*\mathcal{O}(V))|_{t(U_i)} \cong (t(U_i),\alpha_*\mathcal{O}(U_i))$

for each $i$. Given your last paragraph, it sounds like the first of these points is your difficulty. Let $Y$ be a nonempty irreducible closed subset $Y\subseteq U_i$. For each $j$, $Y\cap U_j$ is (when nonempty) a nonempty irreducible closed subset of $U_i\cap U_j$ (since an open subset of an irreducible is irreducible). The intersection $U_i\cap U_j$ is an affine open subset of $U_j$, and it's not hard to see (look at the pre-image of the corresponding prime ideals!) that $Y\cap U_j$ extends in a natural way to an irreducible closed subset of $U_j$. These extensions glue for varying $j$ to give an irreducible closed subset of $V$, since a locally irreducible subset of a (connected) space is irreducible. This furnishes the map $t(U_i)\to t(V)$ (which, in particular, I think addresses the issue you raise in the last paragraph).

It remains to see that this is an open subset and gives an open cover of $t(V)$, and to prove the above isomorphism. So now try from here...

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  • $\begingroup$ Thanks so much for your answer! I think I now understand your map, but I'm still trying to show the image of $t(U_i)$ is open. By saying this map addresses the issue I was having, do you mean that the image will equal $[t(U_{i}^c)]^c$? Or do I need a different approach? $\endgroup$
    – abourdon
    Commented Jul 2, 2011 at 12:47
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    $\begingroup$ The image is as your describe. To see this, I suggest that you first try to prove that the inclusion of $t(U_i)$ into $t(V)$ I described (because I was working in the context of a more general argument to show that $t(V)$ is a scheme) is more simply described as "taking the closure". From this description it's not difficult to show that the image of $t(U)$ is $t(U^c)^c$. Basically, if $X$ is a nonempty irreducible closed subset of $U$, its closure cannot be contained in $U^c$, so it is an element of $t(U^c)^c$ by definition. Conversely, if $X$ is closed irreducible in $V$ not containing ... $\endgroup$
    – Ramsey
    Commented Jul 2, 2011 at 16:02
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    $\begingroup$ rather, not contained in, $U^c$, then $X\cap U$ is nonempty irreducible closed in $U$. Now show that this $X$ is the closure of $X\cap U$ and you're done! $\endgroup$
    – Ramsey
    Commented Jul 2, 2011 at 16:04
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I can't comment yet, so this is a brief comment about some intuition, not the exact answer. At least on affine schemes, the points (i.e. prime ideals) are in one-to-one correspondence with irreducible closed subsets of the affine scheme (I have just read very little EGA, so please excuse my ignorance if this holds generally for any scheme - which is actually great). This really explains the "somewhat unintuitive" construction of the map.

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  • $\begingroup$ With regard to generalizing this: look at Hartshorne's exercises I.1.6 and II.2.7. $\endgroup$
    – Dylan Moreland
    Commented Jul 4, 2011 at 1:45
  • $\begingroup$ @Dylan: Thanks a lot for your help. I think you probably mean II.2.9, nice generalization! $\endgroup$
    – Poldavian
    Commented Jul 6, 2011 at 3:37

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