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The Stacks project

Lemma 5.24.5. Let \mathcal{I} be a cofiltered index category. Let i \mapsto X_ i be a diagram of spectral spaces such that for a : j \to i in \mathcal{I} the corresponding map f_ a : X_ j \to X_ i is spectral. Then the inverse limit X = \mathop{\mathrm{lim}}\nolimits X_ i is a spectral topological space and the projection maps p_ i : X \to X_ i are spectral.

Proof. The limit X = \mathop{\mathrm{lim}}\nolimits X_ i exists (Lemma 5.14.1) and is quasi-compact by Lemma 5.24.1.

Denote p_ i : X \to X_ i the projection. Because \mathcal{I} is cofiltered we can apply Lemma 5.14.2. Hence a basis for the topology on X is given by the opens p_ i^{-1}(U_ i) for U_ i \subset X_ i open. Since a basis for the topology of X_ i is given by the quasi-compact open, we conclude that a basis for the topology on X is given by p_ i^{-1}(U_ i) with U_ i \subset X_ i quasi-compact open. A formal argument shows that

p_ i^{-1}(U_ i) = \mathop{\mathrm{lim}}\nolimits _{a : j \to i} f_ a^{-1}(U_ i)

as topological spaces. Since each f_ a is spectral the sets f_ a^{-1}(U_ i) are closed in the constructible topology of X_ j and hence p_ i^{-1}(U_ i) is quasi-compact by Lemma 5.24.1. Thus X has a basis for the topology consisting of quasi-compact opens.

Any quasi-compact open U of X is of the form U = p_ i^{-1}(U_ i) for some i and some quasi-compact open U_ i \subset X_ i (see Lemma 5.24.4). Given U_ i \subset X_ i and U_ j \subset X_ j quasi-compact open, then p_ i^{-1}(U_ i) \cap p_ j^{-1}(U_ j) = p_ k^{-1}(U_ k) for some k and quasi-compact open U_ k \subset X_ k. Namely, choose k and morphisms k \to i and k \to j and let U_ k be the intersection of the pullbacks of U_ i and U_ j to X_ k. Thus we see that the intersection of two quasi-compact opens of X is quasi-compact open.

Finally, let Z \subset X be irreducible and closed. Then p_ i(Z) \subset X_ i is irreducible and therefore Z_ i = \overline{p_ i(Z)} has a unique generic point \xi _ i (because X_ i is a spectral space). Then f_ a(\xi _ j) = \xi _ i for a : j \to i in \mathcal{I} because \overline{f_ a(Z_ j)} = Z_ i. Hence \xi = \mathop{\mathrm{lim}}\nolimits \xi _ i is a point of X. Claim: \xi \in Z. Namely, if not we can find a quasi-compact open containing \xi disjoint from Z. This would be of the form p_ i^{-1}(U_ i) for some i and quasi-compact open U_ i \subset X_ i. Then \xi _ i \in U_ i but p_ i(Z) \cap U_ i = \emptyset which contradicts \xi _ i \in \overline{p_ i(Z)}. So \xi \in Z and hence \overline{\{ \xi \} } \subset Z. Conversely, every z \in Z is in the closure of \xi . Namely, given a quasi-compact open neighbourhood U of z we write U = p_ i^{-1}(U_ i) for some i and quasi-compact open U_ i \subset X_ i. We see that p_ i(z) \in U_ i hence \xi _ i \in U_ i hence \xi \in U. Thus \xi is a generic point of Z. We omit the proof that \xi is the unique generic point of Z (hint: show that a second generic point has to be equal to \xi by showing that it has to map to \xi _ i in X_ i since by spectrality of X_ i the irreducible Z_ i has a unique generic point). This finishes the proof. \square


Comments (2)

Comment #3648 by Brian Conrad on

In the final paragraph, it should also be explained (with an extra sentence or two) that is also unique as the generic point of (such as by showing that if is a generic point of then each has closure equal to as defined there and so by spectrality of we have for every , forcing as desired).

Comment #3744 by on

OK, in this case I decided to omit the proof of uniqueness, but I gave a hint to the reader. See changes here.


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