
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. $\square$

Comment #3648 by Brian Conrad on

In the final paragraph, it should also be explained (with an extra sentence or two) that $\xi$ is also unique as the generic point of $Z$ (such as by showing that if $\eta$ is a generic point of $Z$ then each $p_i(\eta)$ has closure equal to $Z_i$ as defined there and so by spectrality of $X_i$ we have $p_i(\eta)=\xi_i$ for every $i$, forcing $\eta=\xi$ 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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