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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