The Stacks project

Lemma 32.2.2. Let $I$ be a directed set. Let $(S_ i, f_{ii'})$ be an inverse system of schemes over $I$. If all the morphisms $f_{ii'} : S_ i \to S_{i'}$ are affine, then the limit $S = \mathop{\mathrm{lim}}\nolimits _ i S_ i$ exists in the category of schemes. Moreover,

  1. each of the morphisms $f_ i : S \to S_ i$ is affine,

  2. for an element $0 \in I$ and any open subscheme $U_0 \subset S_0$ we have

    \[ f_0^{-1}(U_0) = \mathop{\mathrm{lim}}\nolimits _{i \geq 0} f_{i0}^{-1}(U_0) \]

    in the category of schemes.

Proof. Choose an element $0 \in I$. Note that $I$ is nonempty as the limit is directed. For every $i \geq 0$ consider the quasi-coherent sheaf of $\mathcal{O}_{S_0}$-algebras $\mathcal{A}_ i = f_{i0, *}\mathcal{O}_{S_ i}$. Recall that $S_ i = \underline{\mathop{\mathrm{Spec}}}_{S_0}(\mathcal{A}_ i)$, see Morphisms, Lemma 29.11.3. Set $\mathcal{A} = \mathop{\mathrm{colim}}\nolimits _{i \geq 0} \mathcal{A}_ i$. This is a quasi-coherent sheaf of $\mathcal{O}_{S_0}$-algebras, see Schemes, Section 26.24. Set $S = \underline{\mathop{\mathrm{Spec}}}_{S_0}(\mathcal{A})$. By Morphisms, Lemma 29.11.5 we get for $i \geq 0$ morphisms $f_ i : S \to S_ i$ compatible with the transition morphisms. Note that the morphisms $f_ i$ are affine by Morphisms, Lemma 29.11.11 for example. By Lemma 32.2.1 above we see that for any affine open $U_0 \subset S_0$ the inverse image $U = f_0^{-1}(U_0) \subset S$ is the limit of the system of opens $U_ i = f_{i0}^{-1}(U_0)$, $i \geq 0$ in the category of schemes.

Let $T$ be a scheme. Let $g_ i : T \to S_ i$ be a compatible system of morphisms. To show that $S = \mathop{\mathrm{lim}}\nolimits _ i S_ i$ we have to prove there is a unique morphism $g : T \to S$ with $g_ i = f_ i \circ g$ for all $i \in I$. For every $t \in T$ there exists an affine open $U_0 \subset S_0$ containing $g_0(t)$. Let $V \subset g_0^{-1}(U_0)$ be an affine open neighbourhood containing $t$. By the remarks above we obtain a unique morphism $g_ V : V \to U = f_0^{-1}(U_0)$ such that $f_ i \circ g_ V = g_ i|_{U_ i}$ for all $i$. The open sets $V \subset T$ so constructed form a basis for the topology of $T$. The morphisms $g_ V$ glue to a morphism $g : T \to S$ because of the uniqueness property. This gives the desired morphism $g : T \to S$.

The final statement is clear from the construction of the limit above. $\square$


Comments (1)

Comment #9577 by on

Between the first and second paragraph of the proof, for clarity, I suggested explaining that because is filtered, the morphisms for induce compatible morphisms for any .


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