## 69.23 Descending finite type spaces

This section continues the theme of Section 69.11 in the spirit of the results discussed in Section 69.7. It is also the analogue of Limits, Section 32.22 for algebraic spaces.

Situation 69.23.1. Let $S$ be a scheme, for example $\mathop{\mathrm{Spec}}(\mathbf{Z})$. Let $B = \mathop{\mathrm{lim}}\nolimits _{i \in I} B_ i$ be the limit of a directed inverse system of Noetherian spaces over $S$ with affine transition morphisms $B_{i'} \to B_ i$ for $i' \geq i$.

Lemma 69.23.2. In Situation 69.23.1. Let $X \to B$ be a quasi-separated and finite type morphism of algebraic spaces. Then there exists an $i \in I$ and a diagram

69.23.2.1
$$\label{spaces-limits-equation-good-diagram} \vcenter { \xymatrix{ X \ar[r] \ar[d] & W \ar[d] \\ B \ar[r] & B_ i } }$$

such that $W \to B_ i$ is of finite type and such that the induced morphism $X \to B \times _{B_ i} W$ is a closed immersion.

Proof. By Lemma 69.11.6 we can find a closed immersion $X \to X'$ over $B$ where $X'$ is an algebraic space of finite presentation over $B$. By Lemma 69.7.1 we can find an $i$ and a morphism of finite presentation $X'_ i \to B_ i$ whose pull back is $X'$. Set $W = X'_ i$. $\square$

Lemma 69.23.3. In Situation 69.23.1. Let $X \to B$ be a quasi-separated and finite type morphism of algebraic spaces. Given $i \in I$ and a diagram

$\vcenter { \xymatrix{ X \ar[r] \ar[d] & W \ar[d] \\ B \ar[r] & B_ i } }$

as in (69.23.2.1) for $i' \geq i$ let $X_{i'}$ be the scheme theoretic image of $X \to B_{i'} \times _{B_ i} W$. Then $X = \mathop{\mathrm{lim}}\nolimits _{i' \geq i} X_{i'}$.

Proof. Since $X$ is quasi-compact and quasi-separated formation of the scheme theoretic image of $X \to B_{i'} \times _{B_ i} W$ commutes with étale localization (Morphisms of Spaces, Lemma 66.16.3). Hence we may and do assume $W$ is affine and maps into an affine $U_ i$ étale over $B_ i$. Then

$B_{i'} \times _{B_ i} W = B_{i'} \times _{B_ i} U_ i \times _{U_ i} W = U_{i'} \times _{U_ i} W$

where $U_{i'} = B_{i'} \times _{B_ i} U_ i$ is affine as the transition morphisms are affine. Thus the lemma follows from the case of schemes which is Limits, Lemma 32.22.3. $\square$

Lemma 69.23.4. In Situation 69.23.1. Let $f : X \to Y$ be a morphism of algebraic spaces quasi-separated and of finite type over $B$. Let

$\vcenter { \xymatrix{ X \ar[r] \ar[d] & W \ar[d] \\ B \ar[r] & B_{i_1} } } \quad \text{and}\quad \vcenter { \xymatrix{ Y \ar[r] \ar[d] & V \ar[d] \\ B \ar[r] & B_{i_2} } }$

be diagrams as in (69.23.2.1). Let $X = \mathop{\mathrm{lim}}\nolimits _{i \geq i_1} X_ i$ and $Y = \mathop{\mathrm{lim}}\nolimits _{i \geq i_2} Y_ i$ be the corresponding limit descriptions as in Lemma 69.23.3. Then there exists an $i_0 \geq \max (i_1, i_2)$ and a morphism

$(f_ i)_{i \geq i_0} : (X_ i)_{i \geq i_0} \to (Y_ i)_{i \geq i_0}$

of inverse systems over $(B_ i)_{i \geq i_0}$ such that such that $f = \mathop{\mathrm{lim}}\nolimits _{i \geq i_0} f_ i$. If $(g_ i)_{i \geq i_0} : (X_ i)_{i \geq i_0} \to (Y_ i)_{i \geq i_0}$ is a second morphism of inverse systems over $(B_ i)_{i \geq i_0}$ such that such that $f = \mathop{\mathrm{lim}}\nolimits _{i \geq i_0} g_ i$ then $f_ i = g_ i$ for all $i \gg i_0$.

Proof. Since $V \to B_{i_2}$ is of finite presentation and $X = \mathop{\mathrm{lim}}\nolimits _{i \geq i_1} X_ i$ we can appeal to Proposition 69.3.10 as improved by Lemma 69.4.5 to find an $i_0 \geq \max (i_1, i_2)$ and a morphism $h : X_{i_0} \to V$ over $B_{i_2}$ such that $X \to X_{i_0} \to V$ is equal to $X \to Y \to V$. For $i \geq i_0$ we get a commutative solid diagram

$\xymatrix{ X \ar[d] \ar[r] & X_ i \ar[r] \ar@{..>}[d] \ar@/_2pc/[dd] |!{[d];[ld]}\hole & X_{i_0} \ar[d]^ h \\ Y \ar[r] \ar[d] & Y_ i \ar[r] \ar[d] & V \ar[d] \\ B \ar[r] & B_ i \ar[r] & B_{i_0} }$

Since $X \to X_ i$ has scheme theoretically dense image and since $Y_ i$ is the scheme theoretic image of $Y \to B_ i \times _{B_{i_2}} V$ we find that the morphism $X_ i \to B_ i \times _{B_{i_2}} V$ induced by the diagram factors through $Y_ i$ (Morphisms of Spaces, Lemma 66.16.6). This proves existence.

Uniqueness. Let $E_ i \to X_ i$ be the equalizer of $f_ i$ and $g_ i$ for $i \geq i_0$. We have $E_ i = Y_ i \times _{\Delta , Y_ i \times _{B_ i} Y_ i, (f_ i, g_ i)} X_ i$. Hence $E_ i \to X_ i$ is a monomorphism of finite presentation as a base change of the diagonal of $Y_ i$ over $B_ i$, see Morphisms of Spaces, Lemmas 66.4.1 and 66.28.10. Since $X_ i$ is a closed subspace of $B_ i \times _{B_{i_0}} X_{i_0}$ and similarly for $Y_ i$ we see that

$E_ i = X_ i \times _{(B_ i \times _{B_{i_0}} X_{i_0})} (B_ i \times _{B_{i_0}} E_{i_0}) = X_ i \times _{X_{i_0}} E_{i_0}$

Similarly, we have $X = X \times _{X_{i_0}} E_{i_0}$. Hence we conclude that $E_ i = X_ i$ for $i$ large enough by Lemma 69.6.10. $\square$

Remark 69.23.5. In Situation 69.23.1 Lemmas 69.23.2, 69.23.3, and 69.23.4 tell us that the category of algebraic spaces quasi-separated and of finite type over $B$ is equivalent to certain types of inverse systems of algebraic spaces over $(B_ i)_{i \in I}$, namely the ones produced by applying Lemma 69.23.3 to a diagram of the form (69.23.2.1). For example, given $X \to B$ finite type and quasi-separated if we choose two different diagrams $X \to V_1 \to B_{i_1}$ and $X \to V_2 \to B_{i_2}$ as in (69.23.2.1), then applying Lemma 69.23.4 to $\text{id}_ X$ (in two directions) we see that the corresponding limit descriptions of $X$ are canonically isomorphic (up to shrinking the directed set $I$). And so on and so forth.

Lemma 69.23.6. Notation and assumptions as in Lemma 69.23.4. If $f$ is flat and of finite presentation, then there exists an $i_3 > i_0$ such that for $i \geq i_3$ we have $f_ i$ is flat, $X_ i = Y_ i \times _{Y_{i_3}} X_{i_3}$, and $X = Y \times _{Y_{i_3}} X_{i_3}$.

Proof. By Lemma 69.7.1 we can choose an $i \geq i_2$ and a morphism $U \to Y_ i$ of finite presentation such that $X = Y \times _{Y_ i} U$ (this is where we use that $f$ is of finite presentation). After increasing $i$ we may assume that $U \to Y_ i$ is flat, see Lemma 69.6.12. As discussed in Remark 69.23.5 we may and do replace the initial diagram used to define the system $(X_ i)_{i \geq i_1}$ by the system corresponding to $X \to U \to B_ i$. Thus $X_{i'}$ for $i' \geq i$ is defined as the scheme theoretic image of $X \to B_{i'} \times _{B_ i} U$.

Because $U \to Y_ i$ is flat (this is where we use that $f$ is flat), because $X = Y \times _{Y_ i} U$, and because the scheme theoretic image of $Y \to Y_ i$ is $Y_ i$, we see that the scheme theoretic image of $X \to U$ is $U$ (Morphisms of Spaces, Lemma 66.30.12). Observe that $Y_{i'} \to B_{i'} \times _{B_ i} Y_ i$ is a closed immersion for $i' \geq i$ by construction of the system of $Y_ j$. Then the same argument as above shows that the scheme theoretic image of $X \to B_{i'} \times _{B_ i} U$ is equal to the closed subspace $Y_{i'} \times _{Y_ i} U$. Thus we see that $X_{i'} = Y_{i'} \times _{Y_ i} U$ for all $i' \geq i$ and hence the lemma holds with $i_3 = i$. $\square$

Lemma 69.23.7. Notation and assumptions as in Lemma 69.23.4. If $f$ is smooth, then there exists an $i_3 > i_0$ such that for $i \geq i_3$ we have $f_ i$ is smooth.

Lemma 69.23.8. Notation and assumptions as in Lemma 69.23.4. If $f$ is proper, then there exists an $i_3 \geq i_0$ such that for $i \geq i_3$ we have $f_ i$ is proper.

Proof. By the discussion in Remark 69.23.5 the choice of $i_1$ and $W$ fitting into a diagram as in (69.23.2.1) is immaterial for the truth of the lemma. Thus we choose $W$ as follows. First we choose a closed immersion $X \to X'$ with $X' \to Y$ proper and of finite presentation, see Lemma 69.12.1. Then we choose an $i_3 \geq i_2$ and a proper morphism $W \to Y_{i_3}$ such that $X' = Y \times _{Y_{i_3}} W$. This is possible because $Y = \mathop{\mathrm{lim}}\nolimits _{i \geq i_2} Y_ i$ and Lemmas 69.10.2 and 69.6.13. With this choice of $W$ it is immediate from the construction that for $i \geq i_3$ the algebraic space $X_ i$ is a closed subspace of $Y_ i \times _{Y_{i_3}} W \subset B_ i \times _{B_{i_3}} W$ and hence proper over $Y_ i$. $\square$

Lemma 69.23.9. In Situation 69.23.1 suppose that we have a cartesian diagram

$\xymatrix{ X^1 \ar[r]_ p \ar[d]_ q & X^3 \ar[d]^ a \\ X^2 \ar[r]^ b & X^4 }$

of algebraic spaces quasi-separated and of finite type over $B$. For each $j = 1, 2, 3, 4$ choose $i_ j \in I$ and a diagram

$\xymatrix{ X^ j \ar[r] \ar[d] & W^ j \ar[d] \\ B \ar[r] & B_{i_ j} }$

as in (69.23.2.1). Let $X^ j = \mathop{\mathrm{lim}}\nolimits _{i \geq i_ j} X^ j_ i$ be the corresponding limit descriptions as in Lemma 69.23.4. Let $(a_ i)_{i \geq i_5}$, $(b_ i)_{i \geq i_6}$, $(p_ i)_{i \geq i_7}$, and $(q_ i)_{i \geq i_8}$ be the corresponding morphisms of inverse systems contructed in Lemma 69.23.4. Then there exists an $i_9 \geq \max (i_5, i_6, i_7, i_8)$ such that for $i \geq i_9$ we have $a_ i \circ p_ i = b_ i \circ q_ i$ and such that

$(q_ i, p_ i) : X^1_ i \longrightarrow X^2_ i \times _{b_ i, X^4_ i, a_ i} X^3_ i$

is a closed immersion. If $a$ and $b$ are flat and of finite presentation, then there exists an $i_{10} \geq \max (i_5, i_6, i_7, i_8, i_9)$ such that for $i \geq i_{10}$ the last displayed morphism is an isomorphism.

Proof. According to the discussion in Remark 69.23.5 the choice of $W^1$ fitting into a diagram as in (69.23.2.1) is immaterial for the truth of the lemma. Thus we may choose $W^1 = W^2 \times _{W^4} W^3$. Then it is immediate from the construction of $X^1_ i$ that $a_ i \circ p_ i = b_ i \circ q_ i$ and that

$(q_ i, p_ i) : X^1_ i \longrightarrow X^2_ i \times _{b_ i, X^4_ i, a_ i} X^3_ i$

is a closed immersion.

If $a$ and $b$ are flat and of finite presentation, then so are $p$ and $q$ as base changes of $a$ and $b$. Thus we can apply Lemma 69.23.6 to each of $a$, $b$, $p$, $q$, and $a \circ p = b \circ q$. It follows that there exists an $i_9 \in I$ such that

$(q_ i, p_ i) : X^1_ i \to X^2_ i \times _{X^4_ i} X^3_ i$

is the base change of $(q_{i_9}, p_{i_9})$ by the morphism by the morphism $X^4_ i \to X^4_{i_9}$ for all $i \geq i_9$. We conclude that $(q_ i, p_ i)$ is an isomorphism for all sufficiently large $i$ by Lemma 69.6.10. $\square$

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