## 76.12 The finite part of a morphism

Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. For an algebraic space or a scheme $T$ over $S$ consider pairs $(a, Z)$ where

76.12.0.1
\begin{equation} \label{spaces-more-groupoids-equation-finite-conditions} \begin{matrix} a : T \to Y\text{ is a morphism over }S,
\\ Z \subset T \times _ Y X\text{ is an open subspace}
\\ \text{such that }\text{pr}_0|_ Z : Z \to T\text{ is finite.}
\end{matrix} \end{equation}

Suppose $h : T' \to T$ is a morphism of algebraic spaces over $S$ and $(a, Z)$ is a pair as in (76.12.0.1) over $T$. Set $a' = a \circ h$ and $Z' = (h \times \text{id}_ X)^{-1}(Z) = T' \times _ T Z$. Then $(a', Z')$ is a pair as in (76.12.0.1) over $T'$. This follows as finite morphisms are preserved under base change, see Morphisms of Spaces, Lemma 64.45.5. Thus we obtain a functor

76.12.0.2
\begin{equation} \label{spaces-more-groupoids-equation-finite} \begin{matrix} (X/Y)_{fin} :
& (\mathit{Sch}/S)^{opp}
& \longrightarrow
& \textit{Sets}
\\ & T
& \longmapsto
& \{ (a, Z)\text{ as above}\}
\end{matrix} \end{equation}

For applications we are mainly interested in this functor $(X/Y)_{fin}$ when $f$ is separated and locally of finite type. To get an idea of what this is all about, take a look at Remark 76.12.6.

Lemma 76.12.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Then we have

The presheaf $(X/Y)_{fin}$ satisfies the sheaf condition for the fppf topology.

If $T$ is an algebraic space over $S$, then there is a canonical bijection

\[ \mathop{Mor}\nolimits _{\mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}/S)_{fppf})}(T, (X/Y)_{fin}) = \{ (a, Z)\text{ satisfying 04PC}\} \]

**Proof.**
Let $T$ be an algebraic space over $S$. Let $\{ T_ i \to T\} $ be an fppf covering (by algebraic spaces). Let $s_ i = (a_ i, Z_ i)$ be pairs over $T_ i$ satisfying 76.12.0.1 such that we have $s_ i|_{T_ i \times _ T T_ j} = s_ j|_{T_ i \times _ T T_ j}$. First, this implies in particular that $a_ i$ and $a_ j$ define the same morphism $T_ i \times _ T T_ j \to Y$. By Descent on Spaces, Lemma 71.6.2 we deduce that there exists a unique morphism $a : T \to Y$ such that $a_ i$ equals the composition $T_ i \to T \to Y$. Second, this implies that $Z_ i \subset T_ i \times _ Y X$ are open subspaces whose inverse images in $(T_ i \times _ T T_ j) \times _ Y X$ are equal. Since $\{ T_ i \times _ Y X \to T \times _ Y X\} $ is an fppf covering we deduce that there exists a unique open subspace $Z \subset T \times _ Y X$ which restricts back to $Z_ i$ over $T_ i$, see Descent on Spaces, Lemma 71.6.1. We claim that the projection $Z \to T$ is finite. This follows as being finite is local for the fpqc topology, see Descent on Spaces, Lemma 71.10.23.

Note that the result of the preceding paragraph in particular implies (1).

Let $T$ be an algebraic space over $S$. In order to prove (2) we will construct mutually inverse maps between the displayed sets. In the following when we say “pair” we mean a pair satisfying conditions 76.12.0.1.

Let $v : T \to (X/Y)_{fin}$ be a natural transformation. Choose a scheme $U$ and a surjective étale morphism $p : U \to T$. Then $v(p) \in (X/Y)_{fin}(U)$ corresponds to a pair $(a_ U, Z_ U)$ over $U$. Let $R = U \times _ T U$ with projections $t, s : R \to U$. As $v$ is a transformation of functors we see that the pullbacks of $(a_ U, Z_ U)$ by $s$ and $t$ agree. Hence, since $\{ U \to T\} $ is an fppf covering, we may apply the result of the first paragraph that deduce that there exists a unique pair $(a, Z)$ over $T$.

Conversely, let $(a, Z)$ be a pair over $T$. Let $U \to T$, $R = U \times _ T U$, and $t, s : R \to U$ be as above. Then the restriction $(a, Z)|_ U$ gives rise to a transformation of functors $v : h_ U \to (X/Y)_{fin}$ by the Yoneda lemma (Categories, Lemma 4.3.5). As the two pullbacks $s^*(a, Z)|_ U$ and $t^*(a, Z)|_ U$ are equal, we see that $v$ coequalizes the two maps $h_ t, h_ s : h_ R \to h_ U$. Since $T = U/R$ is the fppf quotient sheaf by Spaces, Lemma 62.9.1 and since $(X/Y)_{fin}$ is an fppf sheaf by (1) we conclude that $v$ factors through a map $T \to (X/Y)_{fin}$.

We omit the verification that the two constructions above are mutually inverse.
$\square$

Lemma 76.12.2. Let $S$ be a scheme. Consider a commutative diagram

\[ \xymatrix{ X' \ar[rr]_ j \ar[rd] & & X \ar[ld] \\ & Y } \]

of algebraic spaces over $S$. If $j$ is an open immersion, then there is a canonical injective map of sheaves $j : (X'/Y)_{fin} \to (X/Y)_{fin}$.

**Proof.**
If $(a, Z)$ is a pair over $T$ for $X'/Y$, then $(a, j(Z))$ is a pair over $T$ for $X/Y$.
$\square$

Lemma 76.12.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is locally of finite type. Let $X' \subset X$ be the maximal open subspace over which $f$ is locally quasi-finite, see Morphisms of Spaces, Lemma 64.34.7. Then $(X/Y)_{fin} = (X'/Y)_{fin}$.

**Proof.**
Lemma 76.12.2 gives us an injective map $(X'/Y)_{fin} \to (X/Y)_{fin}$. Morphisms of Spaces, Lemma 64.34.7 assures us that formation of $X'$ commutes with base change. Hence everything comes down to proving that if $Z \subset X$ is an open subspace such that $f|_ Z : Z \to Y$ is finite, then $Z \subset X'$. This is true because a finite morphism is locally quasi-finite, see Morphisms of Spaces, Lemma 64.45.8.
$\square$

Lemma 76.12.4. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $T$ be an algebraic space over $S$, and let $(a, Z)$ be a pair as in 76.12.0.1. If $f$ is separated, then $Z$ is closed in $T \times _ Y X$.

**Proof.**
A finite morphism of algebraic spaces is universally closed by Morphisms of Spaces, Lemma 64.45.9. Since $f$ is separated so is the morphism $T \times _ Y X \to T$, see Morphisms of Spaces, Lemma 64.4.4. Thus the closedness of $Z$ follows from Morphisms of Spaces, Lemma 64.40.6.
$\square$

Lemma 76.12.7. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The diagonal of $(X/Y)_{fin} \to Y$

\[ (X/Y)_{fin} \longrightarrow (X/Y)_{fin} \times _ Y (X/Y)_{fin} \]

is representable (by schemes) and an open immersion and the “absolute” diagonal

\[ (X/Y)_{fin} \longrightarrow (X/Y)_{fin} \times (X/Y)_{fin} \]

is representable (by schemes).

**Proof.**
The second statement follows from the first as the absolute diagonal is the composition of the relative diagonal and a base change of the diagonal of $Y$ (which is representable by schemes), see Spaces, Section 62.3. To prove the first assertion we have to show the following: Given a scheme $T$ and two pairs $(a, Z_1)$ and $(a, Z_2)$ over $T$ with identical first component satisfying 76.12.0.1 there is an open subscheme $V \subset T$ with the following property: For any morphism of schemes $h : T' \to T$ we have

\[ h(T') \subset V \Leftrightarrow \Big(T' \times _ T Z_1 = T' \times _ T Z_2 \text{ as subspaces of }T' \times _ Y X\Big) \]

Let us construct $V$. Note that $Z_1 \cap Z_2$ is open in $Z_1$ and in $Z_2$. Since $\text{pr}_0|_{Z_ i} : Z_ i \to T$ is finite, hence proper (see Morphisms of Spaces, Lemma 64.45.9) we see that

\[ E = \text{pr}_0|_{Z_1}\left(Z_1 \setminus Z_1 \cap Z_2)\right) \cup \text{pr}_0|_{Z_2}\left(Z_2 \setminus Z_1 \cap Z_2)\right) \]

is closed in $T$. Now it is clear that $V = T \setminus E$ works.
$\square$

Lemma 76.12.8. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Suppose that $U$ is a scheme, $U \to Y$ is an étale morphism and $Z \subset U \times _ Y X$ is an open subspace finite over $U$. Then the induced morphism $U \to (X/Y)_{fin}$ is étale.

**Proof.**
This is formal from the description of the diagonal in Lemma 76.12.7 but we write it out since it is an important step in the development of the theory. We have to check that for any scheme $T$ over $S$ and a morphism $T \to (X/Y)_{fin}$ the projection map

\[ T \times _{(X/Y)_{fin}} U \longrightarrow T \]

is étale. Note that

\[ T \times _{(X/Y)_{fin}} U = (X/Y)_{fin} \times _{((X/Y)_{fin} \times _ Y (X/Y)_{fin})} (T \times _ Y U) \]

Applying the result of Lemma 76.12.7 we see that $T \times _{(X/Y)_{fin}} U$ is represented by an open subscheme of $T \times _ Y U$. As the projection $T \times _ Y U \to T$ is étale by Morphisms of Spaces, Lemma 64.39.4 we conclude.
$\square$

Lemma 76.12.9. Let $S$ be a scheme. Let

\[ \xymatrix{ X' \ar[d] \ar[r] & X \ar[d] \\ Y' \ar[r] & Y } \]

be a fibre product square of algebraic spaces over $S$. Then

\[ \xymatrix{ (X'/Y')_{fin} \ar[d] \ar[r] & (X/Y)_{fin} \ar[d] \\ Y' \ar[r] & Y } \]

is a fibre product square of sheaves on $(\mathit{Sch}/S)_{fppf}$.

**Proof.**
It follows immediately from the definitions that the sheaf $(X'/Y')_{fin}$ is equal to the sheaf $Y' \times _ Y (X/Y)_{fin}$.
$\square$

Lemma 76.12.10. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. If $f$ is separated and locally quasi-finite, then there exists a scheme $U$ étale over $Y$ and a surjective étale morphism $U \to (X/Y)_{fin}$ over $Y$.

**Proof.**
Note that the assertion makes sense by the result of Lemma 76.12.7 on the diagonal of $(X/Y)_{fin}$, see Spaces, Lemma 62.5.10. Let $V$ be a scheme and let $V \to Y$ be a surjective étale morphism. By Lemma 76.12.9 the morphism $(V \times _ Y X/V)_{fin} \to (X/Y)_{fin}$ is a base change of the map $V \to Y$ and hence is surjective and étale, see Spaces, Lemma 62.5.5. Hence it suffices to prove the lemma for $(V \times _ Y X/V)_{fin}$. (Here we implicitly use that the composition of representable, surjective, and étale transformations of functors is again representable, surjective, and étale, see Spaces, Lemmas 62.3.2 and 62.5.4, and Morphisms, Lemmas 29.9.2 and 29.34.3.) Note that the properties of being separated and locally quasi-finite are preserved under base change, see Morphisms of Spaces, Lemmas 64.4.4 and 64.27.4. Hence $V \times _ Y X \to V$ is separated and locally quasi-finite as well, and by Morphisms of Spaces, Proposition 64.50.2 we see that $V \times _ Y X$ is a scheme as well. Thus we may assume that $f : X \to Y$ is a separated and locally quasi-finite morphism of schemes.

Pick a point $y \in Y$. Pick $x_1, \ldots , x_ n \in X$ points lying over $y$. Pick an étale neighbourhood $a : (U, u) \to (Y, y)$ and a decomposition

\[ U \times _ S X = W \amalg \ \coprod \nolimits _{i = 1, \ldots , n} \ \coprod \nolimits _{j = 1, \ldots , m_ j} V_{i, j} \]

as in More on Morphisms, Lemma 37.36.5. Pick any subset

\[ I \subset \{ (i, j) \mid 1 \leq i \leq n, \ 1 \leq j \leq m_ i\} . \]

Given these choices we obtain a pair $(a, Z)$ with $Z = \bigcup _{(i, j) \in I} V_{i, j}$ which satisfies conditions 76.12.0.1. In other words we obtain a morphism $U \to (X/Y)_{fin}$. The construction of this morphism depends on all the things we picked above, so we should really write

\[ U(y, n, x_1, \ldots , x_ n, a, I) \longrightarrow (X/Y)_{fin} \]

This morphism is étale by Lemma 76.12.8.

Claim: The disjoint union of all of these is surjective onto $(X/Y)_{fin}$. It is clear that if the claim holds, then the lemma is true.

To show surjectivity we have to show the following (see Spaces, Remark 62.5.2): Given a scheme $T$ over $S$, a point $t \in T$, and a map $T \to (X/Y)_{fin}$ we can find a datum $(y, n, x_1, \ldots , x_ n, a, I)$ as above such that $t$ is in the image of the projection map

\[ U(y, n, x_1, \ldots , x_ n, a, I) \times _{(X/Y)_{fin}} T \longrightarrow T. \]

To prove this we may clearly replace $T$ by $\mathop{\mathrm{Spec}}(\overline{\kappa (t)})$ and $T \to (X/Y)_{fin}$ by the composition $\mathop{\mathrm{Spec}}(\overline{\kappa (t)}) \to T \to (X/Y)_{fin}$. In other words, we may assume that $T$ is the spectrum of an algebraically closed field.

Let $T = \mathop{\mathrm{Spec}}(k)$ be the spectrum of an algebraically closed field $k$. The morphism $T \to (X/Y)_{fin}$ is given by a pair $(T \to Y, Z)$ satisfying conditions 76.12.0.1. Here is a picture:

\[ \xymatrix{ & Z \ar[d] \ar[r] & X \ar[d] \\ \mathop{\mathrm{Spec}}(k) \ar@{=}[r] & T \ar[r] & Y } \]

Let $y \in Y$ be the image point of $T \to Y$. Since $Z$ is finite over $k$ it has finitely many points. Thus there exist finitely many points $x_1, \ldots , x_ n \in X$ such that the image of $Z$ in $X$ is contained in $\{ x_1, \ldots , x_ n\} $. Choose $a : (U, u) \to (Y, y)$ adapted to $y$ and $x_1, \ldots , x_ n$ as above, which gives the diagram

\[ \xymatrix{ W \amalg \ \coprod \nolimits _{i = 1, \ldots , n} \ \coprod \nolimits _{j = 1, \ldots , m_ j} V_{i, j} \ar[d] \ar[r] & X \ar[d] \\ U \ar[r] & Y. } \]

Since $k$ is algebraically closed and $\kappa (y) \subset \kappa (u)$ is finite separable we may factor the morphism $T = \mathop{\mathrm{Spec}}(k) \to Y$ through the morphism $u = \mathop{\mathrm{Spec}}(\kappa (u)) \to \mathop{\mathrm{Spec}}(\kappa (y)) = y \subset Y$. With this choice we obtain the commutative diagram:

\[ \xymatrix{ Z \ar[d] \ar[r] & W \amalg \ \coprod \nolimits _{i = 1, \ldots , n} \ \coprod \nolimits _{j = 1, \ldots , m_ j} V_{i, j} \ar[d] \ar[r] & X \ar[d] \\ \mathop{\mathrm{Spec}}(k) \ar[r] & U \ar[r] & Y } \]

We know that the image of the left upper arrow ends up in $\coprod V_{i, j}$. Recall also that $Z$ is an open subscheme of $\mathop{\mathrm{Spec}}(k) \times _ Y X$ by definition of $(X/Y)_{fin}$ and that the right hand square is a fibre product square. Thus we see that

\[ Z \subset \coprod \nolimits _{i = 1, \ldots , n}\ \coprod \nolimits _{j = 1, \ldots , m_ j} \mathop{\mathrm{Spec}}(k) \times _ U V_{i, j} \]

is an open subscheme. By construction (see More on Morphisms, Lemma 37.36.5) each $V_{i, j}$ has a unique point $v_{i, j}$ lying over $u$ with purely inseparable residue field extension $\kappa (u) \subset \kappa (v_{i, j})$. Hence each scheme $\mathop{\mathrm{Spec}}(k) \times _ U V_{i, j}$ has exactly one point. Thus we see that

\[ Z = \coprod \nolimits _{(i, j) \in I} \mathop{\mathrm{Spec}}(k) \times _ U V_{i, j} \]

for a unique subset $I \subset \{ (i, j) \mid 1 \leq i \leq n, \ 1 \leq j \leq m_ i\} $. Unwinding the definitions this shows that

\[ U(y, n, x_1, \ldots , x_ n, a, I) \times _{(X/Y)_{fin}} T \]

with $I$ as found above is nonempty as desired.
$\square$

Proposition 76.12.11. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is separated and locally of finite type. Then $(X/Y)_{fin}$ is an algebraic space. Moreover, the morphism $(X/Y)_{fin} \to Y$ is étale.

**Proof.**
By Lemma 76.12.3 we may replace $X$ by the open subscheme which is locally quasi-finite over $Y$. Hence we may assume that $f$ is separated and locally quasi-finite. We will check the three conditions of Spaces, Definition 62.6.1. Condition (1) follows from Lemma 76.12.1. Condition (2) follows from Lemma 76.12.7. Finally, condition (3) follows from Lemma 76.12.10. Thus $(X/Y)_{fin}$ is an algebraic space. Moreover, that lemma shows that there exists a commutative diagram

\[ \xymatrix{ U \ar[rr] \ar[rd] & & (X/Y)_{fin} \ar[ld] \\ & Y } \]

with horizontal arrow surjective and étale and south-east arrow étale. By Properties of Spaces, Lemma 63.16.3 this implies that the south-west arrow is étale as well.
$\square$

Lemma 76.12.14. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is separated, flat, and locally of finite presentation. In this case

$(X/Y)_{fin} \to Y$ is separated, representable, and étale, and

if $Y$ is a scheme, then $(X/Y)_{fin}$ is (representable by) a scheme.

**Proof.**
Since $f$ is in particular separated and locally of finite type (see Morphisms of Spaces, Lemma 64.28.5) we see that $(X/Y)_{fin}$ is an algebraic space by Proposition 76.12.11. To prove that $(X/Y)_{fin} \to Y$ is separated we have to show the following: Given a scheme $T$ and two pairs $(a, Z_1)$ and $(a, Z_2)$ over $T$ with identical first component satisfying 76.12.0.1 there is a closed subscheme $V \subset T$ with the following property: For any morphism of schemes $h : T' \to T$ we have

\[ h \text{ factors through } V \Leftrightarrow \Big(T' \times _ T Z_1 = T' \times _ T Z_2 \text{ as subspaces of }T' \times _ Y X\Big) \]

In the proof of Lemma 76.12.7 we have seen that $V = T' \setminus E$ is an open subscheme of $T'$ with closed complement

\[ E = \text{pr}_0|_{Z_1}\left(Z_1 \setminus Z_1 \cap Z_2)\right) \cup \text{pr}_0|_{Z_2}\left(Z_2 \setminus Z_1 \cap Z_2)\right). \]

Thus everything comes down to showing that $E$ is also open. By Lemma 76.12.4 we see that $Z_1$ and $Z_2$ are closed in $T' \times _ Y X$. Hence $Z_1 \setminus Z_1 \cap Z_2$ is open in $Z_1$. As $f$ is flat and locally of finite presentation, so is $\text{pr}_0|_{Z_1}$. This is true as $Z_1$ is an open subspace of the base change $T' \times _ Y X$, and Morphisms of Spaces, Lemmas 64.28.3 and Lemmas 64.30.4. Hence $\text{pr}_0|_{Z_1}$ is open, see Morphisms of Spaces, Lemma 64.30.6. Thus $\text{pr}_0|_{Z_1}\left(Z_1 \setminus Z_1 \cap Z_2)\right)$ is open and it follows that $E$ is open as desired.

We have already seen that $(X/Y)_{fin} \to Y$ is étale, see Proposition 76.12.11. Hence now we know it is locally quasi-finite (see Morphisms of Spaces, Lemma 64.39.5) and separated, hence representable by Morphisms of Spaces, Lemma 64.51.1. The final assertion is clear (if you like you can use Morphisms of Spaces, Proposition 64.50.2).
$\square$

Variant: Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $\sigma : Y \to X$ be a section of $f$. For an algebraic space or a scheme $T$ over $S$ consider pairs $(a, Z)$ where

76.12.14.1
\begin{equation} \label{spaces-more-groupoids-equation-finite-conditions-variant} \begin{matrix} a : T \to Y\text{ is a morphism over }S,
\\ Z \subset T \times _ Y X\text{ is an open subspace}
\\ \text{such that }\text{pr}_0|_ Z : Z \to T\text{ is finite and}
\\ (1_ T, \sigma \circ a) : T \to T \times _ Y X\text{ factors through }Z.
\end{matrix} \end{equation}

We will denote $(X/Y, \sigma )_{fin}$ the subfunctor of $(X/Y)_{fin}$ parametrizing these pairs.

Lemma 76.12.15. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $\sigma : Y \to X$ be a section of $f$. Consider the transformation of functors

\[ t : (X/Y, \sigma )_{fin} \longrightarrow (X/Y)_{fin}. \]

defined above. Then

$t$ is representable by open immersions,

if $f$ is separated, then $t$ is representable by open and closed immersions,

if $(X/Y)_{fin}$ is an algebraic space, then $(X/Y, \sigma )_{fin}$ is an algebraic space and an open subspace of $(X/Y)_{fin}$, and

if $(X/Y)_{fin}$ is a scheme, then $(X/Y, \sigma )_{fin}$ is an open subscheme of it.

**Proof.**
Omitted. Hint: Given a pair $(a, Z)$ over $T$ as in (76.12.0.1) the inverse image of $Z$ by $(1_ T, \sigma \circ a) : T \to T \times _ Y X$ is the open subscheme of $T$ we are looking for.
$\square$

## Comments (2)

Comment #4014 by Matthieu Romagny on

Comment #4124 by Johan on