Definition 28.39.1. Let $f : X \to S$ be a morphism of schemes. We say $f$ is *proper* if $f$ is separated, finite type, and universally closed.

## 28.39 Proper morphisms

The notion of a proper morphism plays an important role in algebraic geometry. An important example of a proper morphism will be the structure morphism $\mathbf{P}^ n_ S \to S$ of projective $n$-space, and this is in fact the motivating example leading to the definition.

The morphism from the affine line with zero doubled to the affine line is of finite type and universally closed, so the separation condition is necessary in the definition above. In the rest of this section we prove some of the basic properties of proper morphisms and of universally closed morphisms.

Lemma 28.39.2. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:

The morphism $f$ is universally closed.

There exists an open covering $S = \bigcup V_ j$ such that $f^{-1}(V_ j) \to V_ j$ is universally closed for all indices $j$.

**Proof.**
This is clear from the definition.
$\square$

Lemma 28.39.3. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:

The morphism $f$ is proper.

There exists an open covering $S = \bigcup V_ j$ such that $f^{-1}(V_ j) \to V_ j$ is proper for all indices $j$.

**Proof.**
Omitted.
$\square$

Lemma 28.39.4. The composition of proper morphisms is proper. The same is true for universally closed morphisms.

**Proof.**
A composition of closed morphisms is closed. If $X \to Y \to Z$ are universally closed morphisms and $Z' \to Z$ is any morphism, then we see that $Z' \times _ Z X = (Z' \times _ Z Y) \times _ Y X \to Z' \times _ Z Y$ is closed and $Z' \times _ Z Y \to Z'$ is closed. Hence the result for universally closed morphisms. We have seen that “separated” and “finite type” are preserved under compositions (Schemes, Lemma 25.21.12 and Lemma 28.14.3). Hence the result for proper morphisms.
$\square$

Lemma 28.39.5. The base change of a proper morphism is proper. The same is true for universally closed morphisms.

**Proof.**
This is true by definition for universally closed morphisms. It is true for separated morphisms (Schemes, Lemma 25.21.12). It is true for morphisms of finite type (Lemma 28.14.4). Hence it is true for proper morphisms.
$\square$

Lemma 28.39.6. A closed immersion is proper, hence a fortiori universally closed.

**Proof.**
The base change of a closed immersion is a closed immersion (Schemes, Lemma 25.18.2). Hence it is universally closed. A closed immersion is separated (Schemes, Lemma 25.23.8). A closed immersion is of finite type (Lemma 28.14.5). Hence a closed immersion is proper.
$\square$

Lemma 28.39.7. Suppose given a commutative diagram of schemes

with $Y$ separated over $S$.

If $X \to S$ is universally closed, then the morphism $X \to Y$ is universally closed.

If $X$ is proper over $S$, then the morphism $X \to Y$ is proper.

In particular, in both cases the image of $X$ in $Y$ is closed.

**Proof.**
Assume that $X \to S$ is universally closed (resp. proper). We factor the morphism as $X \to X \times _ S Y \to Y$. The first morphism is a closed immersion, see Schemes, Lemma 25.21.10. Hence the first morphism is proper (Lemma 28.39.6). The projection $X \times _ S Y \to Y$ is the base change of a universally closed (resp. proper) morphism and hence universally closed (resp. proper), see Lemma 28.39.5. Thus $X \to Y$ is universally closed (resp. proper) as the composition of universally closed (resp. proper) morphisms (Lemma 28.39.4).
$\square$

The following lemma says that the image of a proper scheme (in a separated scheme of finite type over the base) is proper.

Lemma 28.39.8. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of schemes over $S$. If $X$ is universally closed over $S$ and $f$ is surjective then $Y$ is universally closed over $S$. In particular, if also $Y$ is separated and of finite type over $S$, then $Y$ is proper over $S$.

**Proof.**
Assume $X$ is universally closed and $f$ surjective. Denote $p : X \to S$, $q : Y \to S$ the structure morphisms. Let $S' \to S$ be a morphism of schemes. The base change $f' : X_{S'} \to Y_{S'}$ is surjective (Lemma 28.9.4), and the base change $p' : X_{S'} \to S'$ is closed. If $T \subset Y_{S'}$ is closed, then $(f')^{-1}(T) \subset X_{S'}$ is closed, hence $p'((f')^{-1}(T)) = q'(T)$ is closed. So $q'$ is closed.
$\square$

The proof of the following lemma is due to Bjorn Poonen, see this location.

Lemma 28.39.9. A universally closed morphism of schemes is quasi-compact.

**Proof.**
Let $f : X \to S$ be a morphism. Assume that $f$ is not quasi-compact. Our goal is to show that $f$ is not universally closed. By Schemes, Lemma 25.19.2 there exists an affine open $V \subset S$ such that $f^{-1}(V)$ is not quasi-compact. To achieve our goal it suffices to show that $f^{-1}(V) \to V$ is not universally closed, hence we may assume that $S = \mathop{\mathrm{Spec}}(A)$ for some ring $A$.

Write $X = \bigcup _{i \in I} X_ i$ where the $X_ i$ are affine open subschemes of $X$. Let $T = \mathop{\mathrm{Spec}}(A[y_ i ; i \in I])$. Let $T_ i = D(y_ i) \subset T$. Let $Z$ be the closed set $(X \times _ S T) - \bigcup _{i \in I} (X_ i \times _ S T_ i)$. It suffices to prove that the image $f_ T(Z)$ of $Z$ under $f_ T : X \times _ S T \to T$ is not closed.

There exists a point $s \in S$ such that there is no neighborhood $U$ of $s$ in $S$ such that $X_ U$ is quasi-compact. Otherwise we could cover $S$ with finitely many such $U$ and Schemes, Lemma 25.19.2 would imply $f$ quasi-compact. Fix such an $s \in S$.

First we check that $f_ T(Z_ s) \ne T_ s$. Let $t \in T$ be the point lying over $s$ with $\kappa (t) = \kappa (s)$ such that $y_ i = 1$ in $\kappa (t)$ for all $i$. Then $t \in T_ i$ for all $i$, and the fiber of $Z_ s \to T_ s$ above $t$ is isomorphic to $(X - \bigcup _{i \in I} X_ i)_ s$, which is empty. Thus $t \in T_ s - f_ T(Z_ s)$.

Assume $f_ T(Z)$ is closed in $T$. Then there exists an element $g \in A[y_ i; i \in I]$ with $f_ T(Z) \subset V(g)$ but $t \not\in V(g)$. Hence the image of $g$ in $\kappa (t)$ is nonzero. In particular some coefficient of $g$ has nonzero image in $\kappa (s)$. Hence this coefficient is invertible on some neighborhood $U$ of $s$. Let $J$ be the finite set of $j \in I$ such that $y_ j$ appears in $g$. Since $X_ U$ is not quasi-compact, we may choose a point $x \in X - \bigcup _{j \in J} X_ j$ lying above some $u \in U$. Since $g$ has a coefficient that is invertible on $U$, we can find a point $t' \in T$ lying above $u$ such that $t' \not\in V(g)$ and $t' \in V(y_ i)$ for all $i \notin J$. This is true because $V(y_ i; i \in I, i \not\in J) = \mathop{\mathrm{Spec}}(A[t_ j; j\in J])$ and the set of points of this scheme lying over $u$ is bijective with $\mathop{\mathrm{Spec}}(\kappa (u)[t_ j; j \in J])$. In other words $t' \notin T_ i$ for each $i \notin J$. By Schemes, Lemma 25.17.5 we can find a point $z$ of $X \times _ S T$ mapping to $x \in X$ and to $t' \in T$. Since $x \not\in X_ j$ for $j \in J$ and $t' \not\in T_ i$ for $i \in I \setminus J$ we see that $z \in Z$. On the other hand $f_ T(z) = t' \not\in V(g)$ which contradicts $f_ T(Z) \subset V(g)$. Thus the assumption “$f_ T(Z)$ closed” is wrong and we conclude indeed that $f_ T$ is not closed, as desired. $\square$

Lemma 28.39.10. Suppose given a commutative diagram of schemes

Assume

$X \to S$ is a universally closed (for example proper) morphism, and

$Y \to S$ is separated and locally of finite type.

Then the scheme theoretic image $Z \subset Y$ of $h$ is proper over $S$ and $X \to Z$ is surjective.

**Proof.**
The scheme theoretic image of $h$ is constructed in Section 28.6. Since $f$ is quasi-compact (Lemma 28.39.9) we find that $h$ is quasi-compact (Schemes, Lemma 25.21.14). Hence $h(X) \subset Z$ is dense (Lemma 28.6.3). On the other hand $h(X)$ is closed in $Y$ (Lemma 28.39.7) hence $X \to Z$ is surjective. Thus $Z \to S$ is a proper (Lemma 28.39.8).
$\square$

The target of a separated scheme under a surjective universally closed morphism is separated.

Lemma 28.39.11. Let $S$ be a scheme. Let $f : X \to Y$ be a surjective universally closed morphism of schemes over $S$.

If $X$ is quasi-separated, then $Y$ is quasi-separated.

If $X$ is separated, then $Y$ is separated.

If $X$ is quasi-separated over $S$, then $Y$ is quasi-separated over $S$.

If $X$ is separated over $S$, then $Y$ is separated over $S$.

**Proof.**
Parts (1) and (2) are a consequence of (3) and (4) for $S = \mathop{\mathrm{Spec}}(\mathbf{Z})$ (see Schemes, Definition 25.21.3). Consider the commutative diagram

The left vertical arrow is surjective (i.e., universally surjective). The right vertical arrow is universally closed as a composition of the universally closed morphisms $X \times _ S X \to X \times _ S Y \to Y \times _ S Y$. Hence it is also quasi-compact, see Lemma 28.39.9.

Assume $X$ is quasi-separated over $S$, i.e., $\Delta _{X/S}$ is quasi-compact. If $V \subset Y \times _ S Y$ is a quasi-compact open, then $V \times _{Y \times _ S Y} X \to \Delta _{Y/S}^{-1}(V)$ is surjective and $V \times _{Y \times _ S Y} X$ is quasi-compact by our remarks above. We conclude that $\Delta _{Y/S}$ is quasi-compact, i.e., $Y$ is quasi-separated over $S$.

Assume $X$ is separated over $S$, i.e., $\Delta _{X/S}$ is a closed immersion. Then $X \to Y \times _ S Y$ is closed as a composition of closed morphisms. Since $X \to Y$ is surjective, it follows that $\Delta _{Y/S}(Y)$ is closed in $Y \times _ S Y$. Hence $Y$ is separated over $S$ by the discussion following Schemes, Definition 25.21.3. $\square$

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