# The Stacks Project

## Tag 01KH

### 25.21. Separation axioms

A topological space $X$ is Hausdorff if and only if the diagonal $\Delta \subset X \times X$ is a closed subset. The analogue in algebraic geometry is, given a scheme $X$ over a base scheme $S$, to consider the diagonal morphism $$\Delta_{X/S} : X \longrightarrow X \times_S X.$$ This is the unique morphism of schemes such that $\text{pr}_1 \circ \Delta_{X/S} = \text{id}_X$ and $\text{pr}_2 \circ \Delta_{X/S} = \text{id}_X$ (it exists in any category with fibre products).

Lemma 25.21.1. The diagonal morphism of a morphism between affines is closed.

Proof. The diagonal morphism associated to the morphism $\mathop{\rm Spec}(S) \to \mathop{\rm Spec}(R)$ is the morphism on spectra corresponding to the ring map $S \otimes_R S \to S$, $a \otimes b \mapsto ab$. This map is clearly surjective, so $S \cong S \otimes_R S/J$ for some ideal $J \subset S \otimes_R S$. Hence $\Delta$ is a closed immersion according to Example 25.8.1 $\square$

Lemma 25.21.2. Let $X$ be a scheme over $S$. The diagonal morphism $\Delta_{X/S}$ is an immersion.

Proof. Recall that if $V \subset X$ is affine open and maps into $U \subset S$ affine open, then $V \times_U V$ is affine open in $X \times_S X$, see Lemmas 25.17.2 and 25.17.3. Consider the open subscheme $W$ of $X \times_S X$ which is the union of these affine opens $V \times_U V$. By Lemma 25.4.2 it is enough to show that each morphism $\Delta_{X/S}^{-1}(V \times_U V) \to V \times_U V$ is a closed immersion. Since $V = \Delta_{X/S}^{-1}(V \times_U V)$ we are just checking that $\Delta_{V/U}$ is a closed immersion, which is Lemma 25.21.1. $\square$

Definition 25.21.3. Let $f : X \to S$ be a morphism of schemes.

1. We say $f$ is separated if the diagonal morphism $\Delta_{X/S}$ is a closed immersion.
2. We say $f$ is quasi-separated if the diagonal morphism $\Delta_{X/S}$ is a quasi-compact morphism.
3. We say a scheme $Y$ is separated if the morphism $Y \to \mathop{\rm Spec}(\mathbf{Z})$ is separated.
4. We say a scheme $Y$ is quasi-separated if the morphism $Y \to \mathop{\rm Spec}(\mathbf{Z})$ is quasi-separated.

By Lemmas 25.21.2 and 25.10.4 we see that $\Delta_{X/S}$ is a closed immersion if an only if $\Delta_{X/S}(X) \subset X \times_S X$ is a closed subset. Moreover, by Lemma 25.19.5 we see that a separated morphism is quasi-separated. The reason for introducing quasi-separated morphisms is that nonseparated morphisms come up naturally in studying algebraic varieties (especially when doing moduli, algebraic stacks, etc). But most often they are still quasi-separated.

Example 25.21.4. Here is an example of a non-quasi-separated morphism. Suppose $X = X_1 \cup X_2 \to S = \mathop{\rm Spec}(k)$ with $X_1 = X_2 = \mathop{\rm Spec}(k[t_1, t_2, t_3, \ldots])$ glued along the complement of $\{0\} = \{(t_1, t_2, t_3, \ldots)\}$ (glued as in Example 25.14.3). In this case the inverse image of the affine scheme $X_1 \times_S X_2$ under $\Delta_{X/S}$ is the scheme $\mathop{\rm Spec}(k[t_1, t_2, t_3, \ldots]) \setminus \{0\}$ which is not quasi-compact.

Lemma 25.21.5. Let $X$, $Y$ be schemes over $S$. Let $a, b : X \to Y$ be morphisms of schemes over $S$. There exists a largest locally closed subscheme $Z \subset X$ such that $a|_Z = b|_Z$. In fact $Z$ is the equalizer of $(a, b)$. Moreover, if $Y$ is separated over $S$, then $Z$ is a closed subscheme.

Proof. The equalizer of $(a, b)$ is for categorical reasons the fibre product $Z$ in the following diagram $$\xymatrix{ Z = Y \times_{(Y \times_S Y)} X \ar[r] \ar[d] & X \ar[d]^{(a , b)} \\ Y \ar[r]^-{\Delta_{Y/S}} & Y \times_S Y }$$ Thus the lemma follows from Lemmas 25.18.2, 25.21.2 and Definition 25.21.3. $\square$

Lemma 25.21.6. An affine scheme is separated. A morphism of affine schemes is separated.

Proof. See Lemma 25.21.1. $\square$

Lemma 25.21.7. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:

1. The morphism $f$ is quasi-separated.
2. For every pair of affine opens $U, V \subset X$ which map into a common affine open of $S$ the intersection $U \cap V$ is a finite union of affine opens of $X$.
3. There exists an affine open covering $S = \bigcup_{i \in I} U_i$ and for each $i$ an affine open covering $f^{-1}U_i = \bigcup_{j \in I_i} V_j$ such that for each $i$ and each pair $j, j' \in I_i$ the intersection $V_j \cap V_{j'}$ is a finite union of affine opens of $X$.

Proof. Let us prove that (3) implies (1). By Lemma 25.17.4 the covering $X \times_S X = \bigcup_i \bigcup_{j, j'} V_j \times_{U_i} V_{j'}$ is an affine open covering of $X \times_S X$. Moreover, $\Delta_{X/S}^{-1}(V_j \times_{U_i} V_{j'}) = V_j \cap V_{j'}$. Hence the implication follows from Lemma 25.19.2.

The implication (1) $\Rightarrow$ (2) follows from the fact that under the hypotheses of (2) the fibre product $U \times_S V$ is an affine open of $X \times_S X$. The implication (2) $\Rightarrow$ (3) is trivial. $\square$

Lemma 25.21.8. Let $f : X \to S$ be a morphism of schemes.

1. If $f$ is separated then for every pair of affine opens $(U, V)$ of $X$ which map into a common affine open of $S$ we have
1. the intersection $U \cap V$ is affine.
2. the ring map $\mathcal{O}_X(U) \otimes_{\mathbf{Z}} \mathcal{O}_X(V) \to \mathcal{O}_X(U \cap V)$ is surjective.
2. If any pair of points $x_1, x_2 \in X$ lying over a common point $s \in S$ are contained in affine opens $x_1 \in U$, $x_2 \in V$ which map into a common affine open of $S$ such that (a), (b) hold, then $f$ is separated.

Proof. Assume $f$ separated. Suppose $(U, V)$ is a pair as in (1). Let $W = \mathop{\rm Spec}(R)$ be an affine open of $S$ containing both $f(U)$ and $f(V)$. Write $U = \mathop{\rm Spec}(A)$ and $V = \mathop{\rm Spec}(B)$ for $R$-algebras $A$ and $B$. By Lemma 25.17.3 we see that $U \times_S V = U \times_W V = \mathop{\rm Spec}(A \otimes_R B)$ is an affine open of $X \times_S X$. Hence, by Lemma 25.10.1 we see that $\Delta^{-1}(U \times_S V) \to U \times_S V$ can be identified with $\mathop{\rm Spec}(A \otimes_R B/J)$ for some ideal $J \subset A \otimes_R B$. Thus $U \cap V = \Delta^{-1}(U \times_S V)$ is affine. Assertion (1)(b) holds because $A \otimes_{\mathbf{Z}} B \to (A \otimes_R B)/J$ is surjective.

Assume the hypothesis formulated in (2) holds. Clearly the collection of affine opens $U \times_S V$ for pairs $(U, V)$ as in (2) form an affine open covering of $X \times_S X$ (see e.g. Lemma 25.17.4). Hence it suffices to show that each morphism $U \cap V = \Delta_{X/S}^{-1}(U \times_S V) \to U \times_S V$ is a closed immersion, see Lemma 25.4.2. By assumption (a) we have $U \cap V = \mathop{\rm Spec}(C)$ for some ring $C$. After choosing an affine open $W = \mathop{\rm Spec}(R)$ of $S$ into which both $U$ and $V$ map and writing $U = \mathop{\rm Spec}(A)$, $V = \mathop{\rm Spec}(B)$ we see that the assumption (b) means that the composition $$A \otimes_{\mathbf{Z}} B \to A \otimes_R B \to C$$ is surjective. Hence $A \otimes_R B \to C$ is surjective and we conclude that $\mathop{\rm Spec}(C) \to \mathop{\rm Spec}(A \otimes_R B)$ is a closed immersion. $\square$

Example 25.21.9. Let $k$ be a field. Consider the structure morphism $p : \mathbf{P}^1_k \to \mathop{\rm Spec}(k)$ of the projective line over $k$, see Example 25.14.4. Let us use the lemma above to prove that $p$ is separated. By construction $\mathbf{P}^1_k$ is covered by two affine opens $U = \mathop{\rm Spec}(k[x])$ and $V = \mathop{\rm Spec}(k[y])$ with intersection $U \cap V = \mathop{\rm Spec}(k[x, y]/(xy - 1))$ (using obvious notation). Thus it suffices to check that conditions (2)(a) and (2)(b) of Lemma 25.21.8 hold for the pairs of affine opens $(U, U)$, $(U, V)$, $(V, U)$ and $(V, V)$. For the pairs $(U, U)$ and $(V, V)$ this is trivial. For the pair $(U, V)$ this amounts to proving that $U \cap V$ is affine, which is true, and that the ring map $$k[x] \otimes_{\mathbf{Z}} k[y] \longrightarrow k[x, y]/(xy - 1)$$ is surjective. This is clear because any element in the right hand side can be written as a sum of a polynomial in $x$ and a polynomial in $y$.

Lemma 25.21.10. Let $f : X \to T$ and $g : Y \to T$ be morphisms of schemes with the same target. Let $h : T \to S$ be a morphism of schemes. Then the induced morphism $i : X \times_T Y \to X \times_S Y$ is an immersion. If $T \to S$ is separated, then $i$ is a closed immersion. If $T \to S$ is quasi-separated, then $i$ is a quasi-compact morphism.

Proof. By general category theory the following diagram $$\xymatrix{ X \times_T Y \ar[r] \ar[d] & X \times_S Y \ar[d] \\ T \ar[r]^{\Delta_{T/S}} \ar[r] & T \times_S T }$$ is a fibre product diagram. The lemma follows from Lemmas 25.21.2, 25.17.6 and 25.19.3. $\square$

Lemma 25.21.11. Let $g : X \to Y$ be a morphism of schemes over $S$. The morphism $i : X \to X \times_S Y$ is an immersion. If $Y$ is separated over $S$ it is a closed immersion. If $Y$ is quasi-separated over $S$ it is quasi-compact.

Proof. This is a special case of Lemma 25.21.10 applied to the morphism $X = X \times_Y Y \to X \times_S Y$. $\square$

Lemma 25.21.12. Let $f : X \to S$ be a morphism of schemes. Let $s : S \to X$ be a section of $f$ (in a formula $f \circ s = \text{id}_S$). Then $s$ is an immersion. If $f$ is separated then $s$ is a closed immersion. If $f$ is quasi-separated, then $s$ is quasi-compact.

Proof. This is a special case of Lemma 25.21.11 applied to $g =s$ so the morphism $i = s : S \to S \times_S X$. $\square$

Lemma 25.21.13. Permanence properties.

1. A composition of separated morphisms is separated.
2. A composition of quasi-separated morphisms is quasi-separated.
3. The base change of a separated morphism is separated.
4. The base change of a quasi-separated morphism is quasi-separated.
5. A (fibre) product of separated morphisms is separated.
6. A (fibre) product of quasi-separated morphisms is quasi-separated.

Proof. Let $X \to Y \to Z$ be morphisms. Assume that $X \to Y$ and $Y \to Z$ are separated. The composition $$X \to X \times_Y X \to X \times_Z X$$ is closed because the first one is by assumption and the second one by Lemma 25.21.10. The same argument works for ''quasi-separated'' (with the same references).

Let $f : X \to Y$ be a morphism of schemes over a base $S$. Let $S' \to S$ be a morphism of schemes. Let $f' : X_{S'} \to Y_{S'}$ be the base change of $f$. Then the diagonal morphism of $f'$ is a morphism $$\Delta_{f'} : X_{S'} = S' \times_S X \longrightarrow X_{S'} \times_{Y_{S'}} X_{S'} = S' \times _S (X \times_Y X)$$ which is easily seen to be the base change of $\Delta_f$. Thus (3) and (4) follow from the fact that closed immersions and quasi-compact morphisms are preserved under arbitrary base change (Lemmas 25.17.6 and 25.19.3).

If $f : X \to Y$ and $g : U \to V$ are morphisms of schemes over a base $S$, then $f \times g$ is the composition of $X \times_S U \to X \times_S V$ (a base change of $g$) and $X \times_S V \to Y \times_S V$ (a base change of $f$). Hence (5) and (6) follow from (1) – (4). $\square$

Lemma 25.21.14. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of schemes. If $g \circ f$ is separated then so is $f$. If $g \circ f$ is quasi-separated then so is $f$.

Proof. Assume that $g \circ f$ is separated. Consider the factorization $X \to X \times_Y X \to X \times_Z X$ of the diagonal morphism of $g \circ f$. By Lemma 25.21.10 the last morphism is an immersion. By assumption the image of $X$ in $X \times_Z X$ is closed. Hence it is also closed in $X \times_Y X$. Thus we see that $X \to X \times_Y X$ is a closed immersion by Lemma 25.10.4.

Assume that $g \circ f$ is quasi-separated. Let $V \subset Y$ be an affine open which maps into an affine open of $Z$. Let $U_1, U_2 \subset X$ be affine opens which map into $V$. Then $U_1 \cap U_2$ is a finite union of affine opens because $U_1, U_2$ map into a common affine open of $Z$. Since we may cover $Y$ by affine opens like $V$ we deduce the lemma from Lemma 25.21.7. $\square$

Lemma 25.21.15. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of schemes. If $g \circ f$ is quasi-compact and $g$ is quasi-separated then $f$ is quasi-compact.

Proof. This is true because $f$ equals the composition $(1, f) : X \to X \times_Z Y \to Y$. The first map is quasi-compact by Lemma 25.21.12 because it is a section of the quasi-separated morphism $X \times_Z Y \to X$ (a base change of $g$, see Lemma 25.21.13). The second map is quasi-compact as it is the base change of $f$, see Lemma 25.19.3. And compositions of quasi-compact morphisms are quasi-compact, see Lemma 25.19.4. $\square$

You may have been wondering whether the condition of only considering pairs of affine opens whose image is contained in an affine open is really necessary to be able to conclude that their intersection is affine. Often it isn't!

Lemma 25.21.16. Let $f : X \to S$ be a morphism. Assume $f$ is separated and $S$ is a separated scheme. Suppose $U \subset X$ and $V \subset X$ are affine. Then $U \cap V$ is affine (and a closed subscheme of $U \times V$).

Proof. In this case $X$ is separated by Lemma 25.21.13. Hence $U \cap V$ is affine by applying Lemma 25.21.8 to the morphism $X \to \mathop{\rm Spec}(\mathbf{Z})$. $\square$

On the other hand, the following example shows that we cannot expect the image of an affine to be contained in an affine.

Example 25.21.17. Consider the nonaffine scheme $U = \mathop{\rm Spec}(k[x, y]) \setminus \{(x, y)\}$ of Example 25.9.3. On the other hand, consider the scheme $$\mathbf{GL}_{2, k} = \mathop{\rm Spec}(k[a, b, c, d, 1/ad - bc]).$$ There is a morphism $\mathbf{GL}_{2, k} \to U$ corresponding to the ring map $x \mapsto a$, $y \mapsto b$. It is easy to see that this is a surjective morphism, and hence the image is not contained in any affine open of $U$. In fact, the affine scheme $\mathbf{GL}_{2, k}$ also surjects onto $\mathbf{P}^1_k$, and $\mathbf{P}^1_k$ does not even have an immersion into any affine scheme.

Remark 25.21.18. The category of quasi-compact and quasi-separated schemes $\mathcal{C}$ has the following properties. If $X, Y \in \mathop{\rm Ob}\nolimits(\mathcal{C})$, then any morphism of schemes $f : X \to Y$ is quasi-compact and quasi-separated by Lemmas 25.21.15 and 25.21.14 with $Z = \mathop{\rm Spec}(\mathbf{Z})$. Moreover, if $X \to Y$ and $Z \to Y$ are morphisms $\mathcal{C}$, then $X \times_Y Z$ is an object of $\mathcal{C}$ too. Namely, the projection $X \times_Y Z \to Z$ is quasi-compact and quasi-separated as a base change of the morphism $Z \to Y$, see Lemmas 25.21.13 and 25.19.3. Hence the composition $X \times_Y Z \to Z \to \mathop{\rm Spec}(\mathbf{Z})$ is quasi-compact and quasi-separated, see Lemmas 25.21.13 and 25.19.4.

The code snippet corresponding to this tag is a part of the file schemes.tex and is located in lines 3822–4278 (see updates for more information).

\section{Separation axioms}
\label{section-separation-axioms}

\noindent
A topological space $X$ is Hausdorff if and only if the
diagonal $\Delta \subset X \times X$ is a closed subset.
The analogue in algebraic geometry is, given a scheme $X$ over
a base scheme $S$, to consider the
diagonal morphism
$$\Delta_{X/S} : X \longrightarrow X \times_S X.$$
This is the unique morphism of schemes such that
$\text{pr}_1 \circ \Delta_{X/S} = \text{id}_X$ and
$\text{pr}_2 \circ \Delta_{X/S} = \text{id}_X$ (it exists in
any category with fibre products).

\begin{lemma}
\label{lemma-diagonal-affines-closed}
The diagonal morphism of a morphism between affines is closed.
\end{lemma}

\begin{proof}
The diagonal morphism associated to the morphism
$\Spec(S) \to \Spec(R)$ is the morphism on spectra
corresponding to the ring
map $S \otimes_R S \to S$, $a \otimes b \mapsto ab$.
This map is clearly surjective, so $S \cong S \otimes_R S/J$
for some ideal $J \subset S \otimes_R S$. Hence
$\Delta$ is a closed immersion according to
Example \ref{example-closed-immersion-affines}
\end{proof}

\begin{lemma}
\label{lemma-diagonal-immersion}
\begin{slogan}
The diagonal morphism for relative schemes is an immersion.
\end{slogan}
Let $X$ be a scheme over $S$.
The diagonal morphism $\Delta_{X/S}$ is an immersion.
\end{lemma}

\begin{proof}
Recall that if $V \subset X$ is affine open and maps into
$U \subset S$ affine open, then $V \times_U V$ is affine open
in $X \times_S X$, see Lemmas \ref{lemma-fibre-product-affines}
and \ref{lemma-open-fibre-product}.
Consider the open subscheme $W$ of $X \times_S X$ which
is the union of these affine opens $V \times_U V$.
By Lemma \ref{lemma-closed-local-target} it is enough
to show that each morphism
$\Delta_{X/S}^{-1}(V \times_U V) \to V \times_U V$ is
a closed immersion. Since $V = \Delta_{X/S}^{-1}(V \times_U V)$
we are just checking that $\Delta_{V/U}$ is a closed
immersion, which is Lemma \ref{lemma-diagonal-affines-closed}.
\end{proof}

\begin{definition}
\label{definition-separated}
Let $f : X \to S$ be a morphism of schemes.
\begin{enumerate}
\item We say $f$ is {\it separated} if the diagonal morphism $\Delta_{X/S}$
is a closed immersion.
\item We say $f$ is {\it quasi-separated} if the diagonal morphism
$\Delta_{X/S}$ is a quasi-compact morphism.
\item We say a scheme $Y$ is {\it separated} if the morphism
$Y \to \Spec(\mathbf{Z})$ is separated.
\item We say a scheme $Y$ is {\it quasi-separated} if the morphism
$Y \to \Spec(\mathbf{Z})$ is quasi-separated.
\end{enumerate}
\end{definition}

\noindent
By Lemmas \ref{lemma-diagonal-immersion} and \ref{lemma-immersion-when-closed}
we see that $\Delta_{X/S}$ is a closed immersion if an only if
$\Delta_{X/S}(X) \subset X \times_S X$ is a closed subset. Moreover,
by Lemma \ref{lemma-closed-immersion-quasi-compact} we see that a
separated morphism is quasi-separated. The reason for introducing
quasi-separated morphisms is that nonseparated morphisms come up naturally
in studying algebraic varieties (especially when doing moduli,
algebraic stacks, etc). But most often they are still quasi-separated.

\begin{example}
\label{example-not-quasi-separated}
Here is an example of a non-quasi-separated morphism.
Suppose $X = X_1 \cup X_2 \to S = \Spec(k)$ with
$X_1 = X_2 = \Spec(k[t_1, t_2, t_3, \ldots])$
glued along the complement of $\{0\} = \{(t_1, t_2, t_3, \ldots)\}$
(glued as in Example \ref{example-affine-space-zero-doubled}).
In this case the inverse image of the affine scheme
$X_1 \times_S X_2$ under $\Delta_{X/S}$ is the scheme
$\Spec(k[t_1, t_2, t_3, \ldots]) \setminus \{0\}$
which is not quasi-compact.
\end{example}

\begin{lemma}
\label{lemma-where-are-they-equal}
Let $X$, $Y$ be schemes over $S$.
Let $a, b : X \to Y$ be morphisms of schemes over $S$.
There exists a largest locally closed subscheme
$Z \subset X$ such that $a|_Z = b|_Z$. In fact $Z$ is
the equalizer of $(a, b)$. Moreover, if $Y$ is separated
over $S$, then $Z$ is a closed subscheme.
\end{lemma}

\begin{proof}
The equalizer of $(a, b)$ is for categorical reasons
the fibre product $Z$ in the following diagram
$$\xymatrix{ Z = Y \times_{(Y \times_S Y)} X \ar[r] \ar[d] & X \ar[d]^{(a , b)} \\ Y \ar[r]^-{\Delta_{Y/S}} & Y \times_S Y }$$
Thus the lemma follows from Lemmas
\ref{lemma-base-change-immersion}, \ref{lemma-diagonal-immersion} and
Definition \ref{definition-separated}.
\end{proof}

\begin{lemma}
\label{lemma-affine-separated}
An affine scheme is separated.
A morphism of affine schemes is separated.
\end{lemma}

\begin{proof}
See Lemma \ref{lemma-diagonal-affines-closed}.
\end{proof}

\begin{lemma}
\label{lemma-characterize-quasi-separated}
Let $f : X \to S$ be a morphism of schemes.
The following are equivalent:
\begin{enumerate}
\item The morphism $f$ is quasi-separated.
\item For every pair of affine opens $U, V \subset X$
which map into a common affine open of $S$ the intersection
$U \cap V$ is a finite union of affine opens of $X$.
\item There exists an affine open covering $S = \bigcup_{i \in I} U_i$
and for each $i$ an affine open covering $f^{-1}U_i = \bigcup_{j \in I_i} V_j$
such that for each $i$ and each pair $j, j' \in I_i$ the
intersection $V_j \cap V_{j'}$ is a finite union of affine
opens of $X$.
\end{enumerate}
\end{lemma}

\begin{proof}
Let us prove that (3) implies (1).
By Lemma \ref{lemma-affine-covering-fibre-product}
the covering $X \times_S X = \bigcup_i \bigcup_{j, j'} V_j \times_{U_i} V_{j'}$
is an affine open covering of $X \times_S X$.
Moreover, $\Delta_{X/S}^{-1}(V_j \times_{U_i} V_{j'}) = V_j \cap V_{j'}$.
Hence the implication follows from Lemma \ref{lemma-quasi-compact-affine}.

\medskip\noindent
The implication (1) $\Rightarrow$ (2) follows from the fact
that under the hypotheses of (2) the fibre product
$U \times_S V$ is an affine open of $X \times_S X$.
The implication (2) $\Rightarrow$ (3) is trivial.
\end{proof}

\begin{lemma}
\label{lemma-characterize-separated}
Let $f : X \to S$ be a morphism of schemes.
\begin{enumerate}
\item If $f$ is separated then for every pair of affine
opens $(U, V)$ of $X$ which map into a
common affine open of $S$ we have
\begin{enumerate}
\item the intersection $U \cap V$ is affine.
\item the ring map
$\mathcal{O}_X(U) \otimes_{\mathbf{Z}} \mathcal{O}_X(V) \to \mathcal{O}_X(U \cap V)$
is surjective.
\end{enumerate}
\item If any pair of points $x_1, x_2 \in X$ lying over a common
point $s \in S$ are contained in affine opens $x_1 \in U$,
$x_2 \in V$ which map into a common affine open of $S$ such
that (a), (b) hold, then $f$ is separated.
\end{enumerate}
\end{lemma}

\begin{proof}
Assume $f$ separated. Suppose $(U, V)$ is a pair as in (1).
Let $W = \Spec(R)$ be an affine open of $S$ containing
both $f(U)$ and $f(V)$. Write $U = \Spec(A)$ and
$V = \Spec(B)$ for $R$-algebras $A$ and $B$.
By Lemma \ref{lemma-open-fibre-product} we see that
$U \times_S V = U \times_W V = \Spec(A \otimes_R B)$
is an affine open of $X \times_S X$. Hence, by
Lemma \ref{lemma-closed-subspace-scheme} we see that
$\Delta^{-1}(U \times_S V) \to U \times_S V$
can be identified with $\Spec(A \otimes_R B/J)$
for some ideal $J \subset A \otimes_R B$.
Thus $U \cap V = \Delta^{-1}(U \times_S V)$ is affine.
Assertion (1)(b) holds because
$A \otimes_{\mathbf{Z}} B \to (A \otimes_R B)/J$ is surjective.

\medskip\noindent
Assume the hypothesis formulated in (2) holds.
Clearly the collection of affine opens $U \times_S V$
for pairs $(U, V)$ as in (2) form an affine open covering
of $X \times_S X$ (see e.g.\ Lemma \ref{lemma-affine-covering-fibre-product}).
Hence it suffices to show that each morphism
$U \cap V = \Delta_{X/S}^{-1}(U \times_S V) \to U \times_S V$
is a closed immersion, see Lemma \ref{lemma-closed-local-target}.
By assumption (a) we have $U \cap V = \Spec(C)$ for some ring $C$.
After choosing an affine open $W = \Spec(R)$ of $S$
into which both $U$ and $V$ map and writing $U = \Spec(A)$,
$V = \Spec(B)$ we see that the assumption (b) means
that the composition
$$A \otimes_{\mathbf{Z}} B \to A \otimes_R B \to C$$
is surjective. Hence $A \otimes_R B \to C$ is surjective and
we conclude that $\Spec(C) \to \Spec(A \otimes_R B)$
is a closed immersion.
\end{proof}

\begin{example}
\label{example-projective-line-separated}
Let $k$ be a field. Consider the structure morphism
$p : \mathbf{P}^1_k \to \Spec(k)$ of the projective
line over $k$, see Example \ref{example-projective-line}.
Let us use the lemma above to prove that $p$
is separated. By construction $\mathbf{P}^1_k$ is covered by two
affine opens $U = \Spec(k[x])$ and $V = \Spec(k[y])$
with intersection $U \cap V = \Spec(k[x, y]/(xy - 1))$
(using obvious notation). Thus it suffices to check that
conditions (2)(a) and (2)(b) of Lemma \ref{lemma-characterize-separated}
hold for the pairs of affine opens $(U, U)$, $(U, V)$, $(V, U)$
and $(V, V)$. For the pairs $(U, U)$ and $(V, V)$ this is trivial.
For the pair $(U, V)$ this amounts to proving
that $U \cap V$ is affine, which is true, and that the ring map
$$k[x] \otimes_{\mathbf{Z}} k[y] \longrightarrow k[x, y]/(xy - 1)$$
is surjective. This is clear because any element in the
right hand side can be written as a sum of a polynomial
in $x$ and a polynomial in $y$.
\end{example}

\begin{lemma}
\label{lemma-fibre-product-after-map}
Let $f : X \to T$ and $g : Y \to T$ be morphisms of schemes
with the same target. Let $h : T \to S$ be a morphism of schemes.
Then the induced morphism $i : X \times_T Y \to X \times_S Y$ is
an immersion. If $T \to S$ is separated, then $i$ is a closed
immersion. If $T \to S$ is quasi-separated, then $i$ is a
quasi-compact morphism.
\end{lemma}

\begin{proof}
By general category theory the following diagram
$$\xymatrix{ X \times_T Y \ar[r] \ar[d] & X \times_S Y \ar[d] \\ T \ar[r]^{\Delta_{T/S}} \ar[r] & T \times_S T }$$
is a fibre product diagram. The lemma follows
from Lemmas \ref{lemma-diagonal-immersion},
\ref{lemma-fibre-product-immersion} and
\ref{lemma-quasi-compact-preserved-base-change}.
\end{proof}

\begin{lemma}
\label{lemma-semi-diagonal}
Let $g : X \to Y$ be a morphism of schemes over $S$.
The morphism $i : X \to X \times_S Y$ is an immersion.
If $Y$ is separated over $S$ it is a closed immersion.
If $Y$ is quasi-separated over $S$ it is quasi-compact.
\end{lemma}

\begin{proof}
This is a special case of Lemma \ref{lemma-fibre-product-after-map}
applied to the morphism $X = X \times_Y Y \to X \times_S Y$.
\end{proof}

\begin{lemma}
\label{lemma-section-immersion}
Let $f : X \to S$ be a morphism of schemes.
Let $s : S \to X$ be a section of $f$ (in a formula $f \circ s = \text{id}_S$).
Then $s$ is an immersion.
If $f$ is separated then $s$ is a closed immersion.
If $f$ is quasi-separated, then $s$ is quasi-compact.
\end{lemma}

\begin{proof}
This is a special case of Lemma \ref{lemma-semi-diagonal} applied to
$g =s$ so the morphism $i = s : S \to S \times_S X$.
\end{proof}

\begin{lemma}
\label{lemma-separated-permanence}
Permanence properties.
\begin{enumerate}
\item A composition of separated morphisms is separated.
\item A composition of quasi-separated morphisms is quasi-separated.
\item The base change of a separated morphism is separated.
\item The base change of a quasi-separated morphism is quasi-separated.
\item A (fibre) product of separated morphisms is separated.
\item A (fibre) product of quasi-separated morphisms is quasi-separated.
\end{enumerate}
\end{lemma}

\begin{proof}
Let $X \to Y \to Z$ be morphisms. Assume that $X \to Y$ and
$Y \to Z$ are separated. The composition
$$X \to X \times_Y X \to X \times_Z X$$
is closed because the first one is by assumption and the second
one by Lemma \ref{lemma-fibre-product-after-map}. The same argument
works for quasi-separated'' (with the same references).

\medskip\noindent
Let $f : X \to Y$ be a morphism of schemes over a base $S$.
Let $S' \to S$ be a morphism of schemes. Let $f' : X_{S'} \to Y_{S'}$
be the base change of $f$. Then the diagonal morphism
of $f'$ is a morphism
$$\Delta_{f'} : X_{S'} = S' \times_S X \longrightarrow X_{S'} \times_{Y_{S'}} X_{S'} = S' \times _S (X \times_Y X)$$
which is easily seen to be the base change of $\Delta_f$.
Thus (3) and (4) follow from the fact that
closed immersions and quasi-compact morphisms are preserved
under arbitrary base change (Lemmas
\ref{lemma-fibre-product-immersion} and
\ref{lemma-quasi-compact-preserved-base-change}).

\medskip\noindent
If $f : X \to Y$ and $g : U \to V$ are morphisms of schemes over a base $S$,
then $f \times g$ is the composition of $X \times_S U \to X \times_S V$
(a base change of $g$) and $X \times_S V \to Y \times_S V$ (a base change
of $f$). Hence (5) and (6) follow from (1) -- (4).
\end{proof}

\begin{lemma}
\label{lemma-compose-after-separated}
\begin{slogan}
Separated and quasi-separated morphisms satisfy cancellation.
\end{slogan}
Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of schemes.
If $g \circ f$ is separated then so is $f$.
If $g \circ f$ is quasi-separated then so is $f$.
\end{lemma}

\begin{proof}
Assume that $g \circ f$ is separated.
Consider the factorization $X \to X \times_Y X \to X \times_Z X$
of the diagonal morphism of $g \circ f$.
By Lemma \ref{lemma-fibre-product-after-map}
the last morphism is an immersion. By assumption the image
of $X$ in $X \times_Z X$ is closed. Hence it is also closed
in $X \times_Y X$. Thus we see that $X \to X \times_Y X$
is a closed immersion by Lemma \ref{lemma-immersion-when-closed}.

\medskip\noindent
Assume that $g \circ f$ is quasi-separated.
Let $V \subset Y$ be an affine open which maps into an affine
open of $Z$. Let $U_1, U_2 \subset X$ be affine opens which
map into $V$. Then $U_1 \cap U_2$ is a finite union of affine
opens because $U_1, U_2$ map into a common affine open
of $Z$. Since we may cover $Y$ by affine opens like $V$ we
deduce the lemma from Lemma \ref{lemma-characterize-quasi-separated}.
\end{proof}

\begin{lemma}
\label{lemma-quasi-compact-permanence}
Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of schemes.
If $g \circ f$ is quasi-compact and $g$ is quasi-separated
then $f$ is quasi-compact.
\end{lemma}

\begin{proof}
This is true because $f$ equals the composition
$(1, f) : X \to X \times_Z Y \to Y$. The first map
is quasi-compact by Lemma \ref{lemma-section-immersion}
because it is a section of the quasi-separated morphism $X \times_Z Y \to X$
(a base change of $g$, see Lemma \ref{lemma-separated-permanence}).
The second map is quasi-compact as it
is the base change of $f$, see
Lemma \ref{lemma-quasi-compact-preserved-base-change}.
And compositions of quasi-compact
morphisms are quasi-compact, see Lemma \ref{lemma-composition-quasi-compact}.
\end{proof}

\noindent
You may have been wondering whether the condition
of only considering pairs of affine opens whose image is contained
in an affine open is really necessary to be able to conclude
that their intersection is affine. Often it isn't!

\begin{lemma}
\label{lemma-curiosity}
Let $f : X \to S$ be a morphism.
Assume $f$ is separated and $S$ is a separated scheme.
Suppose $U \subset X$ and $V \subset X$ are affine.
Then $U \cap V$ is affine (and a closed subscheme of $U \times V$).
\end{lemma}

\begin{proof}
In this case $X$ is separated by Lemma \ref{lemma-separated-permanence}.
Hence $U \cap V$ is affine by
applying Lemma \ref{lemma-characterize-separated} to the
morphism $X \to \Spec(\mathbf{Z})$.
\end{proof}

\noindent
On the other hand, the following example shows that we cannot
expect the image of an affine to be contained in an affine.

\begin{example}
\label{example-image-affine-projective}
Consider the nonaffine scheme
$U = \Spec(k[x, y]) \setminus \{(x, y)\}$ of
Example \ref{example-not-affine}. On the other hand, consider the
scheme
$$\mathbf{GL}_{2, k} = \Spec(k[a, b, c, d, 1/ad - bc]).$$
There is a morphism $\mathbf{GL}_{2, k} \to U$ corresponding
to the ring map $x \mapsto a$, $y \mapsto b$. It is easy to see that
this is a surjective morphism, and hence the image is not contained
in any affine open of $U$. In fact, the affine scheme
$\mathbf{GL}_{2, k}$ also surjects onto $\mathbf{P}^1_k$, and
$\mathbf{P}^1_k$ does not even have an immersion into {\it any} affine scheme.
\end{example}

\begin{remark}
\label{remark-quasi-compact-and-quasi-separated}
The category of quasi-compact and quasi-separated schemes $\mathcal{C}$
has the following properties. If $X, Y \in \Ob(\mathcal{C})$, then any
morphism of schemes $f : X \to Y$ is quasi-compact and quasi-separated by
Lemmas \ref{lemma-quasi-compact-permanence} and
\ref{lemma-compose-after-separated}
with $Z = \Spec(\mathbf{Z})$. Moreover, if $X \to Y$ and $Z \to Y$
are morphisms $\mathcal{C}$, then $X \times_Y Z$ is an object of $\mathcal{C}$
too. Namely, the projection $X \times_Y Z \to Z$ is quasi-compact and
quasi-separated as a base change of the morphism $Z \to Y$, see
Lemmas \ref{lemma-separated-permanence} and
\ref{lemma-quasi-compact-preserved-base-change}.
Hence the composition $X \times_Y Z \to Z \to \Spec(\mathbf{Z})$
is quasi-compact and quasi-separated, see
Lemmas \ref{lemma-separated-permanence} and
\ref{lemma-composition-quasi-compact}.
\end{remark}

Comment #2378 by Junyan Xu on February 15, 2017 a 7:51 pm UTC

If a morphism of schemes is universally injective (aka radicial), then the diagonal morphism is surjective, hence have closed image, so the original morphism is separated. However, if a morphism is just injective on the underlying set, the diagonal morphism isn't necessarily surjective, and it seems that one has to use 25.21.8 to prove that the original morphism is separated in this case also.

Comment #2379 by Johan (site) on February 16, 2017 a 12:50 pm UTC

@#2378: I would do this using the valuative criterion. Thanks for the comment; I will add this later and notify you here.

Comment #2433 by Johan (site) on February 17, 2017 a 3:10 pm UTC

@#2378, @#2379: I just realized what I wanted to do does not work because we don't know the morphism is quasi-separated. So now I am no longer sure it is correct that an injective morphism of schemes is separated. Is it?

Comment #2517 by Ammar Y. Kılıç on April 25, 2017 a 4:19 pm UTC

@#2433: I was asking myself the same question (injectivity implies separated?). On math.stackexchange I was referred to [Bosch - Algebraic Geometry and Commutative Algebra, corollary 7.4.10]. Since the notion of separability is local on the target, one reduces to the case where the base scheme $S$ is affine. Then, one notes that the affine opens of the form $U\times_S U$ form an open affine cover of the product $X\times_S X$ (here one needs the injectivity of $X\to S$). (I was surprised that this result is not stated in EGA I.)

Comment #2523 by JL on April 28, 2017 a 8:38 am UTC

In the second line of the proof of Lemma 25.21.8, I think $g(V)$ should be $f(V)$.

Comment #2559 by Johan (site) on May 25, 2017 a 6:19 pm UTC

Thanks to all. I've added this as a lemma in a later section. @#2517: If you want your name correctly spelled in the Stacks project, then send me the latex for your name please. The fix is here.

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