The Stacks project

26.21 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 26.21.1. The diagonal morphism of a morphism between affines is closed.

Proof. The diagonal morphism associated to the morphism $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ 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 26.8.1 $\square$

slogan

Lemma 26.21.2. 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 26.17.2 and 26.17.3. Consider the open subscheme $W$ of $X \times _ S X$ which is the union of these affine opens $V \times _ U V$. By Lemma 26.4.2 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 26.21.1. $\square$

Definition 26.21.3. 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 $Y \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ is separated.

  4. We say a scheme $Y$ is quasi-separated if the morphism $Y \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ is quasi-separated.

By Lemmas 26.21.2 and 26.10.4 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 26.19.5 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 26.21.4. Here is an example of a non-quasi-separated morphism. Suppose $X = X_1 \cup X_2 \to S = \mathop{\mathrm{Spec}}(k)$ with $X_1 = X_2 = \mathop{\mathrm{Spec}}(k[t_1, t_2, t_3, \ldots ])$ glued along the complement of $\{ 0\} = \{ (t_1, t_2, t_3, \ldots )\} $ (glued as in Example 26.14.3). In this case the inverse image of the affine scheme $X_1 \times _ S X_2$ under $\Delta _{X/S}$ is the scheme $\mathop{\mathrm{Spec}}(k[t_1, t_2, t_3, \ldots ]) \setminus \{ 0\} $ which is not quasi-compact.

Lemma 26.21.5. 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 26.18.2, 26.21.2 and Definition 26.21.3. $\square$

Lemma 26.21.6. 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 26.17.4 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 26.19.2.

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 26.21.7. 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 $W = \mathop{\mathrm{Spec}}(R)$ be an affine open of $S$ containing both $f(U)$ and $f(V)$. Write $U = \mathop{\mathrm{Spec}}(A)$ and $V = \mathop{\mathrm{Spec}}(B)$ for $R$-algebras $A$ and $B$. By Lemma 26.17.3 we see that $U \times _ S V = U \times _ W V = \mathop{\mathrm{Spec}}(A \otimes _ R B)$ is an affine open of $X \times _ S X$. Hence, by Lemma 26.10.1 we see that $\Delta ^{-1}(U \times _ S V) \to U \times _ S V$ can be identified with $\mathop{\mathrm{Spec}}(A \otimes _ R B/J)$ 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 26.17.4). 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 26.4.2. By assumption (a) we have $U \cap V = \mathop{\mathrm{Spec}}(C)$ for some ring $C$. After choosing an affine open $W = \mathop{\mathrm{Spec}}(R)$ of $S$ into which both $U$ and $V$ map and writing $U = \mathop{\mathrm{Spec}}(A)$, $V = \mathop{\mathrm{Spec}}(B)$ 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 $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(A \otimes _ R B)$ is a closed immersion. $\square$

Example 26.21.8. Let $k$ be a field. Consider the structure morphism $p : \mathbf{P}^1_ k \to \mathop{\mathrm{Spec}}(k)$ of the projective line over $k$, see Example 26.14.4. Let us use the lemma above to prove that $p$ is separated. By construction $\mathbf{P}^1_ k$ is covered by two affine opens $U = \mathop{\mathrm{Spec}}(k[x])$ and $V = \mathop{\mathrm{Spec}}(k[y])$ with intersection $U \cap V = \mathop{\mathrm{Spec}}(k[x, y]/(xy - 1))$ (using obvious notation). Thus it suffices to check that conditions (2)(a) and (2)(b) of Lemma 26.21.7 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 26.21.9. 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 26.21.2, 26.17.6 and 26.19.3. $\square$

Lemma 26.21.10. 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 26.21.9 applied to the morphism $X = X \times _ Y Y \to X \times _ S Y$. $\square$

Lemma 26.21.11. 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 26.21.10 applied to $g =s$ so the morphism $i = s : S \to S \times _ S X$. $\square$

Lemma 26.21.12. 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 26.21.9. 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 26.17.6 and 26.19.3).

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). $\square$

slogan

Lemma 26.21.13. 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 26.21.9 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 26.10.4.

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 26.21.6. $\square$

Lemma 26.21.14. 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 26.21.11 because it is a section of the quasi-separated morphism $X \times _ Z Y \to X$ (a base change of $g$, see Lemma 26.21.12). The second map is quasi-compact as it is the base change of $g \circ f$, see Lemma 26.19.3. And compositions of quasi-compact morphisms are quasi-compact, see Lemma 26.19.4. $\square$

Lemma 26.21.15. An affine scheme is separated. A morphism from an affine scheme to another scheme is separated.

Proof. Let $U = \mathop{\mathrm{Spec}}(A)$ be an affine scheme. Then $U \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ has closed diagonal by Lemma 26.21.1. Thus $U$ is separated by Definition 26.21.3. If $U \to X$ is a morphism of schemes, then we can apply Lemma 26.21.13 to the morphisms $U \to X \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ to conclude that $U \to X$ is separated. $\square$

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 26.21.16. 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 26.21.12. Hence $U \cap V$ is affine by applying Lemma 26.21.7 to the morphism $X \to \mathop{\mathrm{Spec}}(\mathbf{Z})$. $\square$

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

Example 26.21.17. Consider the nonaffine scheme $U = \mathop{\mathrm{Spec}}(k[x, y]) \setminus \{ (x, y)\} $ of Example 26.9.3. On the other hand, consider the scheme

\[ \mathbf{GL}_{2, k} = \mathop{\mathrm{Spec}}(k[a, b, c, d, 1/ad - bc]). \]

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 26.21.18. The category of quasi-compact and quasi-separated schemes $\mathcal{C}$ has the following properties. If $X, Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, then any morphism of schemes $f : X \to Y$ is quasi-compact and quasi-separated by Lemmas 26.21.14 and 26.21.13 with $Z = \mathop{\mathrm{Spec}}(\mathbf{Z})$. 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 26.21.12 and 26.19.3. Hence the composition $X \times _ Y Z \to Z \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ is quasi-compact and quasi-separated, see Lemmas 26.21.12 and 26.19.4.


Comments (8)

Comment #2378 by Junyan Xu on

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 on

@#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 on

@#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

@#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 is affine. Then, one notes that the affine opens of the form form an open affine cover of the product (here one needs the injectivity of ). (I was surprised that this result is not stated in EGA I.)

Comment #2523 by JL on

In the second line of the proof of Lemma 25.21.8, I think should be .

Comment #2559 by on

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.

Comment #5337 by KL on

In Lemma 26.21.6, I can only see the inclusion . Surely the other inclusion only holds for ? I must be missing something.

Comment #5338 by on

@#5337: Hint: work out what happens in the category of sets.


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