The Stacks project

36.14 Pushouts in the category of schemes, I

In this section we construct pushouts of $Y \leftarrow X \rightarrow X'$ where $X \to Y$ is affine and $X \to X'$ is a thickening. This will actually be an important case for us, hence a detailed discussion is merited. In Section 36.59 we discuss a more interesting and more difficult case. See Categories, Section 4.9 for a general discussion of pushouts in any category.

Lemma 36.14.1. Let $A' \to A$ be a surjection of rings and let $B \to A$ be a ring map. Let $B' = B \times _ A A'$ be the fibre product of rings. Set $S = \mathop{\mathrm{Spec}}(A)$, $S' = \mathop{\mathrm{Spec}}(A')$, $T = \mathop{\mathrm{Spec}}(B)$, and $T' = \mathop{\mathrm{Spec}}(B')$. Then

\[ \vcenter { \xymatrix{ S \ar[r]_ i \ar[d]_ f & S' \ar[d]^{f'} \\ T \ar[r]^{i'} & T' } } \quad \text{corresponding to}\quad \vcenter { \xymatrix{ A & A' \ar[l] \\ B \ar[u] & B' \ar[l] \ar[u] } } \]

is a pushout of schemes.

Proof. By More on Algebra, Lemma 15.6.2 we have $T' = T \amalg _ S S'$ as topological spaces, i.e., the diagram is a pushout in the category of topological spaces. Next, consider the map

\[ ((i')^\sharp , (f')^\sharp ) : \mathcal{O}_{T'} \longrightarrow i'_*\mathcal{O}_ T \times _{g_*\mathcal{O}_ S} f'_*\mathcal{O}_{S'} \]

where $g = i' \circ f = f' \circ i$. We claim this map is an isomorphism of sheaves of rings. Namely, we can view both sides as quasi-coherent $\mathcal{O}_{T'}$-modules (use Schemes, Lemmas 25.24.1 for the right hand side) and the map is $\mathcal{O}_{T'}$-linear. Thus it suffices to show the map is an isomorphism on the level of global sections (Schemes, Lemma 25.7.5). On global sections we recover the identification $B' \to B \times _ A A'$ from statement of the lemma (this is how we chose $B'$).

Let $X$ be a scheme. Suppose we are given morphisms of schemes $m' : S' \to X$ and $n : T \to X$ such that $m' \circ i = n \circ f$ (call this $m$). We get a unique map of topological spaces $n' : T' \to X$ compatible with $m'$ and $n$ as $T' = T \amalg _ S S'$ (see above). By the description of $\mathcal{O}_{T'}$ in the previous paragraph we obtain a unique homomorphism of sheaves of rings

\[ (n')^\sharp : \mathcal{O}_ X \longrightarrow (n')_*\mathcal{O}_{T'} = m'_*\mathcal{O}_ T \times _{m_*\mathcal{O}_ T} n_*\mathcal{O}_ S \]

given by $(m')^\sharp $ and $n^\sharp $. Thus $(n', (n')^\sharp )$ is the unque morphism of ringed spaces $T' \to X$ compatible with $m'$ and $n$. To finish the proof it suffices to show that $n'$ is a morphism of schemes, i.e., a morphism of locally ringed spaces.

Let $t' \in T'$ with image $x \in X$. We have to show that $\mathcal{O}_{X, x} \to \mathcal{O}_{T', t'}$ is local. If $t' \not\in T$, then $t'$ is the image of a unique point $s' \in S'$ and $\mathcal{O}_{T', t'} = \mathcal{O}_{S', s'}$. Namely, $S' \setminus S \to T' \setminus T$ is an isomorphism of schemes as $B' \to A'$ induces an isomorphism $\mathop{\mathrm{Ker}}(B' \to B) = \mathop{\mathrm{Ker}}(A' \to A)$. If $t'$ is the image of $t \in T$, then we know that the composition $\mathcal{O}_{X, x} \to \mathcal{O}_{T', t'} \to \mathcal{O}_{T, t}$ is local and we conclude also. $\square$

Lemma 36.14.2. Consider a commutative diagram of schemes

\[ \xymatrix{ Z \ar[r]_ i \ar[d]_ j & X \ar[d]^ a \\ Y \ar[r]^-b & W } \]

and set $c = a \circ i = b \circ j$. If there exists an fpqc covering $\{ W_ i \to W\} $ such that for all $i$ and $i'$ the diagrams

\[ \xymatrix{ Z \times _{c, W} W_ i \ar[r] \ar[d] & X \times _{a, W} W_ i \ar[d] \\ Y \times _{b, W} W_ i \ar[r] & W_ i } \quad \xymatrix{ Z \times _{c, W} (W_ i \times _ W W_{i'}) \ar[r] \ar[d] & X \times _{a, W} (W_ i \times _ W W_{i'}) \ar[d] \\ Y \times _{b, W} (W_ i \times _ W W_{i'}) \ar[r] & (W_ i \times _ W W_{i'}) } \]

are cocartesian, then so is the original diagram.

Proof. Namely, for a scheme $T$ a morphism $W \to T$ is the same thing as a collection of morphism $W_ i \to T$ which agree on overlaps, see Descent, Lemma 34.10.7. $\square$

Lemma 36.14.3. Let $X \to X'$ be a thickening of schemes and let $X \to Y$ be an affine morphism of schemes. Then there exists a pushout

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

in the category of schemes. Moreover, $Y \subset Y'$ is a thickening, $X = Y \times _{Y'} X'$, and

\[ \mathcal{O}_{Y'} = \mathcal{O}_ Y \times _{f_*\mathcal{O}_ X} f'_*\mathcal{O}_{X'} \]

as sheaves on $|Y| = |Y'|$.

Proof. We first construct $Y'$ as a ringed space. Namely, as topological space we take $Y' = Y$. Denote $f' : X' \to Y'$ the map of topological spaces which equals $f$. As structure sheaf $\mathcal{O}_{Y'}$ we take the right hand side of the equation of the lemma. To see that $Y'$ is a scheme, we have to show that any point has an affine neighbourhood. Since the formation of the fibre product of sheaves commutes with restricting to opens, we may assume $Y$ is affine. Then $X$ is affine (as $f$ is affine) and $X'$ is affine as well (see Lemma 36.2.3). Say $Y \leftarrow X \rightarrow X'$ corresponds to $B \rightarrow A \leftarrow A'$. Set $B' = B \times _ A A'$; this is the global sections of $\mathcal{O}_{Y'}$. As $A' \to A$ is surjective with locally nilpotent kernel we see that $B' \to B$ is surjective with locally nilpotent kernel. Hence $\mathop{\mathrm{Spec}}(B') = \mathop{\mathrm{Spec}}(B)$ (as topological spaces). We claim that $Y' = \mathop{\mathrm{Spec}}(B')$. To see this we will show for $g' \in B'$ with image $g \in B$ that $\mathcal{O}_{Y'}(D(g)) = B'_{g'}$. Namely, by More on Algebra, Lemma 15.5.3 we see that

\[ (B')_{g'} = B_ g \times _{A_ h} A'_{h'} \]

where $h \in A$, $h' \in A'$ are the images of $g'$. Since $B_ g$, resp. $A_ h$, resp. $A'_{h'}$ is equal to $\mathcal{O}_ Y(D(g))$, resp. $f_*\mathcal{O}_ X(D(g))$, resp. $f'_*\mathcal{O}_{X'}(D(g))$ the claim follows.

It remains to show that $Y'$ is the pushout. The discussion above shows the scheme $Y'$ has an affine open covering $Y' = \bigcup W'_ i$ such that the corresponding opens $U'_ i \subset X'$, $W_ i \subset Y$, and $U_ i \subset X$ are affine open. Moreover, if $A'_ i$, $B_ i$, $A_ i$ are the rings corresponding to $U'_ i$, $W_ i$, $U_ i$, then $W'_ i$ corresponds to $B_ i \times _{A_ i} A'_ i$. Thus we can apply Lemmas 36.14.1 and 36.14.2 to conclude our construction is a pushout in the category of schemes. $\square$

In the following lemma we use the fibre product of categories as defined in Categories, Example 4.30.3.

Lemma 36.14.4. Let $X \to X'$ be a thickening of schemes and let $X \to Y$ be an affine morphism of schemes. Let $Y' = Y \amalg _ X X'$ be the pushout (see Lemma 36.14.3). Base change gives a functor

\[ F : (\mathit{Sch}/Y') \longrightarrow (\mathit{Sch}/Y) \times _{(\mathit{Sch}/Y')} (\mathit{Sch}/X') \]

given by $V' \longmapsto (V' \times _{Y'} Y, V' \times _{Y'} X', 1)$ which has a left adjoint

\[ G : (\mathit{Sch}/Y) \times _{(\mathit{Sch}/Y')} (\mathit{Sch}/X') \longrightarrow (\mathit{Sch}/Y') \]

which sends the triple $(V, U', \varphi )$ to the pushout $V \amalg _{(V \times _ Y X)} U'$. Finally, $F \circ G$ is isomorphic to the identity functor.

Proof. Let $(V, U', \varphi )$ be an object of the fibre product category. Set $U = U' \times _{X'} X$. Note that $U \to U'$ is a thickening. Since $\varphi : V \times _ Y X \to U' \times _{X'} X = U$ is an isomorphism we have a morphism $U \to V$ over $X \to Y$ which identifies $U$ with the fibre product $X \times _ Y V$. In particular $U \to V$ is affine, see Morphisms, Lemma 28.11.8. Hence we can apply Lemma 36.14.3 to get a pushout $V' = V \amalg _ U U'$. Denote $V' \to Y'$ the morphism we obtain in virtue of the fact that $V'$ is a pushout and because we are given morphisms $V \to Y$ and $U' \to X'$ agreeing on $U$ as morphisms into $Y'$. Setting $G(V, U', \varphi ) = V'$ gives the functor $G$.

Let us prove that $G$ is a left adjoint to $F$. Let $Z$ be a scheme over $Y'$. We have to show that

\[ \mathop{Mor}\nolimits (V', Z) = \mathop{Mor}\nolimits ((V, U', \varphi ), F(Z)) \]

where the morphism sets are taking in their respective categories. Let $g' : V' \to Z$ be a morphism. Denote $\tilde g$, resp. $\tilde f'$ the composition of $g'$ with the morphism $V \to V'$, resp. $U' \to V'$. Base change $\tilde g$, resp. $\tilde f'$ by $Y \to Y'$, resp. $X' \to Y'$ to get a morphism $g : V \to Z \times _{Y'} Y$, resp. $f' : U' \to Z \times _{Y'} X'$. Then $(g, f')$ is an element of the right hand side of the equation above (details omitted). Conversely, suppose that $(g, f') : (V, U', \varphi ) \to F(Z)$ is an element of the right hand side. We may consider the composition $\tilde g : V \to Z$, resp. $\tilde f' : U' \to Z$ of $g$, resp. $f$ by $Z \times _{Y'} X' \to Z$, resp. $Z \times _{Y'} Y \to Z$. Then $\tilde g$ and $\tilde f'$ agree as morphism from $U$ to $Z$. By the universal property of pushout, we obtain a morphism $g' : V' \to Z$, i.e., an element of the left hand side. We omit the verification that these constructions are mutually inverse.

To prove that $F \circ G$ is isomorphic to the identity we have to show that the adjunction mapping $(V, U', \varphi ) \to F(G(V, U', \varphi ))$ is an isomorphism. To do this we may work affine locally. Say $X = \mathop{\mathrm{Spec}}(A)$, $X' = \mathop{\mathrm{Spec}}(A')$, and $Y = \mathop{\mathrm{Spec}}(B)$. Then $A' \to A$ and $B \to A$ are ring maps as in More on Algebra, Lemma 15.6.4 and $Y' = \mathop{\mathrm{Spec}}(B')$ with $B' = B \times _ A A'$. Next, suppose that $V = \mathop{\mathrm{Spec}}(D)$, $U' = \mathop{\mathrm{Spec}}(C')$ and $\varphi $ is given by an $A$-algebra isomorphism $D \otimes _ B A \to C' \otimes _{A'} A = C'/IC'$. Set $D' = D \times _{C'/IC'} C'$. In this case the statement we have to prove is that $D' \otimes _{B'} B \cong D$ and $D' \otimes _{B'} A' \cong C'$. This is a special case of More on Algebra, Lemma 15.6.4. $\square$

Lemma 36.14.5. Let $X \to X'$ be a thickening of schemes and let $X \to Y$ be an affine morphism of schemes. Let $Y' = Y \amalg _ X X'$ be the pushout (see Lemma 36.14.3). Let $V' \to Y'$ be a morphism of schemes. Set $V = Y \times _{Y'} V'$, $U' = X' \times _{Y'} V'$, and $U = X \times _{Y'} V'$. There is an equivalence of categories between

  1. quasi-coherent $\mathcal{O}_{V'}$-modules flat over $Y'$, and

  2. the category of triples $(\mathcal{G}, \mathcal{F}', \varphi )$ where

    1. $\mathcal{G}$ is a quasi-coherent $\mathcal{O}_ V$-module flat over $Y$,

    2. $\mathcal{F}'$ is a quasi-coherent $\mathcal{O}_{U'}$-module flat over $X'$, and

    3. $\varphi : (U \to V)^*\mathcal{G} \to (U \to U')^*\mathcal{F}'$ is an isomorphism of $\mathcal{O}_ U$-modules.

The equivalence maps $\mathcal{G}'$ to $((V \to V')^*\mathcal{G}', (U' \to V')^*\mathcal{G}', can)$. Suppose $\mathcal{G}'$ corresponds to the triple $(\mathcal{G}, \mathcal{F}', \varphi )$. Then

  1. $\mathcal{G}'$ is a finite type $\mathcal{O}_{V'}$-module if and only if $\mathcal{G}$ and $\mathcal{F}'$ are finite type $\mathcal{O}_ Y$ and $\mathcal{O}_{U'}$-modules.

  2. if $V' \to Y'$ is locally of finite presentation, then $\mathcal{G}'$ is an $\mathcal{O}_{V'}$-module of finite presentation if and only if $\mathcal{G}$ and $\mathcal{F}'$ are $\mathcal{O}_ Y$ and $\mathcal{O}_{U'}$-modules of finite presentation.

Proof. A quasi-inverse functor assigns to the triple $(\mathcal{G}, \mathcal{F}', \varphi )$ the fibre product

\[ (V \to V')_*\mathcal{G} \times _{(U \to V')_*\mathcal{F}} (U' \to V')_*\mathcal{F}' \]

where $\mathcal{F} = (U \to U')^*\mathcal{F}'$. This works, because on affines we recover the equivalence of More on Algebra, Lemma 15.7.5. Some details omitted.

Parts (a) and (b) follow from More on Algebra, Lemmas 15.7.4 and 15.7.6. $\square$

Lemma 36.14.6. In the situation of Lemma 36.14.4. If $V' = G(V, U', \varphi )$ for some triple $(V, U', \varphi )$, then

  1. $V' \to Y'$ is locally of finite type if and only if $V \to Y$ and $U' \to X'$ are locally of finite type,

  2. $V' \to Y'$ is flat if and only if $V \to Y$ and $U' \to X'$ are flat,

  3. $V' \to Y'$ is flat and locally of finite presentation if and only if $V \to Y$ and $U' \to X'$ are flat and locally of finite presentation,

  4. $V' \to Y'$ is smooth if and only if $V \to Y$ and $U' \to X'$ are smooth,

  5. $V' \to Y'$ is ├ętale if and only if $V \to Y$ and $U' \to X'$ are ├ętale, and

  6. add more here as needed.

If $W'$ is flat over $Y'$, then the adjunction mapping $G(F(W')) \to W'$ is an isomorphism. Hence $F$ and $G$ define mutually quasi-inverse functors between the category of schemes flat over $Y'$ and the category of triples $(V, U', \varphi )$ with $V \to Y$ and $U' \to X'$ flat.

Proof. Looking over affine pieces the assertions of this lemma are equivalent to the corresponding assertions of More on Algebra, Lemma 15.7.7. $\square$


Comments (2)

Comment #1255 by on

The reference in the second sentence of the section refers to this section. I guess you wish to refer to 0025?


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