85.27 Descent in terms of simplicial schemes
Cartesian morphisms are defined as follows.
Definition 85.27.1. Let a : Y \to X be a morphism of simplicial schemes. We say a is cartesian, or that Y is cartesian over X, if for every morphism \varphi : [n] \to [m] of \Delta the corresponding diagram
\xymatrix{ Y_ m \ar[r]_ a \ar[d]_{Y(\varphi )} & X_ m \ar[d]^{X(\varphi )}\\ Y_ n \ar[r]^{a} & X_ n }
is a fibre square in the category of schemes.
Cartesian morphisms are related to descent data. First we prove a general lemma describing the category of cartesian simplicial schemes over a fixed simplicial scheme. In this lemma we denote f^* : \mathit{Sch}/X \to \mathit{Sch}/Y the base change functor associated to a morphism of schemes f :Y \to X.
Lemma 85.27.2. Let X be a simplicial scheme. The category of simplicial schemes cartesian over X is equivalent to the category of pairs (V, \varphi ) where V is a scheme over X_0 and
\varphi : V \times _{X_0, d^1_1} X_1 \longrightarrow X_1 \times _{d^1_0, X_0} V
is an isomorphism over X_1 such that (s_0^0)^*\varphi = \text{id}_ V and such that
(d^2_1)^*\varphi = (d^2_0)^*\varphi \circ (d^2_2)^*\varphi
as morphisms of schemes over X_2.
Proof.
The statement of the displayed equality makes sense because d^1_1 \circ d^2_2 = d^1_1 \circ d^2_1, d^1_1 \circ d^2_0 = d^1_0 \circ d^2_2, and d^1_0 \circ d^2_0 = d^1_0 \circ d^2_1 as morphisms X_2 \to X_0, see Simplicial, Remark 14.3.3 hence we can picture these maps as follows
\xymatrix{ & X_2 \times _{d^1_1 \circ d^2_0, X_0} V \ar[r]_-{(d^2_0)^*\varphi } & X_2 \times _{d^1_0 \circ d^2_0, X_0} V \ar@{=}[rd] & \\ X_2 \times _{d^1_0 \circ d^2_2, X_0} V \ar@{=}[ru] & & & X_2 \times _{d^1_0 \circ d^2_1, X_0} V \\ & X_2 \times _{d^1_1 \circ d^2_2, X_0} V \ar[lu]^{(d^2_2)^*\varphi } \ar@{=}[r] & X_2 \times _{d^1_1 \circ d^2_1, X_0} V \ar[ru]_{(d^2_1)^*\varphi } }
and the condition signifies the diagram is commutative. It is clear that given a simplicial scheme Y cartesian over X we can set V = Y_0 and \varphi equal to the composition
V \times _{X_0, d^1_1} X_1 = Y_0 \times _{X_0, d^1_1} X_1 = Y_1 = X_1 \times _{X_0, d^1_0} Y_0 = X_1 \times _{X_0, d^1_0} V
of identifications given by the cartesian structure. To prove this functor is an equivalence we construct a quasi-inverse. The construction of the quasi-inverse is analogous to the construction discussed in Descent, Section 35.3 from which we borrow the notation \tau ^ n_ i : [0] \to [n], 0 \mapsto i and \tau ^ n_{ij} : [1] \to [n], 0 \mapsto i, 1 \mapsto j. Namely, given a pair (V, \varphi ) as in the lemma we set Y_ n = X_ n \times _{X(\tau ^ n_ n), X_0} V. Then given \beta : [n] \to [m] we define V(\beta ) : Y_ m \to Y_ n as the pullback by X(\tau ^ m_{\beta (n)m}) of the map \varphi postcomposed by the projection X_ m \times _{X(\beta ), X_ n} Y_ n \to Y_ n. This makes sense because
X_ m \times _{X(\tau ^ m_{\beta (n)m}), X_1} X_1 \times _{d^1_1, X_0} V = X_ m \times _{X(\tau ^ m_ m), X_0} V = Y_ m
and
X_ m \times _{X(\tau ^ m_{\beta (n)m}), X_1} X_1 \times _{d^1_0, X_0} V = X_ m \times _{X(\tau ^ m_{\beta (n)}), X_0} V = X_ m \times _{X(\beta ), X_ n} Y_ n.
We omit the verification that the commutativity of the displayed diagram above implies the maps compose correctly. We also omit the verification that the two functors are quasi-inverse to each other.
\square
Definition 85.27.3. Let f : X \to S be a morphism of schemes. The simplicial scheme associated to f, denoted (X/S)_\bullet , is the functor \Delta ^{opp} \to \mathit{Sch}, [n] \mapsto X \times _ S \ldots \times _ S X described in Simplicial, Example 14.3.5.
Thus (X/S)_ n is the (n + 1)-fold fibre product of X over S. The morphism d^1_0 : X \times _ S X \to X is the map (x_0, x_1) \mapsto x_1 and the morphism d^1_1 is the other projection. The morphism s^0_0 is the diagonal morphism X \to X \times _ S X.
Lemma 85.27.4. Let f : X \to S be a morphism of schemes. Let \pi : Y \to (X/S)_\bullet be a cartesian morphism of simplicial schemes. Set V = Y_0 considered as a scheme over X. The morphisms d^1_0, d^1_1 : Y_1 \to Y_0 and the morphism \pi _1 : Y_1 \to X \times _ S X induce isomorphisms
\xymatrix{ V \times _ S X & & Y_1 \ar[ll]_-{(d^1_1, \text{pr}_1 \circ \pi _1)} \ar[rr]^-{(\text{pr}_0 \circ \pi _1, d^1_0)} & & X \times _ S V. }
Denote \varphi : V \times _ S X \to X \times _ S V the resulting isomorphism. Then the pair (V, \varphi ) is a descent datum relative to X \to S.
Proof.
This is a special case of (part of) Lemma 85.27.2 as the displayed equation of that lemma is equivalent to the cocycle condition of Descent, Definition 35.34.1.
\square
Lemma 85.27.5. Let f : X \to S be a morphism of schemes. The construction
\begin{matrix} \text{category of cartesian }
\\ \text{schemes over } (X/S)_\bullet
\end{matrix} \longrightarrow \begin{matrix} \text{ category of descent data}
\\ \text{ relative to } X/S
\end{matrix}
of Lemma 85.27.4 is an equivalence of categories.
Proof.
The functor from left to right is given in Lemma 85.27.4. Hence this is a special case of Lemma 85.27.2.
\square
We may reinterpret the pullback of Descent, Lemma 35.34.6 as follows. Suppose given a morphism of simplicial schemes f : X' \to X and a cartesian morphism of simplicial schemes Y \to X. Then the fibre product (viewed as a “pullback”)
f^*Y = Y \times _ X X'
of simplicial schemes is a simplicial scheme cartesian over X'. Suppose given a commutative diagram of morphisms of schemes
\xymatrix{ X' \ar[r]_ f \ar[d] & X \ar[d] \\ S' \ar[r] & S. }
This gives rise to a morphism of simplicial schemes
f_\bullet : (X'/S')_\bullet \longrightarrow (X/S)_\bullet .
We claim that the “pullback” f_\bullet ^* along the morphism f_\bullet : (X'/S')_\bullet \to (X/S)_\bullet corresponds via Lemma 85.27.5 with the pullback defined in terms of descent data in the aforementioned Descent, Lemma 35.34.6.
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