Lemma 106.7.3. Let \mathcal{X} be an algebraic stack over a scheme S whose structure morphism \mathcal{X} \to S is locally of finite presentation. Let A \to B be a flat S-algebra homomorphism. Let x be an object of \mathcal{X} over A. Then T_ x(M) \otimes _ A B = T_ y(M \otimes _ A B).
Proof. Choose a scheme U and a surjective smooth morphism U \to \mathcal{X}. We first reduce the lemma to the case where x lifts to U. Recall that T_ x(M) is the set of isomorphism classes of lifts of x to A[M]. Therefore Lemma 106.7.21 says that the rule
A_1 \mapsto T_{x|_{A_1}}(M \otimes _ A A_1)
is a sheaf on the small étale site of \mathop{\mathrm{Spec}}(A); the tensor product is needed to make A[M] \to A_1[M \otimes _ A A_1] a flat ring map. We may choose a faithfully flat étale ring map A \to A_1 such that x|_{A_1} lifts to a morphism u_1 : \mathop{\mathrm{Spec}}(A_1) \to U, see for example Sheaves on Stacks, Lemma 96.19.10. Write A_2 = A_1 \otimes _ A A_1 and set B_1 = B \otimes _ A A_1 and B_2 = B \otimes _ A A_2. Consider the diagram
\xymatrix{ 0 \ar[r] & T_ y(M \otimes _ A B) \ar[r] & T_{y|_{B_1}}(M \otimes _ A B_1) \ar[r] & T_{y|_{B_2}}(M \otimes _ A B_2) \\ 0 \ar[r] & T_ x(M) \ar[r] \ar[u] & T_{x|_{A_1}}(M \otimes _ A A_1) \ar[r] \ar[u] & T_{x|_{A_2}}(M \otimes _ A A_2) \ar[u] }
The rows are exact by the sheaf condition. We have M \otimes _ A B_ i = (M \otimes _ A A_ i) \otimes _{A_ i} B_ i. Thus if we prove the result for the middle and right vertical arrow, then the result follows. This reduces us to the case discussed in the next paragraph.
Assume that x is the image of a morphism u : \mathop{\mathrm{Spec}}(A) \to U. Observe that T_ u(M) \to T_ x(M) is surjective since U \to \mathcal{X} is smooth and representable by algebraic spaces, see Criteria for Representability, Lemma 97.6.3 (see discussion preceding it for explanation) and More on Morphisms of Spaces, Lemma 76.19.6. Set R = U \times _\mathcal {X} U. Recall that we obtain a groupoid (U, R, s, t, c, e, i) in algebraic spaces with \mathcal{X} = [U/R]. By Artin's Axioms, Lemma 98.21.6 we have an exact sequence
T_{e \circ u}(M) \to T_ u(M) \oplus T_ u(M) \to T_ x(M) \to 0
where the zero on the right was shown above. A similar sequence holds for the base change to B. Thus the result we want follows if we can prove the result of the lemma for T_ u(M) and T_{e \circ u}(M). This reduces us to the case discussed in the next paragraph.
Assume that \mathcal{X} = X is an algebraic space locally of finite presentation over S. Then we have
T_ x(M) = \mathop{\mathrm{Hom}}\nolimits _ A(x^*\Omega _{X/S}, M)
by the discussion in More on Morphisms of Spaces, Section 76.17. By the same token
T_ y(M \otimes _ A B) = \mathop{\mathrm{Hom}}\nolimits _ B(y^*\Omega _{X/S}, M \otimes _ A B)
Since X \to S is locally of finite presentation, we see that \Omega _{X/S} is locally of finite presentation, see More on Morphisms of Spaces, Lemma 76.7.15. Hence x^*\Omega _{X/S} is a finitely presented A-module. Clearly, we have y^*\Omega _{X/S} = x^*\Omega _{X/S} \otimes _ A B. we conclude by More on Algebra, Lemma 15.65.4. \square
[1] This lemma applies: \Delta : \mathcal{X} \to \mathcal{X} \times _ S \mathcal{X} is locally of finite presentation by Morphisms of Stacks, Lemma 101.27.6 and the assumption that \mathcal{X} \to S is locally of finite presentation. Therefore \mathcal{I}_\mathcal {X} \to \mathcal{X} is locally of finite presentation as a base change of \Delta .
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