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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