Lemma 106.16.7. Let \mathcal{X} be a Noetherian algebraic stack with affine diagonal. Let B be a ring. Let F : \mathit{QCoh}(\mathcal{O}_\mathcal {X}) \to \text{Mod}_ B be a right exact tensor functor which commutes with direct sums. Then F comes from a unique morphism \mathop{\mathrm{Spec}}(B) \to \mathcal{X}.
Proof. Choose a surjective smooth morphism g : U \to \mathcal{X} with U = \mathop{\mathrm{Spec}}(A) affine, see Properties of Stacks, Lemma 100.6.2. Apply Lemma 106.16.2 to get the finite type commutative B-algebra C = F(g_{\mathit{QCoh}, *}\mathcal{O}_ U) and the ring map A \to C. By Lemma 106.16.3 the ring map B \to C is universally injective. Consider the algebra
C \otimes _ B C = F(g_{\mathit{QCoh}, *}\mathcal{O}_ U \otimes _{\mathcal{O}_\mathcal {X}} g_{\mathit{QCoh}, *}\mathcal{O}_ U)
Since g is flat, quasi-compact, and quasi-separated by Lemma 106.16.5 we have the first equality in
g_{\mathit{QCoh}, *}\mathcal{O}_ U \otimes _{\mathcal{O}_\mathcal {X}} g_{\mathit{QCoh}, *}\mathcal{O}_ U = f_{\mathit{QCoh}, *}\mathcal{O}_{U \times _\mathcal {X} U} = g_{\mathit{QCoh}, *}(\text{pr}_{2, *}\mathcal{O}_{U \times _\mathcal {X} U})
where f : U \times _\mathcal {X} U \to \mathcal{X} is the obvious morphism and \text{pr}_2 : U \times _\mathcal {X} U \to U is the second projection. The second equality follows from Cohomology of Stacks, Lemma 103.11.5 and f = g \circ \text{pr}_2. Since the diagonal of \mathcal{X} is affine, we see that U \times _\mathcal {X} U = \mathop{\mathrm{Spec}}(R) is affine. Let us use \text{pr}_2 : A \to R to view R as an A-algebra. All in all we obtain
C \otimes _ B C = F(g_{\mathit{QCoh}, *}\mathcal{O}_ U \otimes _{\mathcal{O}_\mathcal {X}} g_{\mathit{QCoh}, *}\mathcal{O}_ U) = F(g_{\mathit{QCoh}, *}(\text{pr}_{2, *}\mathcal{O}_{U \times _\mathcal {X} U})) = R \otimes _ A C
where the final equality follows from the final statement of Lemma 106.16.2. Since A \to R is flat (because \text{pr}_2 is flat as a base change of U \to \mathcal{X}), we conclude that C \otimes _ B C is flat over C. By Lemma 106.16.4 we conclude that B \to C is faithfully flat.
We claim there is a solid commutative diagram
\xymatrix{ \mathop{\mathrm{Spec}}(C \otimes _ B C) \ar@<1ex>[d] \ar@<-1ex>[d] \ar[r] & U \times _\mathcal {X} U \ar@<1ex>[d] \ar@<-1ex>[d] \\ \mathop{\mathrm{Spec}}(C) \ar[d] \ar[r] & U \ar[d] \\ \mathop{\mathrm{Spec}}(B) \ar@{..>}[r] & \mathcal{X} }
The arrow \mathop{\mathrm{Spec}}(C) \to U = \mathop{\mathrm{Spec}}(A) comes from the ring map A \to C in the statement of Lemma 106.16.2. The arrow \mathop{\mathrm{Spec}}(C \otimes _ B C) \to U \times _\mathcal {X} U similarly comes from the ring map R \to C \otimes _ B C. To verify the top square commutes use Lemma 106.16.6; details omitted. We conclude we get the dotted arrow \mathop{\mathrm{Spec}}(B) \to \mathcal{X} by Proposition 106.15.1.
The statement that F is the functor corresponding to pullback by the dotted arrow is also clear from this and the corresponding statement in Lemma 106.16.2. Details omitted. \square
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