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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