Lemma 105.16.7. Let $\mathcal{X}$ be a Noetherian algebraic stack with affine diagonal. Let $B$ be a ring. Let $F : \text{QCoh}(\mathcal{O}_\mathcal {X}) \to \text{Mod}_ B$ 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 99.6.2. Apply Lemma 105.16.2 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 105.16.3 the ring map $B \to C$ is universally injective. Consider the algebra

Since $g$ is flat, quasi-compact, and quasi-separated by Lemma 105.16.5 we have the first equality in

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 102.11.5 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

where the final equality follows from the final statement of Lemma 105.16.2. 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 105.16.4 we conclude that $B \to C$ is faithfully flat.

We claim there is a solid commutative diagram

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 105.16.2. The arrow $\mathop{\mathrm{Spec}}(C \otimes _ B C) \to U \times _\mathcal {X} U$ simlarly comes from the ring map $R \to C \otimes _ B C$. To verify the top square commutes use Lemma 105.16.6; details omitted. We conclude we get the dotted arrow $\mathop{\mathrm{Spec}}(B) \to \mathcal{X}$ by Proposition 105.15.1.

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 105.16.2. Details omitted. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)