The Stacks project

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 $\mathop{\mathrm{Spec}}(C \otimes _ B C) \to U \times _\mathcal {X} U$ simlarly 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. $\square$


Comments (1)

Comment #8541 by Free RDP on

Thank you for sharing this information. IPTV Github – Collection of 8000+ Free Live publicly available IPTV channels from all over the world. Internet Protocol television (IPTV) is the delivery of television content over Internet Protocol (IP) networks. Link - https://rdphostings.com/blog/iptv-github/


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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0GRQ. Beware of the difference between the letter 'O' and the digit '0'.