The Stacks project

105.16 Tensor functors

Let $f : \mathcal{Y} \to \mathcal{X}$ be a morphism of Noetherian algebraic stacks. The pullback functor

\[ f^* : \textit{Coh}(\mathcal{O}_\mathcal {X}) \longrightarrow \textit{Coh}(\mathcal{O}_\mathcal {Y}) \]

is a right exact tensor functor: it is additive, right exact, and commutes with tensor products of coherent modules. We can ask to what extent any right exact tensor functor $F : \textit{Coh}(\mathcal{O}_\mathcal {X}) \to \textit{Coh}(\mathcal{O}_\mathcal {Y})$ comes from a morphism $f : \mathcal{Y} \to \mathcal{X}$. The reader may consult [Hall-Rydh-coherent] for a very general result of this nature. The aim of this section is to give a short proof of Theorem 105.16.8 as an introduction to these ideas.

We begin with some lemmas.

Lemma 105.16.1. Let $\mathcal{X}$ and $\mathcal{Y}$ be Noetherian algebraic stacks. Any right exact tensor functor $F : \textit{Coh}(\mathcal{O}_\mathcal {X}) \to \textit{Coh}(\mathcal{O}_\mathcal {Y})$ extends uniquely to a right exact tensor functor $F : \mathit{QCoh}(\mathcal{O}_\mathcal {X}) \to \mathit{QCoh}(\mathcal{O}_\mathcal {Y})$ commuting with all colimits.

Proof. The existence and uniqueness of the extension is a general fact, see Categories, Lemma 4.26.2. To see that the lemma applies observe that coherent modules on locally Noetherian algebraic stacks are by definition modules of finite presentation, see Cohomology of Stacks, Definition 102.17.2. Hence a coherent module on $\mathcal{X}$ is a categorically compact object of $\mathit{QCoh}(\mathcal{O}_\mathcal {X})$ by Cohomology of Stacks, Lemma 102.13.5. Finally, every quasi-coherent module is a filtered colimit of its coherent submodules by Cohomology of Stacks, Lemma 102.18.1.

Since $F$ is additive, also the extension of $F$ is additive (details omitted). Since $F$ is a tensor functor and since colimits of modules commute with taking tensor products, also the extension of $F$ is a tensor functor (details omitted).

In this paragraph we show the extension commutes with arbitrary direct sums. If $\mathcal{F} = \bigoplus _{j \in J} \mathcal{H}_ j$ with $\mathcal{H}_ j$ quasi-coherent, then $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits _{J' \subset J\text{ finite}} \bigoplus _{j \in J'} \mathcal{H}_ j$. Denoting the extension of $F$ also by $F$ we obtain

\begin{align*} F(\mathcal{F}) & = \mathop{\mathrm{colim}}\nolimits _{J' \subset J\text{ finite}} F(\bigoplus \nolimits _{j \in J'} \mathcal{H}_ j) \\ & = \mathop{\mathrm{colim}}\nolimits _{J' \subset J\text{ finite}} \bigoplus \nolimits _{j \in J'} F(\mathcal{H}_ j) \\ & = \bigoplus \nolimits _{j \in J} F(\mathcal{H}_ j) \end{align*}

Thus $F$ commutes with arbitrary direct sums.

In this paragraph we show that the extension is right exact. Suppose $0 \to \mathcal{F} \to \mathcal{F}' \to \mathcal{F}'' \to 0$ is a short exact sequence of quasi-coherent $\mathcal{O}_\mathcal {X}$-modules. Then we write $\mathcal{F}' = \bigcup \mathcal{F}'_ i$ as the union of its coherent submodules (see reference given above). Denote $\mathcal{F}''_ i \subset \mathcal{F}''$ the image of $\mathcal{F}'_ i$ and denote $\mathcal{F}_ i = \mathcal{F} \cap \mathcal{F}'_ i = \mathop{\mathrm{Ker}}(\mathcal{F}'_ i \to \mathcal{F}''_ i)$. Then it is clear that $\mathcal{F} = \bigcup \mathcal{F}_ i$ and $\mathcal{F}'' = \bigcup \mathcal{F}''_ i$ and that we have short exact sequences

\[ 0 \to \mathcal{F}_ i \to \mathcal{F}_ i' \to \mathcal{F}_ i'' \to 0 \]

Since the extension commutes with filtered colimits we have $F(\mathcal{F}) = \mathop{\mathrm{colim}}\nolimits _{i \in I} F(\mathcal{F}_ i)$, $F(\mathcal{F}') = \mathop{\mathrm{colim}}\nolimits _{i \in I} F(\mathcal{F}'_ i)$, and $F(\mathcal{F}'') = \mathop{\mathrm{colim}}\nolimits _{i \in I} F(\mathcal{F}''_ i)$. Since filtered colimits of sheaves of modules is exact we conclude that the extension of $F$ is right exact.

The proof is finished as a right exact functor which commutes with all coproducts commutes with all colimits, see Categories, Lemma 4.14.12. $\square$

Lemma 105.16.2. Let $\mathcal{X}$ be an 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. Let $g : U \to \mathcal{X}$ be a morphism with $U = \mathop{\mathrm{Spec}}(A)$ affine. Then

  1. $C = F(g_{\mathit{QCoh}, *}\mathcal{O}_ U)$ is a commutative $B$-algebra and

  2. there is a ring map $A \to C$

such that $F \circ g_{\mathit{QCoh}, *} : \text{Mod}_ A \to \text{Mod}_ B$ sends $M$ to $M \otimes _ A C$ seen as $B$-module.

Proof. We note that $g$ is quasi-compact and quasi-separated, see Morphisms of Stacks, Lemma 100.7.8. In Cohomology of Stacks, Proposition 102.11.1 we have constructed the functor $g_{\mathit{QCoh}, *} : \mathit{QCoh}(\mathcal{O}_ U) \to \mathit{QCoh}(\mathcal{O}_\mathcal {X})$. By Cohomology of Stacks, Remarks 102.11.3 and 102.10.6 we obtain a multiplication

\[ \mu : g_{\mathit{QCoh}, *}\mathcal{O}_ U \otimes _{\mathcal{O}_\mathcal {X}} g_{\mathit{QCoh}, *}\mathcal{O}_ U \longrightarrow g_{\mathit{QCoh}, *}\mathcal{O}_ U \]

which turns $g_{\mathit{QCoh}, *}\mathcal{O}_ U$ into a commutative $\mathcal{O}_\mathcal {X}$-algebra. Hence $C = F(g_{\mathit{QCoh}, *}\mathcal{O}_ U)$ is a commutative algebra object in $\text{Mod}_ B$, in other words, $C$ is a commutative $B$-algebra. Observe that we have a map $\kappa : A \to \text{End}_{\mathcal{O}_\mathcal {X}}(g_{\mathit{QCoh}, *}\mathcal{O}_ U)$ such that for any $a \in A$ the diagram

\[ \xymatrix{ g_{\mathit{QCoh}, *}\mathcal{O}_ U \otimes _{\mathcal{O}_\mathcal {X}} g_{\mathit{QCoh}, *}\mathcal{O}_ U \ar[d]_{\kappa (r) \otimes 1} \ar[rr]_-\mu & & g_{\mathit{QCoh}, *}\mathcal{O}_ U \ar[d]^{\kappa (r)} \\ g_{\mathit{QCoh}, *}\mathcal{O}_ U \otimes _{\mathcal{O}_\mathcal {X}} g_{\mathit{QCoh}, *}\mathcal{O}_ U \ar[rr]^-\mu & & g_{\mathit{QCoh}, *}\mathcal{O}_ U } \]

commutes. It follows that we get a map $\kappa ' = F(\kappa ) : A \to \text{End}_ B(C)$ such that $\kappa '(a)(c) c' = \kappa '(a)(cc')$ and of course this means that $a \mapsto \kappa '(a)(1)$ is a ring map $A \to C$.

The morphism $g : U \to \mathcal{X}$ is affine, see Morphisms of Stacks, Lemma 100.9.4. Hence $g_{\mathit{QCoh}, *}$ is exact and commutes with direct sums by Cohomology of Stacks, Lemma 102.13.4. Thus $F \circ g_{\mathit{QCoh}, *} : \text{Mod}_ A \to \text{Mod}_ B$ is a right exact functor which commutes with direct sums and which sends $A$ to $C$. By Functors and Morphisms, Lemma 56.3.1 we see that the functor $F \circ g_{\mathit{QCoh}, *}$ sends an $A$-module $M$ to $M \otimes _ A C$ viewed as a $B$-module. $\square$

Lemma 105.16.3. Notation as in Lemma 105.16.2. Assume $\mathcal{X}$ is Noetherian and $g$ is surjective and flat. Then $B \to C$ is universally injective.

Proof. Consider the natural map $1 : \mathcal{O}_\mathcal {X} \to g_{\mathit{QCoh}, *}\mathcal{O}_ U$ in $\mathit{QCoh}(\mathcal{O}_\mathcal {X})$. Pulling back to $U$ and using adjunction we find that the composition

\[ \mathcal{O}_ U = g^*\mathcal{O}_\mathcal {X} \xrightarrow {g^*1} g^*g_{\mathit{QCoh}, *}\mathcal{O}_ U \to \mathcal{O}_ U \]

is the identity in $\mathit{QCoh}(\mathcal{O}_ U)$. Write $g_{\mathit{QCoh}, *}\mathcal{O}_ U = \mathop{\mathrm{colim}}\nolimits \mathcal{F}_ i$ as a filtered colimit of coherent $\mathcal{O}_\mathcal {X}$-modules, see Cohomology of Stacks, Lemma 102.18.1. For $i$ large enough the map $1 : \mathcal{O}_\mathcal {X} \to g_{\mathit{QCoh}, *}\mathcal{O}_ U$ factors through $\mathcal{F}_ i$, see Cohomology of Stacks, Lemma 102.13.5. Say $s : \mathcal{O}_\mathcal {X} \to \mathcal{F}_ i$ is the factorization. Then

\[ \mathcal{O}_ U \xrightarrow {g^*s} g^*\mathcal{F}_ i \to g^*g_{\mathit{QCoh}, *}\mathcal{O}_ U \to \mathcal{O}_ U \]

is the identity. In other words, we see that $s$ becomes the inclusion of a direct summand upon pullback to $U$. Set $\mathcal{F}_ i^\vee = hom(\mathcal{F}_ i, \mathcal{O}_\mathcal {X})$ with notation as in Cohomology of Stacks, Lemma 102.10.8. In particular there is an evaluation map $ev : \mathcal{F}_ i \otimes _{\mathcal{O}_\mathcal {X}} \mathcal{F}_ i^\vee \to \mathcal{O}_\mathcal {X}$. Evaluation at $s$ defines a map $s^\vee : \mathcal{F}_ i^\vee \to \mathcal{O}_\mathcal {X}$. Dual to the statement about $s$ we see that $g^*(s^\vee )$ is surjective, see see Cohomology of Stacks, Section 102.12 for compatibility of $hom$ and $\otimes $ with restriction to $U$. Since $g$ is surjective and flat, we conclude that $s^\vee $ is surjective (see locus citatus). Since $F$ is right exact, we conclude that $F(\mathcal{F}_ i^\vee ) \to F(\mathcal{O}_\mathcal {X}) = B$ is surjective. Choose $\lambda \in F(\mathcal{F}_ i^\vee )$ mapping to $1 \in B$. Denote $e = F(s)(1) \in F(\mathcal{F}_ i)$ the image of $1$ by the map $F(s) : B = F(\mathcal{O}_\mathcal {X}) \to F(\mathcal{F}_ i)$. Then the map

\[ F(ev) : F(\mathcal{F}_ i) \otimes _ B F(\mathcal{F}_ i^\vee ) = F(\mathcal{F}_ i \otimes _{\mathcal{O}_\mathcal {X}} \mathcal{F}_ i^\vee ) \longrightarrow F(\mathcal{O}_\mathcal {X}) = B \]

sends $e \otimes \lambda $ to $1$ by construction. Hence the map $B \to F(\mathcal{F}_ i)$, $b \mapsto be$ is universally injective because we have the one-sided inverse $F(\mathcal{F}_ i) \to B$, $\xi \mapsto F(ev)(\xi \otimes \lambda )$. Since this is true for all $i$ large enough we conclude. $\square$

Lemma 105.16.4. Let $B \to C$ be a ring map. If

  1. the coprojections $C \to C \otimes _ B C$ are flat and

  2. $B \to C$ is universally injective,

then $B \to C$ is faithfully flat.

Proof. The map $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(B)$ is surjective as $B \to C$ is universally injective. Thus it suffices to show that $B \to C$ is flat which follows from Descent, Theorem 35.4.25. $\square$

The following very simple version of Künneth should become obsoleted when we write a section on Künneth theorems for cohomology of quasi-coherent modues on algebraic stacks.

Lemma 105.16.5. Let $a : \mathcal{Y} \to \mathcal{X}$ and $b : \mathcal{Z} \to \mathcal{X}$ be representable by schemes, quasi-compact, quasi-separated, and flat. Then $a_{\mathit{QCoh}, *}\mathcal{O}_\mathcal {Y} \otimes _{\mathcal{O}_\mathcal {X}} b_{\mathit{QCoh}, *}\mathcal{O}_\mathcal {Z} = f_{\mathit{QCoh}, *}\mathcal{O}_{\mathcal{Y} \times _\mathcal {X} \mathcal{Z}}$ where $f : \mathcal{Y} \times _\mathcal {X} \mathcal{Z} \to \mathcal{X}$ is the obvious morphism.

Proof. We abbreviate $\mathcal{P} = \mathcal{Y} \times _\mathcal {X} \mathcal{Z}$. Since $a \circ \text{pr}_1 = f$ and $b \circ \text{pr}_2 = f$ we obtain maps $a_*\mathcal{O}_\mathcal {Y} \to f_*\mathcal{O}_\mathcal {P}$ and $b_*\mathcal{O}_\mathcal {Z} \to f_*\mathcal{O}_\mathcal {P}$ (using relative pullback maps, see Sites, Section 7.45). Hence we obtain a relative cup product

\[ \mu : a_*\mathcal{O}_\mathcal {Y} \otimes _{\mathcal{O}_\mathcal {X}} b_*\mathcal{O}_\mathcal {Z} \longrightarrow f_*\mathcal{O}_{\mathcal{Y} \times _\mathcal {X} \mathcal{Z}} \]

Applying $Q$ and its compatibility with tensor products (Cohomology of Stacks, Remark 102.10.6) we obtain an arrow $Q(\mu ) : a_{\mathit{QCoh}, *}\mathcal{O}_\mathcal {Y} \otimes _{\mathcal{O}_\mathcal {X}} b_{\mathit{QCoh}, *}\mathcal{O}_\mathcal {Z} \to f_{\mathit{QCoh}, *}\mathcal{O}_{\mathcal{Y} \times _\mathcal {X} \mathcal{Z}}$ in $\mathit{QCoh}(\mathcal{O}_\mathcal {X})$. Next, choose a scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$. It suffices to prove the restriction of $Q(\mu )$ to $U_{\acute{e}tale}$ is an isomorphism, see Cohomology of Stacks, Section 102.12. In turn, by the material in the same section, it suffices to prove the restriction of $\mu $ to $U_{\acute{e}tale}$ is an isomorphism (this uses that the source and target of $\mu $ are locally quasi-coherent modules with the base change property). Moreover, we may compute pushforwards in the étale topology, see Cohomology of Stacks, Proposition 102.8.1. Then finally, we see that $a_*\mathcal{O}_\mathcal {Y}|_{U_{\acute{e}tale}} = (V \to U)_{small, *}\mathcal{O}_ V$ where $V = U \times _\mathcal {X} \mathcal{Y}$. Similarly for $b_*$ and $f_*$. Thus the result follows from the Künneth formula for flat, quasi-compact, quasi-separated morphisms of schemes, see Derived Categories of Schemes, Lemma 36.23.1. $\square$

Lemma 105.16.6. Let $\mathcal{X}$ be an algebraic stack with affine diagonal. Let $B$ be a ring. Let $f_ i : \mathop{\mathrm{Spec}}(B) \to \mathcal{X}$, $i = 1, 2$ be two morphisms. Let $t : f_1^* \to f_2^*$ be an isomorphism of the tensor functors $f_ i^* : \mathit{QCoh}(\mathcal{O}_\mathcal {X}) \to \text{Mod}_ B$. Then there is a $2$-arrow $f_1 \to f_2$ inducing $t$.

Proof. Choose an affine scheme $U = \mathop{\mathrm{Spec}}(A)$ and a surjective smooth morphism $g : U \to \mathcal{X}$, see Properties of Stacks, Lemma 99.6.2. Since the diagonal of $\mathcal{X}$ is affine, we see that $U_ i = \mathop{\mathrm{Spec}}(B) \times _{f_ i, \mathcal{X}, g} U$ is affine. Say $U_ i = \mathop{\mathrm{Spec}}(C_ i)$. Then $C_ i$ is the $B$-algebra endowed with ring map $A \to C_ i$ constructed in Lemma 105.16.2 using the functor $F = f_ i^*$. Therefore $t$ induces an isomorphism $C_1 \to C_2$ of $B$-algebras, compatible with the ring maps $A \to C_1$ and $A \to C_2$. In other words, we have a commutative diagrams

\[ \xymatrix{ U_ i \ar[r] \ar[d] & U \ar[d]^ g \\ \mathop{\mathrm{Spec}}(B) \ar[r]^{f_ i} & \mathcal{X} } \quad \quad \xymatrix{ & U_2 \ar[ld] \ar[d]^{\cong } \ar[rd] \\ \mathop{\mathrm{Spec}}(B) & U_1 \ar[l] \ar[r] & U } \]

This already shows that the objects $f_1$ and $f_2$ of $\mathcal{X}$ over $\mathop{\mathrm{Spec}}(B)$ become isomorphic after the smooth covering $\{ U_1 \to \mathop{\mathrm{Spec}}(B)\} $. To show that this descends to an isomorphism of $f_1$ and $f_2$ over $\mathop{\mathrm{Spec}}(B)$, we have to show that our isomorphism (which comes from the commutative diagrams above) is compatible with the descent data over $U_1 \times _{\mathop{\mathrm{Spec}}(B)} U_1$. For this we observe that $U \times _\mathcal {X} U$ is affine too, that we have the morphism $g' : U \times _\mathcal {X} U \to \mathcal{X}$, and that

\[ U_ i \times _{\mathop{\mathrm{Spec}}(B)} U_ i = \mathop{\mathrm{Spec}}(B) \times _{f_ i, \mathcal{X}, g'} (U \times _\mathcal {X} U) \]

It follows that the isomorphism $C_1 \otimes _ B C_1 \to C_2 \otimes _ B C_2$ coming from the isomorphism $C_1 \to C_2$ is compatible with the morphisms $U_ i \times _{\mathop{\mathrm{Spec}}(B)} U_ i \to U \times _\mathcal {X} U$. Some details omitted. $\square$

Lemma 105.16.7. 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 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

\[ C \otimes _ B C = F(g_{\mathit{QCoh}, *}\mathcal{O}_ U \otimes _{\mathcal{O}_\mathcal {X}} g_{\mathit{QCoh}, *}\mathcal{O}_ U) \]

Since $g$ is flat, quasi-compact, and quasi-separated by Lemma 105.16.5 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 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

\[ C \otimes _ B C = F(g_{\mathit{QCoh}, *}\mathcal{O}_ U \otimes _{\mathcal{O}_\mathcal {X}} g_{\mathit{QCoh}, *}\mathcal{O}_ U) = F(g_{\mathit{QCoh}, *}(\text{pr}_{2, *}\mathcal{O}_{U \times _\mathcal {X} U})) = R \otimes _ A C \]

where the final equality follows from the final statement of Lemma 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

\[ \xymatrix{ \mathop{\mathrm{Spec}}(C \otimes _ B C) \ar@<1ex>[d] \ar@<-1ex>[d] \ar[r] & U \times _\mathcal {X} U \ar@<1ex>[d] \ar@<-1ex>[d] \\ \mathop{\mathrm{Spec}}(C) \ar[d] \ar[r] & U \ar[d] \\ \mathop{\mathrm{Spec}}(B) \ar@{..>}[r] & \mathcal{X} } \]

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 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. $\square$

For a ring $B$ let us denote $\text{Mod}^{fg}_ B$ the category of finitely generated $B$-modules (AKA finite $B$-modules).

Theorem 105.16.8. Let $\mathcal{X}$ be a Noetherian algebraic stack with affine diagonal. Let $B$ be a Noetherian ring. Let $F : \text{Coh}(\mathcal{O}_\mathcal {X}) \to \text{Mod}^{fg}_ B$ be a right exact tensor functor. Then $F$ comes from a unique morphism $\mathop{\mathrm{Spec}}(B) \to \mathcal{X}$.

Proof. By Lemma 105.16.1 we can extend $F$ uniquely to a right exact tensor functor $F : \mathit{QCoh}(\mathcal{O}_\mathcal {X}) \to \text{Mod}_ B$ commuting with all direct susms. Then we can apply Lemma 105.16.7. $\square$


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