Let $X$ be a scheme. Is the category of quasi-coherent (commutative) $\mathcal{O}_X$-algebras cocomplete?

Remark. The same question was asked in MSE last year. Since nobody has answered it, I post this here.

  • 4
    $\begingroup$ Yes, it is the category of algebras over an accessible monad in a locally presentable category. $\endgroup$ – Fernando Muro Aug 20 '13 at 21:35
  • $\begingroup$ @FernandoMuro Sorry I don't understand. Reference? $\endgroup$ – Makoto Kato Aug 20 '13 at 21:37
  • $\begingroup$ Adamek-Rosicky, Borceux... $\endgroup$ – Fernando Muro Aug 20 '13 at 21:40
  • $\begingroup$ @FernandoMuro Thanks. Does Borceux mean his Handbook of categorical algebra? $\endgroup$ – Makoto Kato Aug 20 '13 at 21:45
  • 2
    $\begingroup$ Reference is the Remark in p. 124 of LOCALLY PRESENTABLE AND ACCESSIBLE CATEGORIES. J.Adámek and J.Rosický. Cambridge University Press 1994 YOu consider the monad "Symmetric tensor algebra on module" (this is accessible: Bourbaki, ALgebra III pag. 503: DIRECT LIMIT OF SYMMETRIC ALGEBRAS) Anyway a more direct answere or construction of colimit (in the setting of ringed spaces ecc.) could be interesting. $\endgroup$ – Buschi Sergio Aug 21 '13 at 0:21

It is well-known that the category of commutative algebras over a commutative ring is cocomplete, for example since it's algebraic over the category of sets. For a diagram of quasi-coherent algebras on a scheme, we can construct the colimit on each affine piece and then glue the resulting algebras together. That this works requires some tedious checking. So let me offer another global and categorified argument, which actually proves the affine case and the scheme case simultanuously:

Proposition. If $C$ is any cocomplete symmetric monoidal category (which means in particular that $\otimes$ is cocontinous in each variable), then the category of commutative monoids $\mathsf{CMon}(C)$ is cocomplete.

Note that this Proposition has lots of special cases, for example we can take $C$ to be the category of sets, or abelian groups, or modules over a commutative ring, more generally quasi-coherent modules on a scheme or even algebraic stack, or sheaves of modules on a ringed space or even site, graded modules etc. in order to get the cocompleteness of the categories of commutative monoids, rings, algebras over a commutative ring, more generally quasi-coherent algebras on a scheme or even algebraic stack, or sheaves of algebras on a ringed space or even site, graded-commutative algebras etc. Basically modern category theory tells us that it is not necessary to repeat the same argument in special cases all the time.

Proof: For cocompleteness, it suffices to find 1) an initial object, 2) binary coproducts, 3) directed colimits, 4) coequalizers. In fact, 1)-2) give finite coproducts, which combined with 3) give all coproducts, and it is well-known that colimits can be constructed from coproducts and coequalizers.

1) The initial object is the unit $1$ with the obvious monoid structure. 2) The tensor product (of the underlying objects) of two commutative monoids has a canonical monoid structure and becomes the coproduct. This is well-known, see for example SE/467317. 3) One checks that the forgetful functor to $C$ creates directed colimits.

4) Let $f,g : A \to B$ be homomorphisms of commutative monoids in $C$. By abuse of notation, the underlying objects in $C$ will be denoted the same. Define $f ' : B \otimes A \xrightarrow{B \otimes f} B \otimes B \xrightarrow{*} B$ and likewise $g'$. Let $p : B \to P$ be the coequalizer of $f',g'$ in $C$. Note that $p$ also coequalizes the analoguous morphisms $f'',g'' : A \otimes B \rightrightarrows B$ since $B$ is commutative. Here it is quite instructive to see what happens for example when $C=\mathsf{Ab}$ (if $f,g : A \to B$ are homomorphisms of commutative rings, the coequalizer is the quotient of $B$ not by the subgroup, but rather by the ideal generated by all $f(a)-g(a)$, i.e. the subgroup generated by all $b*(f(a)-g(a))$. This is where the extra $B$ comes from!). I claim that $P$ has the structure of a commutative monoid such that $p$ is the coequalizer of $f$ and $g$ in $\mathsf{CMon}(C)$.

The unit is defined by $1 \to B \to P$. In order to define a multiplication $P \otimes P \to P$ extending the one of $B$, use that $P \otimes P$ is the coequalizer of $(f,g \otimes B) \oplus (B \otimes f,g) : (A \otimes B) \oplus (B \otimes A) \rightrightarrows B \otimes B$ and that $B \otimes B \xrightarrow{*} B \xrightarrow{p} P$ coequalizes these morphisms, since $p$ coequalizes $f'$ and $g'$ as well as $f''$ and $g''$. Since $p$ and therefore $p^{\otimes 2}$ ,$p^{\otimes 3}$ are epimorphisms, the commutativity of the diagrams which assert that $B$ is a commutative monoid imply the commutativity of the corresponding diagrams for $P$. Hence, $P$ is a commutative monoid, and by construction $p$ is a morphism of commutative monoids.

Now let $h : B \to C$ be a morphism of commutative monoids which coequalizes $f$ and $g$. Then $B \otimes h : B \otimes B \to B \otimes C$ coequalizes $B \otimes f$ and $B \otimes g$. Hence, the following commutative diagram

$\begin{array}{c} & & & & B \otimes C & & \\ & & & \nearrow ~~ & & \searrow & \\ B \otimes A & \rightrightarrows & B \otimes B & & \rightarrow & & C \otimes C \\ & & \downarrow & & & & \downarrow \\ & & B & & \rightarrow & & C\end{array}$

shows that $h$ coequalizes $f'$ and $g'$. Hence, there is a unique morphism $\tilde{h} : P \to C$ such that $\tilde{h}p=h$. We are left to show that $\tilde{h}$ preserves the monoid structure. This is clear for the unit. For the multiplication we have to check that a certain diagram starting with $P \otimes P$ commutes. But since $p^{\otimes 2}$ is an epimorphism, this easily reduces to a diagram starting with $B \otimes B$, which comes down to the multiplicativity of $h$ (draw it!). $\square$

EDIT: Another proof can be found in Florian Marty's thesis, Prop. 1.2.14.

  • $\begingroup$ Martin Brandenburg: Very interesting! $\endgroup$ – Buschi Sergio Aug 21 '13 at 9:04
  • $\begingroup$ Thanks! Please wait for a few days before I accept this. $\endgroup$ – Makoto Kato Aug 22 '13 at 22:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.