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This is a follow up question to this question. You can define an $F$-algebra over a category where $F$ is an endofunctor. This lets you define a group object, a ring object, a monoid object, and several other types of objects.

It is true in the case of groups and monoids that a group object in the category of groups is a commutative group. It is true that a monoid object in the category of monoids is a commutative monoid. Does this extend to all $F$-objects you can define?

Is it true that _____ objects in the category of _____ are commutative _____?

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    $\begingroup$ It's an interesting question, and one I've never thought about before. But I wonder what the statement "Is it true that total-order objects in the category of totally ordered sets are commutative total orders?" means. I need to sleep now, so I can't think about this at the moment. $\endgroup$ – Patrick Stevens May 16 '18 at 21:46
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    $\begingroup$ There is a general notion of "commutative" for Lawvere theories (and their multi-sorted generalizations, algebraic theories). I have no idea if it means what you want it to mean as that's pretty unclear. It does reproduce commutative monoids and various other "commutative" structures. $\endgroup$ – Derek Elkins May 16 '18 at 21:50
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    $\begingroup$ I know about getting group objects, ring objects, monoid objects, etc. as $M$-algebras where $M$ is an appropriate monad (also called "triple"), but I'm not aware that they can be obtained just as $F$-algebras for appropriate endofunctors $F$ (without a monad structure). $\endgroup$ – Andreas Blass May 17 '18 at 0:15
  • $\begingroup$ A monoid object in the category of rings is a ring with zero multiplication, so the only ring object in the category of rings is the terminal ring (if you want your rings to have identities). $\endgroup$ – Arnaud D. May 17 '18 at 10:11
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It's very unclear to me what "commutative" ought to mean at this level of generality. Consider, for example, Poisson algebras. What would it mean for a Poisson algebra to be commutative? Does it mean that the algebra multiplication is commutative? That the Poisson bracket is commutative (which means it's zero)? Both? Or what?

The fact that a monoid in the category of monoids is a commutative monoid is the Eckmann-Hilton argument, and it's a shadow of important higher-categorical / homotopical phenomena which are in some sense special to monoids. Namely, what "monoid in monoids" actually gets you is something called an $E_2$ algebra in full generality. In ordinary categories an $E_2$ algebra is just a commutative monoid, but in higher categories it's something more interesting; for example, in the 2-category $\text{Cat}$ it's a braided monoidal category, and in the $\infty$-category of homotopy types it's almost the same thing as a double loop space.

You can keep going and define $E_3$ algebras (monoids in monoids in monoids), $E_4$ algebras (monoids in monoids in monoids in monoids), all the way up to $E_{\infty}$ algebras, which it turns out is the correct way to say "commutative monoid" in this higher setting. The fact that $E_2, E_3, \dots E_{\infty}$ all collapse to "commutative" in an ordinary category is an artifact of there not being enough higher structure to distinguish them.

Independent of all of that, there remains the interesting question of what happens when one has two kinds of structure $A$ and $B$, and asks what "$A$-structures in $B$-structures" looks like in general. One answer is that if $A$ and $B$ are given by Lawvere theories, then there is a Lawvere theory whose models are models of $A$ in models of $B$ called the tensor product Lawvere theory $A \otimes B$; see e.g. Hyland for details, particularly Theorem 3.4. From this perspective what the Eckmann-Hilton argument reveals is that the tensor product of the Lawvere theory of monoids with itself is the Lawvere theory of commutative monoids. But one can consider more complicated tensor products, of which I know basically nothing off the top of my head.

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    $\begingroup$ If I remember correctly, there's a good deal of information relevant to this question in Peter Freyd's "Algebra-valued functors in general and tensor products in partiular" (item [9] in the bibliography of the Hyland-Power paper that you linked to). $\endgroup$ – Andreas Blass May 17 '18 at 0:18
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    $\begingroup$ The paper mentioned by Andreas Blass was studied in the Kan extension seminar last year, which resulted in the following blog post : golem.ph.utexas.edu/category/2017/03/… $\endgroup$ – Arnaud D. May 17 '18 at 9:59
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Something related to what you might be looking for arises in homotopical algebra - particularly when trying to define derived geometric objects. Let $\mathfrak{C}$ be a monoidal model category. Denote by Comm$(\mathfrak{C})$ the subcategory of commutative monoids in $\mathfrak{C}$. Let $k$ be a commutative ring. Let $\mathfrak{C}$ = $k$ - Mod be the symmetric monodical category of $k$-modules. Then Comm$(\mathfrak{C})$ is the category of commutative $k$-algebras. A slightly less trivial case, when char $k$ = $0$, is $\mathfrak{C}$ = C(k), the category of unbounded complexes of $k$-modules. Comm$(\mathfrak{C})$ is the category of commutative differential graded $k$-algebras. Exercise: what is Comm$(\mathfrak{C})$ for $\mathfrak{C}$ the category of simplicial $k$-modules? This is a slightly stronger notion than Qiaochu's above - for $k$ of non-zero characteristic, the homotopy theory of simplicial commutative $k$-algebras is not equivalent to the homotopy theory of $E_{\infty}$-monoids in simplicial $k$-modules.

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