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$\begingroup$

$\newcommand{\pt}{\mathrm{pt}}\newcommand{\N}{\mathbb{N}}\newcommand{\Z}{\mathbb{Z}}$A number of algebraic structures can be defined as monoids in some appropriate monoidal category:

  • A monoid is a monoid in $(\mathsf{Sets},\times,\pt)$;
  • A semiring is a monoid in $(\mathsf{CMon},\otimes_{\N},\N)$;
  • A ring is a monoid in $(\mathsf{Ab},\otimes_\Z,\Z)$;
  • An $R$-algebra is a monoid in $(\mathsf{Mod}_R,\otimes_R,R)$;
  • A graded $R$-algebra is a monoid in $(\mathsf{Gr}_\Z\mathsf{Mod}_R,\otimes_R,R)$;
  • A differential graded $R$-algebra is a monoid in $(\mathsf{Ch}_\bullet(\mathsf{Mod}_R),\otimes_R,R)$.

Is this also the case for differential rings?

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  • $\begingroup$ Differential rings, yes. The rest I bet no. $\endgroup$ Aug 29, 2021 at 21:05
  • $\begingroup$ Do you have a reference for differential rings? (I bet the same, by the way.) $\endgroup$
    – Emily
    Aug 29, 2021 at 21:07
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    $\begingroup$ Instead of a reference, why not take as an exercise to construct the correct monoidal category for differential rings yourself? $\endgroup$
    – mme
    Aug 29, 2021 at 22:09
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    $\begingroup$ @mme I hadn't realised then how simple that would be (mostly because I never really worked why DGAs are monoids in chain complexes). I've recorded it in an answer below and split the rest of the question. Thanks for the suggestion! (And thanks for the confirmation that this can be done, Fernando!) $\endgroup$
    – Emily
    Aug 30, 2021 at 4:57

1 Answer 1

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$\newcommand{\defeq}{\overset{\mathrm{def}}{=}}\newcommand{\id}{\mathrm{id}}\newcommand{\Mod}{\mathrm{Mod}}\newcommand{\pt}{\mathrm{pt}}\newcommand{\N}{\mathbb{N}}\newcommand{\Z}{\mathbb{Z}}\newcommand{\d}{\mathrm{d}}\newcommand{\dAb}{\mathsf{End}(\mathsf{Ab})}$DGAs are monoids in chain complexes. To get differential rings as monoids in some monoidal category, it suffices to remove the grading and the $\d^{2}=0$ condition.

In detail, consider the category $\mathsf{End}(\mathsf{Ab})\defeq\mathsf{Fun}(\mathbf{B}\N,\mathsf{Ab})$ whose

  • Objects are pairs $(A,\d)$ with $A$ an abelian group and $d\colon A\to A$ a morphism of abelian groups.
  • Morphisms $(A,\d_A)\to(B,\d_B)$ are morphisms of abelian groups preserving the derivation, i.e. such that the diagram $$ \require{AMScd} \begin{CD} A @>\d_A>> A\\ @V f V V @VV f V\\ B @>>\d_B> B \end{CD} $$ commutes.

We can then put a monoidal structure $\otimes_\Z$ on $\dAb$ by defining $$(A,\d_A)\otimes(B,\d_B)=(A\otimes_\Z B,\d_A\otimes_\Z1_B+1_A\otimes_\Z\d_B),$$ where the unit is given by the pair $(\Z,\d_\Z)$ with $\d_\Z\overset{\mathrm{def}}{=} 0$. Note that a morphism in $\dAb$ from $(\Z,\d_\Z)$ to $(A,\d_A)$ is just a "constant" element of $A$, i.e. an element with $\d_A a = 0$.

A monoid in $(\dAb,\otimes_\Z,(\Z,\d_\Z))$ will then be a triple $((A,\d),\mu,\eta)$ with

  • $(A,\d)$ an object of $\dAb$; this accounts for the underlying additive abelian group of a differential ring and the derivation $\d$, which is $\Z$-linear;
  • $\mu\colon(A,\d_A)\otimes_\Z(A,\d_A)\to(A,\d_A)$ a morphism of $\dAb$; this accounts for the multipication and the Leibniz rule: asking for the diagram $$ \require{AMScd} \begin{CD} A\otimes_\Z A @>\d_A\otimes_\Z1_A+1_A\otimes_\Z\d_A>> A\otimes_\Z A\\ @V \mu V V @VV \mu V\\ A @>>\d_A> A \end{CD} $$ to commute is equivalent to asking $$\d(ab)=\d(a)b+a\d(b)$$ to hold for all $a,b\in A$;
  • $\eta\colon(\Z,\d_\Z)\to(A,\d_A)$ a morphism of $\dAb$, determining an element $1_A$ of $A$;

such that the usual associativity and unitality diagrams commute, which makes $(A,\mu,\eta)$ into a ring, and together with $\d$, this makes the quadruple $((A,\d),\mu,\eta)$ into a differential ring.

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  • 2
    $\begingroup$ $\newcommand{\N}{\mathbb{N}}\newcommand{\defeq}{\overset{\mathrm{def}}{=}}$Here's a happy coincidence: the original question asked also whether we can view $\psi$-rings as monoids in a monoidal category. As Gabriel C. Drummond-Cole pointed out, we can, and actually this is done in a similar way to the above: while differential rings are monoids in $\mathsf{Fun}(\bf{B}\N,\mathsf{Ab})$ with $\N\defeq(\N,+,0)$, it turns out that $\psi$-rings are monoids in $\mathsf{Fun}\left(\mathbf{B}\N_{\geq1},\mathsf{Ab}\right)$ wiith $\N_{\geq1}\defeq(\N_{\geq1},\cdot,1)$! $\endgroup$
    – Emily
    Aug 31, 2021 at 2:30
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    $\begingroup$ +1 for this (research) exercise. BTW, what are $\mathbf{B}\N$ and $\mathtt{Fun}(\bullet,\bullet)$ ? $\endgroup$ Aug 31, 2021 at 4:39
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    $\begingroup$ @DuchampGérardH.E. Thanks! Here $\mathbf{B}A$ is the delooping of a monoid $A$, the category having only one object $\star$ and with $\mathrm{Hom}_{\mathbf{B}A}(\star,\star)\cong A$, where the composition and identity maps come from the multiplication and unit maps of $A$. The notation $\mathsf{Fun}(1,2)$ denotes the category of functors and natural transformations from $1$ to $2$. In particular, when $1=\mathbf{B}A$, it turns out that a functor $\mathbf{B}A\to\mathcal{C}$ is precisely an object $X$ of $\mathcal{C}$ together with a morphism of monoids $A\to\mathrm{Hom}_{\mathcal{C}}(X,X)$. $\endgroup$
    – Emily
    Aug 31, 2021 at 20:35
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    $\begingroup$ For example, when $A=\mathbb{N}$, such a morphism is precisely an element of $\mathrm{Hom}_{\mathcal{C}}(X,X)$, i.e. an endomorphism of $X$! Similarly, for $A=\mathbb{Z}$, such a morphism corresponds precisely to an invertible element of $\mathrm{Hom}_{\mathcal{C}}(X,X)$, i.e. an automorphism of $X$. $\endgroup$
    – Emily
    Aug 31, 2021 at 20:35
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    $\begingroup$ I find this construction really beautiful. +1 $\endgroup$ Sep 3, 2021 at 9:09

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