$\newcommand{\pt}{\mathrm{pt}}\newcommand{\N}{\mathbb{N}}\newcommand{\Z}{\mathbb{Z}}$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](https://en.wikipedia.org/wiki/Differential_algebra)?



Are differential rings monoids in a monoidal category?
$\newcommand{\defeq}{\overset{\mathrm{def}}{=}}\newcommand{\id}{\mathrm{id}}\newcommand{\Mod}{\mathrm{Mod}}\newcommand{\pt}{\mathrm{pt}}\newcommand{\N}{\mathbb{N}}\newcommand{\Z}{\mathbb{Z}}\newcommand{\d}{\mathrm{d}}\newcommand{\dAb}{\mathsf{End}(\mathsf{Ab})}$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.


