Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: learning: questions

Topic: a ring as a category where the elements are objects

David Egolf (Aug 15 2023 at 16:17):

Working in a commutative ring $R$, we can introduce a relation as follows: for any two elements of the ring $a,b \in R$, we set $a \sim b$ exactly when $a=ub$, where $u$ is an element of the ring that has a multiplicative inverse (it is a "unit"). I was working an exercise where the aim was to show that $\sim$ is an equivalence relationship.

When thinking about this, I found it interesting to introduce a category $R'$ where:

• the objects are the elements of $R$
• there is a morphism $r: a \to b$ exactly when $ra=b$
• composition is by multiplication
• the identity morphisms correspond to multiplication by 1

We have that $a \sim b$ exactly when $a$ and $b$ are isomorphic in $R'$. I believe that isomorphism classes in any category correspond to an equivalence relationship, and so $\sim$ is an equivalence relationship on $R$.

This seems like an interesting way to think about a ring. I was hoping that we could make $R'$ into a monoidal category, by letting $\otimes$ be the $+$ operation available in $R$. However, it seems like $+:R' \times R' \to R'$ doesn't define a functor. I then tried creating a new category $R''$ where we have a morphism $r: a \to b$ exactly when $r + a = b$, in the hopes that I could make it into a monoidal category by letting $\otimes$ be the multiplication operation $\cdot$ from $R$. However, it seems like $\cdot :R'' \times R'' \to R''$ isn't a functor.

Is there some way to define a category from a commutative ring $R$ such that:

• the objects of the category are the elements of the ring
• both operations of the ring, $+$ and $\cdot$ are modelled in this category (e.g. as morphisms, or functors)
• the way in which $\cdot$ distributes over $+$ (e.g. $a \cdot (b+c) = a\cdot b + a \cdot c$) is also modelled in this category?

James Deikun (Aug 15 2023 at 17:07):

This isn't the usual thing, but I think in your original multiplicative category $+$ is a [[cofunctor]] with the distributive law giving the reversed action on arrows.

Simon Burton (Aug 15 2023 at 18:05):

i think i tried something like this for the complex numbers, like, how do you categorify the complex numbers? And the big problem here is if you categorify addition as a coproduct, then you run into trouble because coproducts don't have inverses except in degenerate cases. But there's this paper from Tom Leinster.

Mike Shulman (Aug 15 2023 at 18:12):

There is of course the discrete category on the ring, which is a [[bimonoidal category]], where both addition and multiplication are monoidal structures, neither (co)cartesian.

David Egolf (Aug 15 2023 at 18:38):

Thanks everyone for your responses! The links you all shared will be interesting to take a look at and learn from.

Claudio Pisani (Sep 06 2023 at 09:17):

David Egolf said:

When thinking about this, I found it interesting to introduce a category $R'$ where:

• the objects are the elements of $R$
• there is a morphism $r: a \to b$ exactly when $ra=b$
• composition is by multiplication
• the identity morphisms correspond to multiplication by 1
  

Note that this is the slice category of the underlying monoid of the ring over its unique object
(or, equivalently, the category of elements of the functor represented by its unique object).

David Egolf said:

Is there some way to define a category from a commutative ring $R$ such that:

• the objects of the category are the elements of the ring
• both operations of the ring, $+$ and $\cdot$ are modelled in this category (e.g. as morphisms, or functors)
• the way in which $\cdot$ distributes over $+$ (e.g. $a \cdot (b+c) = a\cdot b + a \cdot c$) is also modelled in this category?

There is an interesting way to see a ring as a cartesian operad, rather than a monoidal category.

First, for a monoid $M$ we can consider the operad (= one-object symmetric multicategory) $M'$
whose arrows are finite families of elements of $M$, with the obvious composition.
The algebras for this operad (the multifunctors $M' \to Set$, to the symmetric multicategory of sets and several variables functions)
are commutative monoids with an action of $M$ on them.

Second, if $R$ is a ring then the operad $R'$ on the underlying monoid has a cartesian structure.

Recall that $Set$ (like any multicategory obtained from a cartesian monoidal category) has a cartesian structure
where contraction and weakening maps are given respectively by precomposing an arrow $X\times X \to Y$ with the diagonal
and an arrow $X\to Y$ with a projection.

In the case of the operad $R'$, contraction and weakening maps are instead given respectively by the sum of the family
and by inserting zero's in a family.
More than that, to give a rig structure on a monoid $M$ amounts to giving a cartesian structure on the operad $M'$.
(Recall that rigs differ from rings in that the existence of negatives is not assumed).

Now, the algebras for this cartesian operad (the multifunctors $R' \to Set$ preserving the cartesian structure)
amount to modules over $R$.

So as a functor $C \to Set$ gives rise to a discrete fibration $D \to C$ over $C$,
an algebra for an operad (or more generally for a multicategory) $O$, gives rise to a discrete fibration of multicategories $D \to O$.
In the case of (the algebra corresponding to) a module over $R$, the "multicategory of elements" $D$
has the elements of the module as objects and linear combinations as arrows:
if $r_1 x_1 + \cdots r_n x_n = y$ we have an arrow $x_1, \cdots, x_n \vdash_{r_1,\cdots,r_n} y$ in $D$.
The sequent style notation is motivated by the fact that $D$ is itself cartesian so that the structural laws hold:
the contraction and weakening rules read
$r x + s x = y \implies (r+s)x = y$ and $r x = z \implies rx+0y = z$.

In particular, if we consider $R$ as a module over itself, we get a cartesian multicategory whose objects are the elements of $R$.

Patrick Nicodemus (Sep 07 2023 at 14:31):

I also felt it was natural to view $R'$ as the category of elements of the functor represented by its unique object, as Claudio suggested.