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There's a kind of construction of an extension of a ring, and that I've seen used e.g. here (although that may not be the best example) which is essentially adding a function into the ring, and taking the closure under the function.

So for every element $r$ in $R_0$ you would adjoin elements $f(r)$ with no relations to make $R_1$, and do this again to $R_1$ to make $R_2$ etc. so that $R_\omega$, the limit of these, is closed under a free unary operation $f$.

I hope I have described this clearly. I would like to know what the terminology here is, and where I can learn more about using this construction and generalisations of it.


E.g. with what I currently know, even defining a quotient by a relation, like $f(0)=1$ seems very messy, as you have to be able to describe all the other elements in the ring that have $f(0)$ as part of their construction. So I want to see how this is usually dealt with, and if there's a convenient way of defining variations like that which makes it easy to work with.

Worse, proving that any two things aren't equal after taking a massive quotient seems a near-impossible task, but I'm sure there are neat ways of doing it that I'm just not aware of.


Edit: @diracdeltafunk provided a nice universal property that makes this easier to work with. I would still like a reference for where to learn more about constructions like this, though.

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Interesting question. Let's use $R \mapsto R\{f\}$ to denote this construction.

Note that there is a function $R\{f\} \to R\{f\}$ defined by $r \mapsto f(r)$. By abuse of notation, we'll also call this function $f$.

Now I claim that the pair $(R\{f\}, f)$ has a universal property. First, I need to define the category in which $(R\{f\}, f)$ lives.

Definition. Let $\mathsf{C}$ be the category

  • whose objects are pairs $(A,a)$ where $A$ is a ring and $a : A \to A$ is some function,
  • whose morphisms $(A,a) \to (B,b)$ are the ring homomorphisms $\varphi : A \to B$ such that $\varphi(a(x)) = b(\varphi(x))$ for all $x \in A$,
  • and whose composition is just ordinary composition of ring homomorphisms.

Proposition. (Universal Property of $(R\{f\},f)$)

$$\operatorname{Hom}_{\mathsf{C}}((R\{f\},f), (A,a)) \cong \operatorname{Hom}_{\mathsf{Rng}}(R,A)$$ naturally in $(A,a)$.

This is relatively easy to prove, and of course uniquely characterizes $(R\{f\}, f)$ as an object of $\mathsf{C}$ up to isomorphism.


The above is a bit ad-hoc, but it is a special case of a more general idea.

Let $\mathsf{X}$ be any category (this was $\mathsf{Rng}$ in the special case above).

Let $P : \mathsf{X}^\text{op} \times \mathsf{X} \to \mathsf{Set}$ be any functor (this was $\operatorname{Hom}_{\mathsf{Set}}$ in the special case above).

Define $\mathsf{E}(\mathsf{X},P)$ to be the category

  • whose objects are pairs $(A,a)$ where $A$ is an object of $\mathsf{X}$ and $a \in P(A,A)$,
  • whose morphisms $(A,a) \to (B,b)$ are the morphisms $\varphi : A \to B$ in $\mathsf{X}$ such that $\varphi_* a = \varphi^* b$,
  • and whose composition is just ordinary composition of morphisms in $\mathsf{X}$.

There is a forgetful functor $U : \mathsf{E}(\mathsf{X},P) \to \mathsf{X}$ given by $(A,a) \mapsto A$.

In the special case above, $U$ has a left adjoint, namely the construction $R \mapsto (R\{f\},f)$. This is precisely what is encoded by the universal property of $(R\{f\},f)$.

In general, I think $U$ may or may not have a left adjoint. But with some assumptions on $\mathsf{X}$ and $P$, this can probably be guaranteed! See e.g. the adjoint functor theorems.


One more note: in this case, $U$ was a forgetful functor of varieties (in the sense of universal algebra). This sort of functor always has a left adjoint, and things can be made even more general than this: you may want to look into locally presentable categories.

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    $\begingroup$ Thank you, that's really neat. Do you know where I can learn more about this? $\endgroup$
    – Zoe Allen
    Commented Mar 12, 2023 at 5:58
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    $\begingroup$ To be honest I just made this up; it was easy to spot as someone who does a lot of category theory. But this whole thing does generalize -- I'll edit my post to add some thoughts about that! $\endgroup$ Commented Mar 12, 2023 at 6:02

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