Categorical characterizations of ring properties

Looking at the interesting list of ring properties that are inherited from a ring $$\mathcal{R}$$ by its polynomial ring $$\mathcal{R}$$[X] and remembering a question I once asked I want to repeat the latter in a more general way:

Can you give ring properties with catchy categorical characterizations like these:

What about being commutative, factorial, Noetherian, Abelian, or an integral domain?

[Note that the property of having a multiplicative identity (i.e. of being unital) doesn't have to be defined, because it's presupposed in the category of rings.]

List of characterizations from the answers below:

• There is a fledgling list at DaRT. Oct 10, 2018 at 13:47
• There are lots of things you can say that make main characters out of our favorite rings. Like $\mathbb Q$ is the smallest ordered field, or $\mathbb R$ is the smallest complete ordered field, or $\mathbb R$ is the largest Archimedian ordered field. Oct 10, 2018 at 15:04
• But this sounds like set theory, not category theory. Oct 10, 2018 at 15:22
• I.e.: What does "smallest"/"largest" mean in categorical terms? And: What does "ordered field", "complete ordered field" and "Archimedian ordered field" mean in categorical terms? Oct 10, 2018 at 15:23
• This means: The same question might be asked with "category theory" replaced by "set theory"? Oct 10, 2018 at 15:24

There is a notion of noetherian object in any category, which is simply that every ascending chain of subobjects is stationary. With this definition, noetherian objects in the category of modules over a ring are simply noetherian modules; however, ideals are not subrings (since in general they do not contain the identity element), so noetherian objects in the category of unital rings are not noetherian rings.

This is easy to fix, at least for the category of commutative rings, however : every ascending chain of ideals $$I_0\subset I_1\subset\dots \subset I_{n}\subset I_{n+1}\subset \dots,$$ in a commutative ring $$R$$ induces an ascending chain of quotient rings $$R/I_0\to R/I_1\to\dots\to R/I_n\to R/I_{n+1}\to \dots$$ and the original chain of ideals is stationary if and only if the chain of quotients is. Now quotient maps are just the surjective maps, which are the same thing as strong or regular epimorphisms in the category of commutative rings (or in any "algebraic" category). Thus noetherian commutative rings are precisely those for which every ascending chain of strong epimorphisms is stationary, which one might call "strongly co-noetherian objects".

For a non-commutative ring, this condition is equivalent to every ascending chain of two-sided ideals being stationary, but it doesn't say anything about left and right ideals.

• This only works within the category of commutative rings, I think? Oct 10, 2018 at 12:56
• Yes, good point. I've edited my answer accordingly. Oct 10, 2018 at 16:21
• You could also use a canonical bijection between two-sided ideals of $R$ and congruences on $R$, which can be described in categorical terms as certain subobjects of $R \times R$ - and this correspondence preserves the containment partial order. Oct 10, 2018 at 18:27

The finitely presented rings are the compact objects of the category of rings. That is, those objects $$R$$ such that the functor $$\hom(R,-)\colon \mathrm{(Ring)}\to\mathrm{(Set)}$$ preserves filtered colimits. In fact, this applies to any variety of algebras, e.g., by Jiří Adámek, Jiří Rosický, Locally Presentable and Accessible Categories, Corollary 3.13.

A ring is isomorphic to the zero ring iff it is a terminal object in the category of rings.

A commutative ring is a field iff any epimorphism from it is either an isomorphism or a morphism to a zero ring. (Since not every epimorphism in the category of rings is surjective, this isn't instantaneous, but see Lemma 10.106.8. Alternatively, one can replace "epimorphism" with "strong epimorphism" given @Arnaud D.'s categorisation of surjective homomorphisms as strong epimorphisms.)

• Is there a similar characterization for skew fields? (Which would allow to drop "commutative".) Oct 10, 2018 at 14:42
• skew field = division ring Oct 10, 2018 at 14:48
• Could you also say a commutative ring $R$ is a field if and only if every morphism with source $R$ either is a monomorphism or has target being a final object? (If $R$ isn't a field, and $a \in R$ is a nonzero nonunit, then the quotient map $R \to R / \langle a \rangle$ would satisfy neither.) Oct 10, 2018 at 18:22
• The zero ring also satisfies your property, so you need to exclude that case. Oct 10, 2018 at 18:55
• As for strong/regular/extremal epimorphisms, they coincide with surjective morphisms in any variety of algebras, or even more generally in any category monadic over sets; monadic functors to the category of sets preserve and reflect them, and in the category of sets they are just the surjections (this requires the axiom of choice, though). Oct 10, 2018 at 19:03