Categorical characterizations of ring properties

Solution 1:

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.

Solution 2:

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.

Solution 3:

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.)