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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:

Hans-Peter Stricker
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    There is a fledgling list at DaRT. – rschwieb Oct 10 '18 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. – rschwieb Oct 10 '18 at 15:04
  • But this sounds like set theory, not category theory. – Hans-Peter Stricker Oct 10 '18 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? – Hans-Peter Stricker Oct 10 '18 at 15:23
  • This means: The same question might be asked with "category theory" replaced by "set theory"? – Hans-Peter Stricker Oct 10 '18 at 15:24
  • I'm not sure why you think it's "set theoretic." There is a category of ordered fields, a category of Archimedean ordered fields, and so on. What I said amounts to $\mathbb Q$ being initial in the category of ordered fields, and $\mathbb R$ being a terminal object in the category of Archimedian fields. I'm not 100% sure a category of complete ordered fields is a thing, but it seems like it should be, and that $\mathbb R$ would be initial. And the category of complete, ordered Archimedian fields of course just has one object $\mathbb R$. – rschwieb Oct 10 '18 at 16:45
  • Classifying things set-theoretically seems like a fruitless line of thought. I can't think of anything set-theoretic that influences rings beyond cardinality. – rschwieb Oct 10 '18 at 16:50
  • Let us continue this discussion in chat. – Hans-Peter Stricker Oct 10 '18 at 17:13
  • I'm not sure if this fits as an answer to the question as given - but we know $R$ is a UFD if and only if every projective module over $R$ of rank 1 is free. – Daniel Schepler Oct 10 '18 at 17:51
  • Maybe using the categorical categorization of fields, you could say $R$ is a local ring if and only if there is a unique effective epimorphism from $R$ to a field, up to isomorphism of the combination? – Daniel Schepler Oct 10 '18 at 17:54
  • And $R$ is an integral domain if and only if there is a monomorphism from $R$ to a field. – Daniel Schepler Oct 10 '18 at 17:56
  • It might also be interesting that Galois theoretic properties can be expressed in terms of ring tensor products - for example, IIRC, $L / K$ is a Galois extension of degree $n$ if and only if $L \otimes_K L \simeq L^{\oplus n}$. – Daniel Schepler Oct 10 '18 at 18:11
  • Related question on MathOverflow: https://mathoverflow.net/q/161358/27465 – Torsten Schoeneberg Oct 10 '18 at 23:12

3 Answers3

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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.

Arnaud D.
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    This only works within the category of commutative rings, I think? – Christopher Oct 10 '18 at 12:56
  • Yes, good point. I've edited my answer accordingly. – Arnaud D. Oct 10 '18 at 16:21
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    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. – Daniel Schepler Oct 10 '18 at 18:27
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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.

Ben
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  • Also see https://qchu.wordpress.com/2015/04/25/compact-objects/ and replace modules by rings for a proof. – Ben Oct 10 '18 at 11:54
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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.)

Christopher
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  • Is there a similar characterization for skew fields? (Which would allow to drop "commutative".) – Hans-Peter Stricker Oct 10 '18 at 14:42
  • skew field = division ring – Hans-Peter Stricker Oct 10 '18 at 14:48
  • If "strong epimorphism" = "surjective homomorphism" holds for general rings (it's not clear from @Arnaud D.'s post and I don't know enough to say), then you can say that a ring is simple iff any strong epimorphism from it is either an isomorphism or a morphism to a zero ring. (Simple meaning no non-trivial two-sided ideals - a simple commutative ring is a field). – Christopher Oct 10 '18 at 14:53
  • @Arnaud D.: Would you mind to comment on that? – Hans-Peter Stricker Oct 10 '18 at 14:56
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    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.) – Daniel Schepler Oct 10 '18 at 18:22
  • Of course, if you "identify" ideals of $R$ with isomorphism classes of effective epimorphisms from $R$, then this would translate to the condition that $R$ has as ideals only $\langle 0 \rangle$ and $\langle 1 \rangle$. (And then, effective epimorphisms, or ideals, are also in bijection with congruences on $R$, which can be categorically described as certain subobjects of $R \times R$.) – Daniel Schepler Oct 10 '18 at 18:24
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    The zero ring also satisfies your property, so you need to exclude that case. – Arnaud D. Oct 10 '18 at 18:55
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    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). – Arnaud D. Oct 10 '18 at 19:03