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For any given $n\in N$, Can we find a non-commutative ring of $n$ elements (with or without identity)?

If not, can we find some condition on $n$ such that a non-commutative ring of $n$ elements exists?

I found some examples here https://ringtheory.herokuapp.com/search/results/?H=43&L=1 but I want the examples for any $n$.

SARTHAK GUPTA
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    Not if $n$ is prime, compare https://math.stackexchange.com/q/1509692/96384, https://math.stackexchange.com/q/1043373/96384. Also not if your ring has an identity and $n$ is square-free, compare https://math.stackexchange.com/q/1825661/96384 (and I think the existence of an identity might be redundant in this case too). – Torsten Schoeneberg Jul 15 '19 at 23:31

1 Answers1

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TL;DR

There are rngs that aren't commutative of every finite order that is not squarefree.

There are rings that aren't commutative of every finite order that is not cubefree.

The smallest a noncommutative ring without identity can be is $4$ elements: see this post for an explanation.

It is obvious how to repeat this feat for any prime $p$ so that you have a noncommutative rng of order $p^2$ for any prime.

Then you have a noncommutative rng for any nonsquarefree number, because you just do the product of you noncommutative rng times $\mathbb Z_n$ for some strategically chosen $n$.

If the order is squarefree, then its underlying abelian group is necessarily cyclic. This implies the ring is commutative since you are just multiplying things of the form $m\cdot c$ where $m$ is a natural number and $c$ is the generator.

Trying to do the same thing for rings with identity, it’s immediately clear that you have noncommutative rings of order $p^3$ for any prime(namely the upper triangular 2 by 2 matrices over finite fields), and so all noncubefree orders are covered.

It is also known that all rings of order $p^2$ (with identity obviously) are commutative.

A la this answer, you can argue that the subgroup of elements of additive order divisible by $p$ is an ideal and that the subgroup of elements of additive not divisible by $p$ is another ideal. Furthermore, they have intersection $\{0\}$ and their sum generates the whole ring, so in fact the representation is a direct sum of ideals. Repeating this, you can eventually decompose the ring into a product of rings of order $p^2$ or $p$ for distinct primes $p$. Since these are all necessarily commutative, the whole ring is commutative.

PS: thanks for using my site.

rschwieb
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  • Every finite ring is a product of its "Sylow subrings" (they're not subrings but hopefully it's clear what is meant by this), which implies that a ring of cubefree order is commutative. – Qiaochu Yuan Jul 16 '19 at 04:51