I'm trying to write some test cases for set of code, and I need an example of a finite ring (with identity) which is not commutative, but has inverses (for non-zero elements) and whose additive identity is distinct from its multiplicative identity.

The reason I need this is because I had some code to detect a finite field which was passing all my test cases, yet I was not checking whether the multiplication is commutative. Having such a test case would make my code more robust.

My test for field is that the set be

  1. a commutative ring with identity,
  2. having distinct multiplicative and additive identities,
  3. for which all non-zero elements have inverses
  • 4
    $\begingroup$ Actually, such an example does not exist because any finite division ring is a field. $\endgroup$
    – Temoi
    Commented Jan 15 at 10:42
  • $\begingroup$ I've tried to post it as an answer, I don't know why it is automatically converted to a comment. $\endgroup$
    – Temoi
    Commented Jan 15 at 10:43
  • $\begingroup$ Perhaps, Matteo, it doesn't have enough characters for the software to accept it as an answer. Try expanding on it a little, and then posting it. (You aren't under an answer ban, are you?) $\endgroup$ Commented Jan 15 at 12:04
  • $\begingroup$ @GerryMyerson Thanks, I'll follow your advice (no, I'm not under an answer ban) $\endgroup$
    – Temoi
    Commented Jan 15 at 13:12
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    $\begingroup$ Hello Jim: with exceedingly few exceptions, rings in which multiplication is commutative are called commutative and not Abelian. I realize this doesn't quite line up with what groups do. Actually the adjective abelian is used for a different condition in rings. At any rate, using commutative instead will help others with the same question find the solutions here. $\endgroup$
    – rschwieb
    Commented Jan 15 at 15:44

1 Answer 1


A ring is called a division ring if any non-zero element is invertible. A field is a commutative division ring.

So what you are asking for is a finite division ring which is not a field. Actually such an example does not exist beacuse any finite division ring is a field.

  • 1
    $\begingroup$ Great to know. That means I don't need to check commutativity if the ring is finite. $\endgroup$
    – Jim Newton
    Commented Jan 15 at 13:16
  • 1
    $\begingroup$ The result that any finite division ring is a field is known as Wedderburn's little theorem. $\endgroup$ Commented Jan 15 at 15:48

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