Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Give an example of a finite, non-commutative ring, which does not have a unity.

I can't think of any thing which fits this question. I was thinking $M_2(\mathbb{R})$ but it has the identity. Any help is appreciated.

share|cite|improve this question
Also $M_2(\Bbb R)$ is not really very finite. – Marc van Leeuwen Dec 12 '13 at 10:25
up vote 8 down vote accepted

$\textbf{Hint:}$ Matrix rings are a good example of non-commutative rings.

share|cite|improve this answer
Don't matrix rings all have unity (the identity matrix)? – William Ballinger Dec 12 '13 at 6:15
Not if you take the ring of matrices over a non-unital ring. – Arthur Dec 12 '13 at 6:17

There are many examples in this spirit: the $n\times n$ matrices over a finite field with bottom row zero.

share|cite|improve this answer
The matrix that looks like the identity matrix except that its final entry is $0$ instead of $1$ seems to be a "left-unity", but it does not work from the right, so the example is valid. – Jeppe Stig Nielsen Dec 12 '13 at 10:46

The easiest example of such a ring is to let $$ S=\{2 n\;|\; n \in \mathbb{Z}\} $$ and then consider the ring $M_n(S)$, the ring of $n \times n$ matrices with elements in $S$ (notice this does not include the identity matrix as $1 \notin S$). To get the finite example, instead, simply take $\mathbb{Z}_n$ instead of the set $S$.

In fact, for every prime $p$, there is a noncommutative ring without unity of order $p^2$. Moreover, is a ring of such order had unity it would also necessarily be commutative.

share|cite|improve this answer
If you take $\Bbb Z/n\Bbb Z$ in the place of $S$, you will have a unit element. You probably wanted $2\Bbb Z/2n\Bbb Z$. – Marc van Leeuwen Dec 12 '13 at 10:27

In the spirit of the answer by massy255: take the rng of strictly upper triangular $n\times n$ matrices over a finite field for $n\geq3$. This rng does not even have a nonzero subrng with a unit.

share|cite|improve this answer

protected by user26857 Oct 24 '15 at 16:11

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.