# Example of a finite non-commutative ring without a unity

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.

• Also $M_2(\Bbb R)$ is not really very finite. – Marc van Leeuwen Dec 12 '13 at 10:25

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

• 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.

• 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 $$2\mathbb{Z}/2n\mathbb{Z}$$ instead of the set $$S$$.

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

• 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.