There are many examples in this spirit: the $n\times n$ matrices over a finite field with bottom row zero.
answered Dec 12, 2013 at 6:19 324 1 1 silver badge 15 15 bronze badges$\begingroup$ 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. $\endgroup$
Commented Dec 12, 2013 at 10:46 $\begingroup$The easiest example of such a ring is to let $$ S=\\> $$ 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 $2\mathbb/2n\mathbb$ 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.
answered Dec 12, 2013 at 6:20 mathematics2x2life mathematics2x2life 13.4k 2 2 gold badges 32 32 silver badges 52 52 bronze badges$\begingroup$ If you take $\Bbb Z/n\Bbb Z$ in the place of $S$, you will have a unit element. You probably wanted $2\Bbb Z/2n\Bbb Z$. $\endgroup$
Commented Dec 12, 2013 at 10:27$\begingroup$ I think this is not always true, since $2\mathbb