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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.

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    $\begingroup$ Also $M_2(\Bbb R)$ is not really very finite. $\endgroup$ Dec 12 '13 at 10:25
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$\textbf{Hint:}$ Matrix rings are a good example of non-commutative rings.

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  • $\begingroup$ Don't matrix rings all have unity (the identity matrix)? $\endgroup$ Dec 12 '13 at 6:15
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    $\begingroup$ Not if you take the ring of matrices over a non-unital ring. $\endgroup$
    – Arthur
    Dec 12 '13 at 6:17
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There are many examples in this spirit: the $n\times n$ matrices over a finite field with bottom row zero.

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  • $\begingroup$ 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. $\endgroup$ Dec 12 '13 at 10:46
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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.

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    $\begingroup$ If you take $\Bbb Z/n\Bbb Z$ in the place of $S$, you will have a unit element. You probably wanted $2\Bbb Z/2n\Bbb Z$. $\endgroup$ Dec 12 '13 at 10:27
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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.

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