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Exercise: Give an example of a non-commutative ring with 64 elements.

Official solution: Let $A = \mathrm{GL}_6(\mathbb{Z}_2)$. We know that $A$ is a non-commutative ring. Since each entry of $a$ is from $\mathbb{Z}_2$, each entry of $a$ has two choices. Since $a$ is a $6 \times 6$ matrix and each entry has two choices, we conclude that $a$ has $2^6$ choices. Thus, $|A| = 64$.

Uh-oh, I may very well believe that this ring $A$ has $64$ elements, but it seems to me, a lot more has to be said about it.

A $6 \times 6$ matrix has $36$ entries, so we have $2^{36}$ possible choices. Then we must exclude matrices which don't have the full rank of 6. That seems to be a nice combinatorial problem...

Can you help me?

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    https://www.proofwiki.org/wiki/Order_of_General_Linear_Group_over_Finite_Field –  Jan 09 '15 at 17:34

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I can't make any sense of that official solution. $\text{GL}_6(\mathbb{Z}_2)$ is a group, not a ring, and it has way more than $64$ elements. $M_6(\mathbb{F}_2)$, as you say, has $2^{36}$ elements.

For a correct example, take $R \times R$ where $R$ is the ring of $2 \times 2$ upper-triangular matrices with entries in $\mathbb{F}_2$ (the smallest noncommutative ring, with $8$ elements). You could also take the group algebra $\mathbb{F}_2[S_3]$ or the product $\mathbb{F}_2 \times \mathbb{F}_2 \times M_2(\mathbb{F}_2)$.

Qiaochu Yuan
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