Noncommutative ring with eight elements and with unity I'm a bit confused about an exercise I read. Namely, T. Y. Lam's A First Course in Noncommutative Rings has the following on page $23$ Ex. $1.10$.
Let $p$ be a fixed prime. Show that there exists a noncommutative ring (with identity) of order $p^3$.
Well, this implies that there is a noncommutative ring with identity of order $8<16$, which contradicts with benh's answer given in Smallest non-commutative ring with unity . So have I understood correctly that there is a mistake in Lam's exercise?
 A: Indeed, there is a noncommutative ring with unity of order $p^3$ for all prime $p$. Though I have not read benh's proof, it must have an error of some form. I will not bother to repeat the argument. See this paper for many various constructions related to your question as well as the proof that there is a noncommutative ring with unity of order $p^3$ for all prime $p$. 
A: $(n-1) \times (n-1)$ matrices over ${\mathbb Z}_p$ of the form
$a I + C e^T$ where $a \in {\mathbb Z}_p$,  $C \in ({\mathbb Z}_p)^{n-1}$ (as a column vector) and $e^T = [1,0,\ldots,0]$ form a noncommutative ring (with identity) of order
$p^n$.  Thus for $n=4$ the matrices are of the form
$$ \pmatrix{a_1 & 0 & 0\cr
            a_2 & a_3 & 0\cr
            a_4 & 0 & a_3\cr}$$
Here  $p$ is an integer $\ge 2$ (not necessarily prime) and $n$ is an integer $\ge 3$.  It is noncommutative because e.g.
$$  \pmatrix{0 & 0 & \ldots\cr 0 & 1 &\ldots\cr
            \ldots & \ldots & \ldots} \pmatrix{0 & 0 & \ldots\cr 1 & 0 &\ldots\cr
            \ldots & \ldots & \ldots}= \pmatrix{0 & 0 & \ldots\cr 1 & 0 &\ldots\cr
            \ldots & \ldots & \ldots}$$
while
$$  \pmatrix{0 & 0 & \ldots\cr 1 & 0 &\ldots\cr
            \ldots & \ldots & \ldots} \pmatrix{0 & 0 & \ldots\cr 0 & 1 &\ldots\cr
            \ldots & \ldots & \ldots}= \pmatrix{0 & 0 & \ldots\cr 0 & 0 &\ldots\cr
            \ldots & \ldots & \ldots}$$
