Smallest non-commutative ring with unity 
Find the smallest non-commutative ring with unity. (By smallest it means it has the least cardinal.) 

I tried rings of size 4 and I found no such ring.
 A: When one thinks of a noncommutative ring with unity (at least I), tend to think of how I can create such a ring with $M_n(R)$, the ring of $n \times n$ matrices over the ring $R$. The smallest such ring you can create is $R=M_2(\mathbb{F}_2)$. Of course, $|R|=16$. Now it is a matter if you can find a even smaller ring than this. Of course, the subring of upper/lower triangular matrices of $R$ is a subring of order $8$ which is a noncommutative ring with unity. This is indeed the smallest such rings. 
In fact, there is a noncommutative ring with unity of order $p^3$ for all primes $p$. See this paper for this and many other interesting/useful constructions. 
A: Here is another approach: Let $R$ be a non-zero ring with identity detone its center by $Z(R)$. It can be easily proved that: If $\frac {R}{Z(R)}$ is a cyclic group (with additive structure) then $R$ is a commutative ring. 
Now since $0,1 \in Z(R)$ then for any ring $R$ with $|R|< 8$ we have $|\frac{R}{Z(R)}| \leq 3$ which implies that $R$ is commutative (since any group of order $1,2$ or $3$ is cyclic) Therefore the smallest non-commutative with identity must have at least $8$ elements and there are such rings of course as it is mentioned in other solutions. The thing I would like to add is that there is no other example other than the ring of upper triangular matrices over $\Bbb{Z}_2$ scince we have the following theorem: 
Let $p$ be a prime number and let $R$ be a non-commutative ring with identity and suppose $|R| = p^3$ then $R$ is isomorphic to the ring of upper triangular matrices over $\Bbb{Z}_p$. 
Also I would like to add that in a similar way we can show that the smallest non-commutative ring (not necessary with identity) has order $4$. As an example consider $R = \{ \begin{pmatrix}a & b \\ 0 & 0 \end{pmatrix} \; | \; a, b \in \Bbb{Z}_2\}$.
A: In an earlier version of this post I caused an accident by giving a wrong answer. 
Thanks to those who pointed out the error! Here is a little update:
The ring $M_2(\Bbb F_2)$ of $2 \times 2$-matrices with entries in $\Bbb F_2$ is a non-commutative ring with 16 elements, because $$\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix} \neq \begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}$$
It has a subring of order $8$, namely the upper triangular matrices which are non-commutative by the example above.
We show that $8$ is minimal:
Let $R$ be a finite ring with unity with $n$ elements.
We need two preliminaries:
1.) If the additive group is cyclic, then $R$ is commutative.
Proof: If the additive group of $R$ is cyclic, we can choose $1$ as a generator: If we have $0 = 1+\dots+1 = m \cdot 1$ for some $m\in \Bbb N$, then $0=(m\cdot 1) \cdot g = m\cdot(1\cdot g)=m\cdot g$ so the additive order of $1$ is maximal. Thus, the multiplication table of $R$ is determined by $1 \cdot 1 = 1$, showing $R \cong \Bbb Z/n \Bbb Z$ and $R$ is commutative.
2.) All rings of order $4$ are commutative.
Proof: As a general result, all rings with order equal to a squared prime are commutative: Ring of order $p^2$ is commutative.
Thus any ring of order $1,2,3,5,6,7$ is ruled out by 1.) using the Sylow-theorems and $4$ is ruled out by 2.).
So $8$ is the minimal cardinality a non-commutative ring can have.
A: As others have noted, the upper triangular $2\times 2$ matrices with entries in $\mathbb{F}_2$ is a smaller ring. Here is one way to see that it is the smallest: cyclicity of the additive group, as remarked above, rules out the orders $1,2,3,5,6,7$.
As for $4$, it is not hard to see that any ring with unity of order $p^2$ is commutative: either it is cyclic or the additive subgroup generated by $1$ is of order $p$; hence there is an element $x$ that is not in this subgroup, whence $R$ must be a quotient of $\mathbb{Z}/p[t]$ via $t\mapsto x$ since the $x$ element necessarily commutes with all elements of the additive subgroup $\langle 1\rangle$.
