Non commutative ring without unity. Find order of smallest non-commutative ring.
According to me, the order should be 4. 
2×2 matrix with 2 non-zero entries over finite field Z2 and bottom row is zero.This matrix has no unity.
Is this a right answer or am i doing something wrong?
Please help.Thanks.
 A: Yes, your example (pieced together from the original post and the comments) $\begin{bmatrix}F_2&F_2\\0&0\end{bmatrix}$ is a noncommutative rng without identity of order $4$.
Since $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ acts as an identity on the left, it would have to be equal to any candidate two-sided identity, if it existed, but $\begin{bmatrix}0&1\\0&0\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}\neq \begin{bmatrix}0&1\\0&0\end{bmatrix}$
Obviously you can't have such a ring with $1$ or $2$ elements.
If such a ring had $3$ elements, say $\{0,a,b\}$, all distinct, then $a+b$ must be somewhere in this set. $a+b\in \{a,b\}$ creates a contradiction, so $a+b=0$ necessarily, so that $b=-a$.  But multiplication in $\{-a,0,a\}$ necessarily commutes.
What you gave is actually a nice representation for one of my favorite examples of semigroup rings. You take the semigroup $S=\{a,b\}$ with multiplication defined by $a^2=ba=a$ and $b^2=ab=b$, and make the semigroup ring $F_2[S]$. I think this is the same ring as if we had taken $a=\begin{bmatrix}1&0\\0&0\end{bmatrix}$ and $b=\begin{bmatrix}1&1\\0&0\end{bmatrix}$.
