# 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.

• I’m not sure what your ring elements are. Are these in the ring? What is $\begin{pmatrix} 1 & 1\\ 0& 0\end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix}?$ – Dan Robertson Nov 29 '17 at 13:55
• Also what is your definition of a (noncommitative) ring? You can’t have a noncommutative ring with no unity if your definition requires the existence of a unit. Often a ring with no unit is called a rng (a ring with no ‘i’). – Dan Robertson Nov 29 '17 at 13:58
• Non commutative ring means it is not commutative under multiplication.I'm talking about the 2×2 matrix with only 2 non zero entries at any place and element are from finite field Z2. – Seeker Nov 29 '17 at 14:03
• @user378511 This ring has a unit, but is not closed under addition. – lisyarus Nov 29 '17 at 14:07
• @lisyarus take bottom row of the matrix zero. – Seeker Nov 29 '17 at 14:08

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 $b=\begin{bmatrix}1&1\\0&0\end{bmatrix}$.

• So the question should be asked what is the order of smallest non-commutative "rng"? Right? – Seeker Nov 29 '17 at 14:41
• @user378511 "rng=ring possibly without identity". If I was given free reign with the question I would pose it as "what's the smallest ring that isn't commutative and doesn't have identity?" but that's just me. How you wrote it is understandable. – rschwieb Nov 29 '17 at 14:44
• Oh yeah!This sounds more understandable to me. Next time I'll be more precise with the language ;) – Seeker Nov 29 '17 at 14:49