Let $p$ be a prime number. Is there a non-commutative reduced ring of order $p^2$? (Note that any ring of order $p^2$ with identity is commutative). 



Non-commutative reduced rings of order $p^2$
Let $A$ be a reduced ring of order $p^n$. Then $(pa)^n = p^n a^n = 0$ for all $a \in A$ so $pA = 0$ and $A$ is an algebra over $\mathbb{F}\_p$. 


Adjoining an identity element to $A$ if necessary, we may view $A$ as an ideal (of codimension at most $1$) in a finite dimensional unital $\mathbb{F}\_p$-algebra $B$. Let $J$ be the radical of $B$ and $C = B/J$.


Since $A$ is reduced and $J$ is nil, $A \cap J = 0$ so $A$ is isomorphic to an ideal in $C$. Write the finite semisimple ring $C$ as a direct sum of simple rings $C = S\_1\oplus \cdots \oplus S\_m$; then $A \cap S\_i$ is either $S\_i$ or $0$ for all $i$. Assume without loss of generality that $S\_i \subseteq A$ for $i=1,\ldots, k$ and $A \cap S\_i = 0$ for $i > k$. Then $A$ contains $S\_1 \oplus \cdots \oplus S\_k$.


We can find unique central idempotents $e\_i \in S\_i$ such that $S\_i = e\_iC$ and $1 = e\_1 + \cdots + e\_m$. Because $A$ is an ideal in $C$, if $a \in A$ and $i > k$ then $e\_i a \in A \cap e\_i C = A \cap S\_i = 0$ so $a = e\_1a + \cdots + e\_ka \in S\_1 \oplus \cdots \oplus S\_k$.


Hence $A = S\_1 \oplus \cdots \oplus S\_k$ is a finite direct sum of finite simple rings. By Artin-Weddernburn, each $S\_i$ is a matrix ring over a finite division ring. These division rings are commutative by another theorem of Wedderburn, and the matrix sizes involved must all be $1 \times 1$ because otherwise $A$ would not be reduced. Therefore $A$ is commutative, and actually is isomorphic to a finite direct product of finite fields.


Note that if the reduced assumption on $A$ is dropped, then any non-zero one-sided ideal in the matrix ring $M\_2(\mathbb{F}\_p)$ gives an example of an associative, non-unital, non-commutative ring of order $p^2$.


