TITLE: Non-commutative reduced rings of order $p^2$
QUESTION [10 upvotes]: 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).

REPLY [3 votes]: A more conceptual path to essentially the same solution:
Lam, a first course in noncommutative rings, pag. 207: (12.7) Theorem. A nonzero ring $R$ is reduced iff $R$ is a subdirect product of domains. [Yes, it works also for unitless rings]
Finite subdirect product of rings are direct products: Chinese remainder theorem. [This is an optional step, not needed to obtain only commutativity but used to obtain the full structure of the ring]
Finite domains are division rings (since injective endofunctions of a finite set are surjective)
Now apply Wedderburn (finite division rings are Galois fields) [Kaplansky, Jacobson, Herstein et. al. have many generalizations of Wedderburn's theorem (whose proof, however, uses Wedderburn's result, plus steps related to the first stem in this proof)].
With the same proof (without Wedderburn):
reduced (right) artinian rings are exactly the finite direct products of (possibly skew) fields.