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I need to find an example of two non-isomorphic rings of cardinality 16 that are non-commutative. What is the best approach to such problem?

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Are there non-commutative rings of cardinality 16 but characteristics 2 and 4? – Hurkyl Dec 10 '12 at 0:03
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Here is an answer using two rings with identity.

The ring of two by two matrices $R=M_2(\Bbb{Z}_2)$ is a simple ring with 16 elements which is not commutative. ($\Bbb{Z}_2$ is meant to denote the field of two elements, here.)

Let $S$ be the subring of $R$ consisting of the upper triangular matrices, that is, the matrices whose $(2,1)$ entry is zero. This is an 8 element ring with identity. To form a ring with 16 elements, just form $T=S\oplus \Bbb{Z}_2$. The product is not commutative since $S$ is not commutative.

$T$, unlike $R$, is nonsimple because it is a direct product of rings, so they are nonisomorphic.

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Here is what I think works:

Ring 1: Take $\text{Mat}_2(\mathbb{F}_2)$

Ring 2: Take $R^2$ where $R$ is the noncommutative ring of order $4$.

They are no isomorphic because Ring 1 is unital and Ring 2 is not.

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