Easy example of a "non associative" ring

Question

Is there an easy example (one that an undergraduate math education student who learned the definition of a ring 5 minutes ago can understand) of a set who satisfies all the conditions of a ring except for associativity of multiplication?

Answer 1

You can get the ring of vectors with $\times$ because for all $a$, $b$ and $c$ we obtain: $$(a\times b)\times c+(b\times c)\times a+(c\times a)\times b=0.$$

Answer 2

Let $R$ be a ring. The commutator $$[·~·]\colon R × R → R,~(x,y) ↦ xy - yx$$ then defines an operation on $R$ which is trivial if and only if $R$ is commutative.

It is easy to check that this operation satisfies distributive laws. However, it has no unit in general.

If $R$ is non-commutative, this operation tends to be non-associative. For $x, y, z ∈ R$, $$[[x~y]~z] = (xy - yx)z - z(xy - yx) ≠ x(yz - zy) - (yz - zy) x = [x~[y~z]]$$ if and only if $$yxz + zxy ≠ xzy + yzx.$$

For example, if $x$ and $y$ are a symmetric pair of zero divisors, that is $xy = yx = 0$ and $z$ isolates them, the operation becomes non-associative. To specify further, take $x = [\begin{smallmatrix}1 & 0 \\ 0 & 0\end{smallmatrix}]$, $y = [\begin{smallmatrix}0 & 0 \\ 0 & 1\end{smallmatrix}]$ and $z = [\begin{smallmatrix}0 & 1 \\ 1 &0\end{smallmatrix}]$. Then $yxz + zyx = 0$, but $xzy + yzx ≠ 0$. (This assumes the student knows how to multiply matrices and recognizes matrix rings as rings.)

Answer 3

Here's an example of a non-associative ring.

Let's call it $R$.

Start with the free Abelian group on two generators $x$ and $y$.

Define an operation $*$ in the following way.

• $x*x = y$
• $x*y = 0$
• $y*x = 0$
• $y*y = x$

Additionally, note that the following two equations hold, which can be used to define $*$ on terms headed by $-$, noting that $a + b$ is equivalent to $a - (0 - b)$.

• $(a - b) * c = (a * c) - (b * c)$
• $a * (b - c) = (a * b) - (a * c)$

We can use the triple $(x, y, y)$ to witness the non-associativity of $R$.

$$ x * (y * y) = x * x = y $$ $$ (x * y) * y = 0 * y = 0 $$