Easy examples of right alternative rings that are not alternative

Question

By a ring I mean a $\mathbb{Z}$-module $A$ equipped with a binary operation $*:A\times A\rightarrow A$ that is $\mathbb{Z}$-bilinear and denoted by $(x,y)\mapsto xy$.

Let $A$ be a ring.

1. $A$ is said to be right alternative if $(xy)y=x(yy)$ for any $x,y\in A$.

2. $A$ is said to be left alternative if $(xx)y=x(xy)$ for any $x,y\in A$.

3. $A$ is said to be flexible if $(xy)x=x(yx)$ for any $x,y\in A$.

I know that, if we assume two of the properties above, then the remaining property holds.

A ring is said to be alternative if all three properties above hold.

However, I do not know an easy example of a right alternative ring that is not alternative.

Of course I can try the free right alternative ring, but are there easier examples? There is another very complicated example, provided by Mikheev in the article Simple Right Alternative Rings, that is a simple right alternative ring that is not alternative.

I know that every right alternative ring is power-associative, where a ring $A$ is said to be power-associative if for every $x\in A$ the subring $A(x)$ generated by $x$ is associative. Therefore, in a power-associative ring, we can define powers of an element as usual.

I know that, by a result from Mikheev, a right alternative ring satisfies the identity $(x,x,y)^4=0$, where $(x,y,z)$ stands for $(xy)z-x(yz)$. In particular, every right alternative ring without nonzero nilpotent elements is alternative.

I know that, by a result from Albert, every finite-dimensional right alternative algebra over a field of characteristic $\neq2$, which does not contain nil-ideals, is alternative

I also know that, by another result from Albert, every semisimple, right alternative algebra over a field of characteristic $0$ is alternative.

Answer 1

Let $R$ be a suitable commutative base ring, e.g. $R=\mathbb{F}_2$ is fine, but $\mathbb{Z}$ or $\mathbb{Q}$ should do as well.

Look at the free $R$-module $A=\langle x,y,z\rangle_R$ of rank 3. It suffices to give the multiplication on a basis, e.g. by a multiplication table:

a*b	x	y	z
x	x	z	0
y	y	0	0
z	0	0	0

where $a$ is from the first column and $b$ from the top row.

$A$ is not left alternative: $(xx)y=xy=z\neq 0=xz=x(xy)$.

$A$ is right alternative: can be checked by a computer.

In fact I used a little MAGMA-script to come up with the example. One can run it on the MAGMA online calculator:

k := GF(2);
// generate the set of all structure constants
M := RModule(k, 27);
for x in M do 
// generate all rank=3 k-algebras until we have found a suitable one
A := Algebra< k,3 | ElementToSequence(x) >;
// check right alternative
b1 := forall{<x, y>: x, y in A | (x*y)*y eq x*(y*y)};
// check left alternative
b2 := forall{<x, y>: x, y in A | (x*x)*y eq x*(x*y)};
// if right but not left alternative: put out structure constants and break
if (b1 and not(b2)) then 
b1;
b2;
x;
break;
end if;
end for;

$A$ has 8 elements and as an $R=\mathbb{F}_2$-algebra is also a $\mathbb{Z}$-algebra. In this sense it is easy. It is not simple however, as $\langle z\rangle$ is a two-sided ideal.

If you add and IsSimple(A) in the if-conditional above, MAGMA produces such an algebra, which is simple:

a*b	x	y	z
x	z	x	y
y	0	y	0
z	y	0	0

Remark: $A$ is not unital. If unitarity is required add another basis element $1_A$ and extend the definition of the product in the obvious way.