# Easy examples of right alternative rings that are not alternative

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.

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.