Is this 3D algebra $T$ power-associative?

Question

Before reading this question it is essential that you understand power associativity

https://en.wikipedia.org/wiki/Power_associativity

In particular a commutative algebra does not necc imply a power-associative algebra.

Now Consider a commutative 3D algebra $T$ where the nonreal units $x,y$ satisfy

$$x^2 = A_1 + A_2 x + A_3 y $$ $$xy = B_1 + B_2 x + B_3 y$$ $$y^2 = C_1 + C_2 x + C_3 y$$

where all the parameters $A_1,A_2,..$ are real ofcourse.

So for instance we get

$$(a + b x + cy)^2 = a^2 + b^2 A_1 + 2bc B_1 + c^2 C_1 + ( 2ab + b^2 A_2 + 2bc B_2 + c^2 C_2) x + (2ac + b^2 A_3 + 2bc B_3 + c^2 C_3 ) y$$

And ofcourse there are some restrictions for the parameters $A_1,A_2,...$ ; the $x,y$ are not solvable in complex numbers ( such as $x = 1 +i, y = 2 $ ) and $x \neq y$ and more generally $x,y$ and the reals are linearly independant , because otherwise it is not really a 3D algebra. This implies that there are no numbers $i$ in the algebra that satisfy $i^2 = -1$.

I know that if $T$ is associative it is power-associative ofcourse, but that is trivial.

Now it is a known theorem that if an algebra is commutative and if for every element $p$ of that algebra $ p^4 = p^2 p^2 = p p^3 $ then the algebra is power-associative.

See also the link from the top.

This is a powerful tool.

Can we easily classify the condition for power-associative ?

In principle we just need to compute $p^2 p^2 - p p^3 $ and see if it is always $0$.

But that is alot of work.

Maybe there is an easier way.

Or maybe we can simplify things ?

addendum

For example

$$x^2 = 1 + 2 x + 3 y $$ $$xy = 4 + 5 x + 6 y$$ $$y^2 = 7 + 8 x + 9 y$$

Is not power-associative.

We have for instance $x^2 x^2 \neq x x^3 $

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Update :

It is relatively easy to show that One requires

$$B_2,A_3,B_3,C_3 \neq 0$$

for the dimension to be actually $3$, aka the linear independance of $x,y$ I mentioned above.

I am also aware that this algebra is not closed under square roots.

Just informing.

edit 1

(removed due to mistake)

EDIT 2

It turns out that by substition $x/q - v = x' , y/s - w = y'$,

we can make $A_2 = C_3 = 0$ , $B_1 = 1$.

This greatly simplifies the system !

So in essense we are studying

$$x^2 = A_1 + A_3 y $$ $$xy = 1 + B_2 x + B_3 y$$ $$y^2 = C_1 + C_2 x$$

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Remark : Im thinking about the connection to nilpotenty ... ($v^n = 0$), assuming one exists.

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Answer 1

Edit: I'm still quite confused by the wording, and this is not a complete answer but to correct my earlier mistakes while keep the meaningful parts.

Due to commutativity, for any $t\in T$, $t(tt)=(tt)t$, hence $t^3$ can be defined without ambiguity. The condition for power-associative by Albert requires $t^2t^2=t^3t$, in which case, $t^4$ can be defined without ambiguity. (We should not take $p^4=p^2p^2=p^3p$ as the condition, rather the condition is just $p^2p^2=p^3p$ and then use it to define $p^4$.)

I don't know an easy method to verify the condtion but here is a pretty easy example of power-associative but not associative algebra: $x^2=y^2=0, xy=yx=1$. One can check by hands this is indeed power-associative, or by Sage:

Build T

class T:
    def __init__(self, a, b, c):
        try:
            self.a = a.simplify_full()
            self.b = b.simplify_full()
            self.c = c.simplify_full()
        except:
            self.a, self.b, self.c = a, b, c
    def __str__(self):
        return str([self.a, self.b, self.c])
    def __eq__(self, B):
        return self.a==B.a and self.b==B.b and self.c == B.c
    def __mul__(self, B):
        return T(self.a*B.a + self.b*B.c + self.c*B.b, self.a*B.b + self.b*B.a, self.a*B.c + self.c*B.a)

Test

one = T(1, 0, 0)
x = T(0, 1, 0)
y = T(0, 0, 1)
zero = T(0, 0, 0)
assert(one*one==one and one*x==x and one*y==y)
assert(x*one==x and x*x==zero and x*y==one)
assert(y*one==y and y*y==zero and y*x==one)

Verification

var('a b c')
A = T(a, b, c)
A_sq = A*A
A_cub = A_sq*A
A4 = A_sq*A_sq
A4_ = A_cub*A
print(A4)
print(A4_)
assert(A4_ == A_cub*A)

Answer 2

After alot of algebra it turns out that if $x^2 \neq y^2$ we get that power-associative implies that the algebra is completely associative.

The shortcut to showing this is that if $x^2 \neq y^2$ and we can show $x$ and $y$ are power-associative, we get the same equations required for associative.

Hence power-associative implies that the algebra is completely associative.

The special case $x^2 = y^2 = 0$ has been given by the answer of "Just a user".

The other cases $x^2 = y^2 \neq 0$ reduces to lower dimensions such as complex numbers or $x= y$, so it is not really 3D.