Does $(x^2)(x^3) \neq 0$ imply $(x)(x^3) \neq 0$?

Question

Consider a commutative unital algebra $A$ of finite dimension $n>3$ over the reals. The product is defined such that elements are generated with real number coefficients $(a_0, \dotsc, a_n) $ for some basis $\{1, i_1, \dotsc, i_n \}$.

Also we require $A$ to be a non-associative algebra. In fact not even power-associative. (otherwise the question has a trivial answer)

Define "F-nilpotent" as there is a nonzero element $x$ in $A$ such that $(x^2) (x^3) = 0$

Conjecture $C$ :

If the algebra $A$ is not "F-nilpotent", there is no nonzero element $x$ in $A$ such that $(x)(x^3) = 0$.

How to prove or disprove it ? Any counterexamples ?

Answer 1

The answer to,

Does $x^2x^3\ne 0$ imply $xx^3\ne 0$ for elements $x$ of a unital, commutative (non-associative) algebra $A$ of finite dimension $n>3$?

is no. Here is a counterexample.

Let $A$ be the unital, commutative, non-associative algebra of dimension $4$ over a field $F$ (you can take $F=\mathbb{R}$ for your question) with basis $1,x,x^2,x^3$ where the multiplication is as in the table

\begin{array}{c|ccc} & 1 & x & x^2 & x^3 \\ \hline 1 & 1 & x & x^2 & x^3 \\ x & x & x^2 & x^3 & 0 \\ x^2 & x^2 & x^3 & 0 & 1 \\ x^3 & x^3 & 0 & 1 & 0 \end{array}

Then $xx^3=0$ but $x^2x^3=1\ne 0$.

You have also asked a different question, which I understand is

Suppose there is no non-zero $x\in A$ satisfying $x^2x^3=0$, where $A$ is a unital, commutative (non-associative) algebra of finite dimension $n>3$. Does it follow that there is no non-zero $y\in A$ such that $yy^3= 0$?

This question is different to the one in the title, of course. It is also harder, as you note. I will have a think about this question too.