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Is the following statement interesting or even trivial ?

• For every $n$ - dimensional associative algebra $A$ over a field $F$ there is a $n$ n +1$- dimensional nonassociative algebra$V_A$over$F$with the following properties :$1.V_A$is non commutative and non power associative !$2.A$is isomorphic to$N(V_A)$, where$N(V_A)$is the nucleus of$V_A$.$3.$If$n$n +1$ is odd then $Z(V_A ) = N(V_A)$, where $Z(V_A)$ is the center of $V_A$.

Ps - Sorry guys I have changed the formulation few times.The last change was due to a typo, I meant $V_A$ to be of dimension $n + 1$ ! I will now stop, thanks for all the replies...

Thank you

Is the following statement true interesting or even trivial ?

• For every $n$ - dimensional associative algebra $A$ over a field $F$ there is a $n$ - dimensional nonassociative algebra $V_A$ over $F$ with the following properties :

$1.$ $V_A$ is non commutative and non power associative !

$2.$ $A$ is isomorphic to $N(V_A)$, where $N(V_A)$ is the nucleus of $V_A$.

$3.$ If $n$ is odd then $Z(V_A ) = N(V_A)$, where $Z(V_A)$ is the center of $V_A$.

Thank you

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Is the following statement true or even trivial ?

• For every $n$ - dimensional associative algebra $A$ over a field $F$ there is a $n$ - dimensional nonassociative algebra $V_A$ such that over $F$ with the following properties :

$1.$ $V_A$ is non commutative and non power associative !

$2.$ $A$ is isomorphic to $N(V_A)$, where $N(V_A)$ is the nucleus of $V_A$.

Thank you

Ps - In recent days I have been doing some constructions for fun, wich led me to the statement above.Indeed if there are no flaws in my constructions then the statement is true !

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