Linking associative algebras with nonassociative algebras - MathOverflow

Linking associative algebras with nonassociative algebras

2011-11-29T21:21:43Z

Is the following statement interesting or even trivial ?

For every $n$ - dimensional associative algebra $A$ over a field $F$ there is a $n +1$ - 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 $n +1$ 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

Answer by DamienC for Linking associative algebras with nonassociative algebras

2011-11-29T21:32:41Z

It seems to be trivial: take $V_A:=A$.

EDIT: as it has been reformulated, the question has to be answered negatively now.

If you require $A$ and $V_A$ to have the same dimension $n$, and you ask that there exists a triple $(x,y,z) $ in $V_A$ such that $(xy)z\neq x(yz)$, then $N(V_A)\subsetneq V_A$ and thus $dim(N(V_A))<n$ . So there is no hope to have $A=N(V_A)$ even at the level of vector spaces.

Answer by Noah S for Linking associative algebras with nonassociative algebras

2011-11-30T00:51:12Z

I'm still not convinced the question isn't trivial. Let $A$ be an associative algebra over a field $F$, and let $N$ be any nonassociative (in Serfo's particular sense of the word) $F$-algebra disjoint from $A$ with trivial nucleus. Then the Cartesian product $A\times N$ naturally inherits the structure of an $F$-algebra, and with this structure $A\times N$ is nonassociative and has nucleus (canonically?) isomorphic to $A$.

Am I missing something?

EDIT: This answer no longer (I think) applies to the question as it has been edited.