Linking associative algebras with nonassociative algebras - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T01:54:45Z http://mathoverflow.net/feeds/question/82220 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/82220/linking-associative-algebras-with-nonassociative-algebras Linking associative algebras with nonassociative algebras Serfo 2011-11-29T21:21:43Z 2011-12-01T11:06:57Z <p>Is the following statement interesting or even trivial ?</p> <ul> <li>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 :</li> </ul> <p>$1.$ $V_A$ is non commutative and non power associative !</p> <p>$2.$ $A$ is isomorphic to $N(V_A)$, where $N(V_A)$ is the nucleus of $V_A$.</p> <p>$3.$ If $n +1$ is odd then $Z(V_A ) = N(V_A)$, where $Z(V_A)$ is the center of $V_A$.</p> <p>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...</p> <p>Thank you</p> http://mathoverflow.net/questions/82220/linking-associative-algebras-with-nonassociative-algebras/82221#82221 Answer by DamienC for Linking associative algebras with nonassociative algebras DamienC 2011-11-29T21:32:41Z 2011-12-01T11:06:57Z <p>It seems to be trivial: take $V_A:=A$. </p> <p>EDIT: as it has been reformulated, the question has to be answered negatively now. </p> <p>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 <code>$dim(N(V_A))&lt;n$</code>. So there is no hope to have $A=N(V_A)$ even at the level of vector spaces. </p> http://mathoverflow.net/questions/82220/linking-associative-algebras-with-nonassociative-algebras/82234#82234 Answer by Noah S for Linking associative algebras with nonassociative algebras Noah S 2011-11-30T00:51:12Z 2011-11-30T03:17:54Z <p>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 $A$.</p> <p>Am I missing something?</p> <p>EDIT: This answer no longer (I think) applies to the question as it has been edited.</p>