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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 $n +1$ - 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

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    $\begingroup$ This is simply not an interesting statement. If $V_A$ would have any reasonable positive property rather than just non(power)associativity and noncommutativity, it might be of some use. Negative properties hardly interest anyone here (which is the reason why people misunderstood your "nonassociative" as "not necessarily associative" first!). $\endgroup$
    – darij grinberg
    Nov 30, 2011 at 1:56
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    $\begingroup$ Dear Serfo, I would like to make two side comments: 1. when you change the formulation of the question in a sensible way, please keep track of the original formulation so that people who read the answers that were posted before your edits understand the discussion. 2. it seems that you have posted a different but similar question a few days ago under a different pseudo (mathoverflow.net/questions/81878/…) without paying so much attention to the answer that was given. I might be totally wrong but it sounds a bit unfriendly. $\endgroup$
    – DamienC
    Nov 30, 2011 at 8:10
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    $\begingroup$ @darij: I strongly disagree with what you said. On the contrary, requiring non-associativity often limits the possibilities enormously! For instance, consider alternative division rings. If they are associative, then there are heaps of them, too many to classify. But if you impose non-associativity, there is only one class: the 8-dimensional Cayley-Dickson algebras (a.k.a. octonion algebras). $\endgroup$
    – Tom De Medts
    Nov 30, 2011 at 8:51
  • $\begingroup$ What if $n=0$ and $A$ is the zero algebra? Your statement seems to be false in this case. $\endgroup$
    – S. Carnahan ♦
    Dec 1, 2011 at 8:18

2 Answers 2

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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.

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  • $\begingroup$ Dear Damien thanks for the reply but let say $V_A : \neq A$. Don`t forget the assumption is $A$ associative while $V_A$ non associative. $\endgroup$
    – Serfo
    Nov 29, 2011 at 21:52
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    $\begingroup$ Then the question does not make sens for me. It seems to me that associative algebras are examples of non-associative algebras. To be short, you have a functor $N:non-As-alg\to As-alg$ and you are looking for an adjoint to it (which is nothing but the forgetful functor $As-alg\to non-As-alg$). $\endgroup$
    – DamienC
    Nov 29, 2011 at 22:17
  • $\begingroup$ Dear Damien let say I am old school guy ( I do not use categorical concepts at all ) ! "It seems to me that associative algebras are examples of non-associative algebras " Well allow me to use the following " old school set theoretic " definition : - An algebra over a field $F$ is associative if $a(bc) = (ab)c$ for all $a,b, c \in A$, otherwise $A$ is non associative ( i.e, $a(bc) =(ab)c $ does not hold in general ! $\endgroup$
    – Serfo
    Nov 29, 2011 at 22:55
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    $\begingroup$ @Serfo: my understanding is that people generally use "non-associative algebra" to mean "algebra that is not necessarily associative." $\endgroup$
    – Qiaochu Yuan
    Nov 30, 2011 at 0:11
  • $\begingroup$ DamienC, I don't think an adjoint is what is being sought here. $\endgroup$
    – Mariano Suárez-Álvarez
    Nov 30, 2011 at 0:54
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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.

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  • $\begingroup$ I should add that I am assuming, here, that algebras need not be unital; otherwise there is no algebra with trivial nucleus. Even then, I think, some similarly straightforward construction would probably still work (maybe adjoin new nonassociating objects $x, y, z$, and ensure that no new element of the algebra is in the nucleus?). $\endgroup$
    – Noah Schweber
    Nov 30, 2011 at 0:55

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