I saw somewhere (I appreciate if anyone has any references to proof of this fact) that if $A$ is a finite dimensional associative algebra such that $\textrm{dim}(A)<n$, then $A$ satisfies the standard identity of degree $n$: $$s_n(x_1,\dots, x_n)=\sum\limits_{\sigma\in S_n}\textrm{sgn}(\sigma)x_{\sigma(1)}\cdots x_{\sigma(n)},$$ where $S_n$ is the symmetric group of degree $n$.

Is there a non-associative version of this result? I mean, if $A$ is a finite-dimensional (not necessarily associative) algebra, then $A$ satisfies some kind of "non-associative" standard polynomials? For example, maybe: $$P_1(x_1)=x_1=s_1(x_1),$$ $$P_2(x_1,x_2)=x_1x_2-x_2x_1=s_2(x_1,x_2),$$ $$P_3(x_1,x_2,x_3)=x_1(x_2x_3)-(x_1x_2)x_3-x_1(x_3x_2)+(x_1x_3)x_2-x_2(x_1x_3)+(x_2x_1)x_3+x_2(x_3x_1)-(x_2x_3)x_1+x_3(x_1x_2)-(x_3x_1)x_2-x_3(x_2x_1)+(x_3x_2)x_1=x_1(x_2x_3-x_3x_2)+x_2(x_3x_1-x_1x_3)+x_3(x_1x_2-x_2x_1)+(x_3x_2-x_2x_3)x_1+(x_1x_3-x_3x_1)x_2+(x_2x_1-x_1x_2)x_3$$ and etc... Thanks in advance for answers!

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    $\begingroup$ This is the Amitsur-Levitzki identity. Some basic searching around this name yields the paper Minimal Identities of Octonian Algebras, although this seems to be a particular case of your question. $\endgroup$
    – Mark
    Nov 24 '19 at 6:03
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    $\begingroup$ The Amitsur-Levistski is quite different. It says that $M_n(k)$ satisfies the identity with $s_{2n}$, although its dimension is $n^2$. $\endgroup$ Nov 24 '19 at 18:39
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    $\begingroup$ @LSpice You're right. Thanks! $\endgroup$ Nov 25 '19 at 3:06

The first question, why the standard identity $s_n$ holds in any $< n$-dim associative algebra is easy: This is a multilinear identity, so in order to check it for arbitrary $x_1,...,x_n$, it is enough to check it for basic elements $x_1,...,x_n$. Since the dim $<n$, two of the $x_1,...,x_n$ are equal, which makes $s_n=0$.

In the non-associative case the situation is slightly more complicated. For every elements $x_1,....,x_n$ of the algebra denote $x_1\cdot ...\cdot x_n=(...(x_1x_2)x_3)...)x_n$ (one can choose any other arrangement of parentheses. Then $s_n$ is defined as usual, and the proof that $s_n=0$ is an identity is as above.


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