Let us consider the matrix algebra. $Mat\_n(\mathbb{C})$. The Amitsur-Levitzki identity states that for any matrices $X\_1, X\_2, ..., X\_{2n} \in Mat\_n(\mathbb{C})$ the sum $\Sigma\_{\sigma \in S\_{2n}} sgn(\sigma)X\_{\sigma(1)}...X\_{\sigma(2n)}$ vanishes identically.


Is it true that any identity (noncommutative polynomial, which always vanishes) between the matrices from $Mat\_n(\mathbb{C})$ follows from Amitsur-Levitzki in some sense$^\*$?


$\*$ - I assume the following precise meaning: let us consider the oper of "noncommutative polynomial mappings" with the substitution as the composition of operations. It naturally (also non-linearly) acts on any associative algebra (as we can evaluate an $r$-tuple noncommutative polynomial of $k$ variables on the $k$-tuple of elements of an algebra to obtain an $r$-tuple). Is it true that the polynomials that act trivially on the $Mat\_n(\mathbb{C})$ are precisely the ideal, generated by the Amitsur-Levitski polynomial?



Is the Amitsur-Levitzki identity essentially unique?
The answer to your question is "no", as explained by Anton Klyachko in his answer. Let me refer you to a remarkable statement of Razmyslov and Procesi that describes all identities. They proved (independently) that in fact all identities of $Mat\_n(\mathbb{Q})$ follow, in a sense, from the Cayley--Hamilton theorem. To be more precise:


Let us first define the notion of an *algebra with trace* as a vector space $A$ over a field $F$ that has a bilinear product, and a linear functional $t\colon A\to F$ satisfying $t(AB)=t(BA)$. Of course we can consider the free algebra with trace generated by a set $X$; it is the free associative algebra generated by $X$ over the ring of polynomials in $t(m)$, where $m$ is a cyclic word in $X$ (a cyclic group orbit on words), and $t$ is extended to this algebra by an obvious rule $t(t(m\_1)m\_2)=t(m\_1)t(m\_2)$. This gives us a language to discuss we can talk about identities of algebras with trace. (Alternatively, one can use the language of operads to discuss that; I prefer that latter language but choose to write a more "classical" definition here).


Next, let us define the $n$-th Cayley-Hamilton identity of an algebra with trace as the identity
 $$
CH\_n=\sum\_{\sigma\in S\_{n+1}}X\_{i\_1^{(1)}}\cdots X\_{i\_{k\_1}^{(1)}}t(X\_{i\_1^{(2)}}\cdots X\_{i\_{k\_2}^{(2)}})\cdots t(X\_{i\_1^{(l)}}\cdots X\_{i\_{k\_l}^{(l)}})=0
 $$
where $\sigma$ has the disjoint cycle decomposition $(0,i\_1^{(1)},\ldots,i\_{k\_1}^{(1)})(i\_1^{(2)},\ldots,i\_{k\_2}^{(2)})\cdots(i\_1^{(l)},\ldots,i\_{k\_l}^{(l)})$. (This is a full multilinearisation of the Cayley--Hamilton theorem $\chi\_A(A)=0$).


Theorem (Procesi [1], Razmyslov[2]). Every identity of the matrix algebra viewed as an algebra with trace is a consequence of the identity $CH\_n=0$.


[1] C.Procesi, The invariant theory of n × n matrices, Advances in Mathematics
Volume 19, Issue 3, March 1976, Pages 306–381,
<http://www.sciencedirect.com/science/article/pii/000187087690027X>


[2] Yu. P. Razmyslov, Trace identities of full matrix algebras over a field of characteristic zero, 
Izv. Akad. Nauk SSSR Ser. Mat., 1974, Volume 38, Issue 4, Pages 723–756, <http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=im&paperid=1989&option_lang=eng>


Of course, every usual identity is a particular case of a trace identity, so in principle this theorem also classifies all usual identities.


