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I recently have started studyng about free algebras and I want to know what kind of methods or approaches are there to prove that free algebra $A$ is nilpotent with nilpotency index $n.$

Let me remind that algebra $A$ is called nilpotent with niplotency index $n$, if for some small $n$ we have $A^n=0.$

My question is that if there is a free algebra $A$ that satisfied an identity $f = 0$ what kind of approaches are there to prove algebra $A$ is nilpotent(if it is nilpotent). I know only one method: if $A$ is nilpotent with nilpotency index $n$, then we can calculate all the elements in degree $n$ obtained from $f$. If these elements are $0$ then $A$ is niplotent.

Johny
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    For the case of free (nilpotent) Lie algebras, see here. – Dietrich Burde May 09 '23 at 13:19
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    A bit of clarification on which algebras are meant here (and what kind of nilpotence) would be nice. A nilpotent associative algebra need not have its nilindex (=maximal nilpotency index of an element) equal to its nilpotency index: $\Bbb F_2[x,y]/(x^2,y^2)$ has all elements square zero, but is nilpotent only of index $3$, as $xy \not = 0$. – Amateur_Algebraist Oct 01 '23 at 18:37

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