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Prove that a finite-dimensional algebra $A$ over a field $K$ of characteristic zero having a basis consisting of nilpotent elements $\{e_1,...,e_n\}$ is nilpotent.

My approach: Let $m_i$ be the smallest positive integer such that $e_i^{m_i}=0$. Let $m:=m_1+m_2+\cdots+m_n$ and let $a_1,...,a_m$ be any $m$ elements from $A$. There exist $\lambda_{i1},...,\lambda_{in}$ such that $a_i=\lambda_{i1}e_1+\cdots+\lambda_{in}e_n$ for each $1 \leq i \leq m$.

Expand $a_1\cdots a_m$ in terms of the basis to get each term in the sum (expansion) has the form $\lambda e_1^{t_1}\cdots e_n^{t_n}$ where $\lambda \in K$ and $t_1+\cdots+t_n=m$ with $t_i \geq0$. Observe there exists $j$ such that $t_j \geq m_j$ otherwise it would contradict to $t_1+\cdots+t_n=m$, so each term in the expansion of $a_1\cdots a_m$ is zero; this implies $A^m=0$, therefore $A$ is nilpotent with nilpotency class at most $m$.

My query: I realized now that my proof relies on the commutativity of $A$, however $A$ is not assumed to be commutative. How to used the condition $\operatorname{Char}(A)=0$ for the general case?

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    It would help to know if this is a problem in a book, or in whatever the context you found it...2012-09-05
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    This is an important result in the foundations of noncommutative algebra and it is true without the commutativity hypothesis and without assuming $\mathrm{char}(K)=0$. I can write out a fairly short proof, but if you were assigned this in a class then I suspect you have recently covered some tools that make this doable, and it would help to know what you have recently been learning. Assigning this to someone who hasn't learned any algebra strikes me as unreasonable. (Unless you were only meant to do the commutative case, in which case your solution works.)2012-09-05
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    Yes, I should have added that good context to include would be any important theorems leading up to this problem :) Thanks to David Speyer's comment for reminding me...2012-09-05
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    The statement is from the second paragraph here: http://www.encyclopediaofmath.org/index.php/Nilpotent_algebra2012-09-05
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    Dear user, You should be able to adapt [this answer](http://math.stackexchange.com/a/133749/221) to prove your assertion. (Take $V$ to be $A$ itself, acting on itself by left multiplication.) Regards,2012-09-05
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    Thanks for the hint! That answer was based on the hypothesis that every element of $A$ is nilpotent. But in the question $A$ has only a nil basis, which does not imply each element is also nilpotent.2012-09-06

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