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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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    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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(An algebra without identity! Somewhat unusual...)

It looks OK, but at this step in your argument: $\lambda e_1^{t_1}...e_n^{t_n}$ it looks like you are assuming commutativity, which I don't know if you intended or not. You would need commutativity to lump the like factors into powers this way.

If you were not supposed to assume commutativity, then maybe the zero characteristic is important for an alternative proof.

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    This fact was proved by Wedderburn which wasn't assuming $char(K)=0$. It seems one needs to use upper triangular matrices to prove the weaker case?2012-09-06