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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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user31899
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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... – rschwieb Sep 05 '12 at 12:41
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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.) – David E Speyer Sep 05 '12 at 12:50
  • 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... – rschwieb Sep 05 '12 at 13:42
  • The statement is from the second paragraph here: http://www.encyclopediaofmath.org/index.php/Nilpotent_algebra – user31899 Sep 05 '12 at 22:11
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    Dear user, You should be able to adapt this answer to prove your assertion. (Take $V$ to be $A$ itself, acting on itself by left multiplication.) Regards, – Matt E Sep 05 '12 at 22:25
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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. – user31899 Sep 06 '12 at 00:02

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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.

rschwieb
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    Thanks for pointing my mistake out! The algebra may not be commutative, I have been introduced Jacobson radical ideal in the lectures. – user31899 Sep 05 '12 at 22:09
  • 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? – user31899 Sep 06 '12 at 06:34