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Let $A$ be an associative finite dimensional simple algebra over a field $K$ of charicteristic zero. Prove that $A$ can not have a basis consists of nilpotent elements.

$Remark$: The statement is true if $A$ is algebraically closed, since by first structure theorem $A$ is isomorphic to a matrix space over a division algebra, for example $Mat_{n \times n}(D)$; but all division algebras over an algebraiclly closed field is the field itself, so one may prove by contradiction by examine $Mat_{m \times m}(K)$ and get a contradiction based on trace analysis.

However if $K$ is NOT algebraically closed then one can't work on matrices over $K$? Is there a way to prove the assertion by merely using $char(K)=0$ but WITHOUT using anything related to algebraically closeness? (also without extend $K$ in any form: I know one can consider $\bar A =A \otimes_{K} \bar K$ as an algebra over algebraically closed field $\bar K$ and deduce $A \otimes 1$ is nilpotent based on the nilpotency of $\bar A$, however this still uses algebraic closeness, which I intend to avoid)

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What about base change? Take a basis over your non-algebraically closed field $K$ and consider the $\overline K$-algebra $A\otimes_K\overline K$. What can you say about the original basis? – M Turgeon Sep 7 '12 at 0:40
Thanks! But in that way we still have to use the algebraically closeness of $\bar K$, which would imply $A \otimes 1$ is nilpotent. I wonder is there a way which does not use algebraic closeness at all.. – user31899 Sep 7 '12 at 0:50
up vote 5 down vote accepted

The statement of your previous question answers this: if $A$ is non-zero and has a nilpotent basis, then $A^2$ is a proper ideal of $A$ (since $A^n = 0$ for large $n$), and hence if $A$ is simple then $A^2 = 0$. Thus $A$ is just a $K$-vector space with the zero multiplication, and any proper subspace of $A$ is a proper ideal. The conclusion is that $A$ must be one-dimensional (and indeed, a one-dimensional $K$-vector space with the zero multiplication is a simple algebra, at least according to appropriate definitions of simple and algebra).

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Yeah, I knew this question sounded familiar :) From the author's using Wedderburn theory in his post, I thought he was assuming the ring had an identity, which would preclude the ring from being nil... – rschwieb Sep 7 '12 at 1:26
To be honest the aim of this new question is to get another proof of my previous question without assuming or extend $K$ to be algebraically closed :) So the proof is circular in some sense.. – user31899 Sep 7 '12 at 1:28

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