# Simple algebra over a field of char zero

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
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).
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