Characterize finite dimensional algebras without nilpotent elements

Solution 1:

Suppose that $A$ is a finite dimensional associative algebra with $1$ which has no nilpotent elements.

The Jacobson radical $J(A)$ of $A$ is then zero, because $J(A)$ is a nilpotent ideal of $A$, so its elements are themselves nilpotent. This implies that $A$ is a semisimple algebra and Wedderburn's theorem tells us then that $A$ is isomorphic to a finite direct product of matrix algebras $\prod_{i=1}^rM_{n_i}(D_i)$ for some $n_1,\dots,n_r\geq1$ and some finite dimensional division $K$-algebras $D_1,\dots,D_r$.

Now, if $n>1$, then $M_n(D)$ has non-zero nilpotents whatever $D$ is, so we must have $n_1=\cdots=n_r=1$.

We thus conclude that $A$ is a direct product of finitely many finite dimensional division $K$-algebras.

(If we assume further that $K$ is algebraically closed, then there are no non-trivial finite dimensional division $K$-algebras and we must have $A=K\times\cdots\times K$.)