# Characterize finite dimensional algebras without nilpotent elements

Characterize all finite dimensional algebras (may not be commutative) over a field $K$ without nilpotent elements.

My condition: Let $A$ be any algebra (may not be finite dimensional), then it's easy to prove that $A$ has no nilpotent elements iff the equation $x^2=0$ has a unique solution (the trivial solution).

But is there a more explicit characterization of these finite dimensional algebras?

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You want them to be with unit and associative? – Mariano Suárez-Alvarez Sep 6 '12 at 6:29
Rings that have no nonzero nilpotent elements are usually called reduced rings. There is a short wiki article on them. – rschwieb Sep 6 '12 at 12:46

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 $A=K\times\cdots\times K$.)

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Thanks! Just another query:Is all semi-simple algebras unital? In other words if the Jacobson ideal of $A$ is trivial then $A$ must be unital? – user31899 Sep 6 '12 at 7:46
@user31899 No, they do not necessarily have identity, but that is the usual assumption. Actually it seems like conditions that imply a ring has unity are few and far between. (I'm thinking only of these sorts of "internal" conditions while writing this.) – rschwieb Sep 6 '12 at 12:44
@rschwieb, there are conditions implying the presence of idempotents, though. There's a famous one by Albert, the statement of which now I forget. – Mariano Suárez-Alvarez Sep 6 '12 at 17:37
@all Didn't find the result I remembered, but found two interesting results anyway: Kaplansky: "A compact dual ring has identity." Hirano, Poon & Tsutsui: "If $R$ is commutative, and $R^2=R$ and all ideals are weakly prime, then $R$ has an identity." – rschwieb Sep 6 '12 at 19:23