Proposition 5

— Let $A$ be a commutative algebra of finite degree over $K$. For $A$ to be reduced it is necessary and sufficient that there exist extensions $L_1, \ldots, L_s$ of finite degree over $K$ such that $A$ is $K$-isomorphic to $L_1 \times \cdots \times L_s$.

The stated condition is clearly sufficient.

Conversely, assume that $A$ is reduced; arguing by induction on the degree of $A$ we see that it suffices to prove that if $A$ is not a field, there exist two non-zero algebras $A_1$ and $A_2$ such that $A$ is isomorphic to $A_1 \times A_2$, or also that there exists in $A$ an idempotent different from 0 and 1.

Suppose from now on that $A$ is reduced and is not a field. Among the ideals of $A$ different from 0 and $A$ let $a$ be an ideal whose dimension as vector $K$-space is minimal. For any $x \neq 0$ in $a$ we have $x^2 \neq 0$, because $A$ is reduced, whence $a' \neq \{0\}$. We have $a^2 \subset a$ and by the minimality of $a$ we find that $a' = a$. By Lemma 4 there exists an idempotent $e$ such that $a = Ae$, and we have $e \neq 0, e \neq 1$ because $a$ is distinct from 0 and $A$.