Corollary 2

Let A be a commutative algebra of finite degree over a perfect field K, and let $\mathfrak{r}$ be its radical. There exists a unique subalgebra S of A such that $A = S\oplus \mathfrak{r}$. Moreover, S is isomorphic to a product of finitely many extensions of K of finite degree.

The K-algebra $A/\mathfrak{r}$ is semisimple (VIII, p. 173, Proposition 1) and has finite degree; it is absolutely semisimple because the field K is perfect (VIII, p. 232, Theorem 1). Since the ideal $\mathfrak{r}$ is nilpotent and A is commutative, the existence and uniqueness of S then follow from Corollary 1. Since S is semisimple and commutative and has finite degree, the last assertion is a consequence of Proposition 3 of VIII, p. 137.