# How to show that $A=(A/\operatorname{rad}A)\oplus \operatorname{rad}A$ using Wedderburn-Malcev theorem?

Let $A$ be a $K$-algebra and $K$ an algebraically closed field. How to show that $A=(A/\operatorname{rad}A)\oplus \operatorname{rad}A$ using Wedderburn-Malcev theorem?

Thank you very much.

Edit: This question is from Line -14 of page 64 of the book Elements of Representation Theory of Associative Algebras, volume 1. It is said that $A$ is a split extension of $A/\operatorname{rad}A$ by $\operatorname{rad}A$. I attached the page of the book. The Wedderburn-Malcev theorem (in the version for algebraically closed fields as stated in the mentioned book) says that $A=B\oplus \operatorname{rad} A$ and the restriction of the canonical homomorphism $A\twoheadrightarrow A/\operatorname{rad} A$ on $B$ is an isomorphism.
By definition $A$ is a split extension of $A/\operatorname{rad}A$ if there is a split surjective algebra morphism $A\twoheadrightarrow A/\operatorname{rad}A$ whose kernel is a nilpotent ideal of $A$. That the Jacobson radical is nilpotent in this case is well-known and Wedderburn-Malcev tells you that the morphism $B\hookrightarrow A$ has a right inverse (the restriction of the canonical homomorphism $A\twoheadrightarrow A/\operatorname{rad}A$ followed by a claimed isomorphism $A/\operatorname{rad} A\to B$).
If you have an algebraic closed field $K$ then $A/rad(A)$ is isomorphic to full matrix algebras over $K$. Therefore the radical factor structure is separable and your claim follows by W-M-T.