0
$\begingroup$

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. enter image description here

$\endgroup$
1
$\begingroup$

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 $A/\operatorname{rad} A\to B$).

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I'm not sure I see what this answer adds to the (already two-year-old) accepted answer... $\endgroup$ – Pierre-Guy Plamondon Jul 21 '15 at 8:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.