# Show that a ring is isomorphic finite direct product of matrix rings over a field

Let $$F$$ be an algebraically closed field and let $$R$$ be a finite-dimensional $$F$$-algebra and suppose that $$R$$ has no nonzero nil ideals. Show that $$R$$ is isomorphic to a finite direct product of matrix rings over $$F$$.

I think it might be useful to use the Artin-Wedderburn theorem. I know that if $$R$$ is a finite-dimensional $$F$$-algebra then $$R$$ is an artinian ring, but I'm not sure how to prove this.

However, if I can prove that claim, by the Artin-Wedderburn theorem, $$R$$ is isomorphic to a direct product of matrix rings over division rings, but these division rings might not be $$F$$, so I was wondering how I could fix this?

For your first question, a right ideal is also an $$F$$ subspace. A proper containment of subspaces must experience a drop of at least one dimension. This cannot continue indefinitely.
For your second question, you should probably double check your version of the A-W theorem. Usually for algebras it’s stated as a product of matrix rings “…over division rings that are finite dimensional extensions of $$F$$”. It’s not hard to show an algebraically closed field cannot have (proper) extensions like that.