I'm embarassed to ask this question, but the literature on noncommutative rings seems to give this a berth as if it was absolutely trivial and not worth discussing, and I can't prove it, so all I can do is ask it here...


Let $A$ be a finite-dimensional $k$-algebra, where $k$ is a field. Is it true that the Jacobson radical equals the intersection of all maximal two-sided ideals? (The latter intersection is known as the Brown-McCoy radical of $A$.)


If yes, a short proof (the more self-contained, the better) would be great.


(This is again for use in coalgebra theory.)



Jacobson radical = intersection of all maximal two-sided ideals
Yes, this is true; it's essentially just a restatement of Artin-Wedderburn. All you need to do is note that by Artin-Wedderburn, a finite dimensional algebra with trivial Jacobson radical is a sum of matrix algebras over division rings (where it's obvious that the intersection of all maximal ideals is trivial); for an arbitrary ring, kill the Jacobson radical, and apply the result to see you've killed the intersection of maximal ideals.


**EDIT**: Kevin makes a good point, which is that there are basically two parts of Artin-Wedderburn: 


1. Showing that a semi-simple Artinian ring (in the sense of trivial Jacobson radical) is a direct sum of simple rings (in the sense of no proper two-sided ideals).
2. Showing that every simple Artinian ring is a matrix ring over a division ring.


You only need 1. for this fact. On the other hand, if I wanted to use this fact in a paper, I would say something like "the Jacobson radical of a finite-dimensional $k$-algebra is the intersection of its maximal two-sided ideals; this follows from Artin-Wedderburn."


Of course, you could cite this MO page.


