Jacobson radical = intersection of all maximal two-sided ideals 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.)
 A: 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: 


*

*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).

*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.
A: Just for the record, here is an example of a (necessarily infinite dimensional) $k$-algebra $A$ where the Jacobson radical is not equal to the intersection of all maximal two-sided ideals.
Let $k$ be a field of characteristic $0$ and let $A = U(\mathfrak{sl}_2) / \langle C \rangle $ where $C = ef + fe + \frac{1}{2}h^2$ is the Casimir element. The image of the augmentation ideal of $U(\mathfrak{sl}_2)$ in $A$ is the unique maximal two-sided ideal of $A$, but $A$ acts faithfully on the Verma module of highest weight $-2$ so $A$ is primitive and its Jacobson radical is zero.
