TITLE: Jacobson radical = intersection of all maximal two-sided ideals
QUESTION [9 upvotes]: 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.)

REPLY [5 votes]: 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.