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.)

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– Alex Dec 25 '10 at 18:25
The counterexample in the Wikipedia article is bullshit (cf. mathoverflow.net/questions/775/… ). If I had any expert to check with, I'd fix it. – darij grinberg Dec 25 '10 at 18:33
I wont venture so far as to say its bs in general, but it certainly doesnt apply to the k algebra case, since a simple algebra always splits over a finite extension of the base field – Adam Gal Dec 25 '10 at 18:58

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.

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This shows that Jacobson contains Brown-McCoy. Is the other containment obvious? – Kevin Ventullo Dec 25 '10 at 21:25
For Artin rings, the other containment follows from the fact that the Jacobson radical is nilpotent (which is clear by the Jordan-Hölder decomposition - here the Artinianity(sp?) comes in). – darij grinberg Dec 25 '10 at 21:58
Thanks, Ben, this does it, although I hoped to avoid Artin-Wedderburn. Is it really a restatement, i. e., equivalent to Artin-Wedderburn? – darij grinberg Dec 25 '10 at 22:03
I think you can avoid using the description of simple $k$-algebras: you should only really need semisimplicity of $A/\text{rad}(A)$ -- once you know that then $A$ will be a (finite) direct sum of its minimal two-sided ideals, so then the maximal two-sided ideals will be the direct sum of all but one of these, so you're done. – Kevin McGerty Dec 25 '10 at 22:23
You mean $A\diagup \mathrm{rad}\left(A\right)$ rather than $A$, right? (Otherwise, nice comment; will have a closer look at it.) – darij grinberg Dec 25 '10 at 22:57

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.

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