Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.)

share|improve this question
en.wikipedia.org/wiki/Jacobson_radical :( –  mathic 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
add comment

2 Answers

up vote 4 down vote accepted

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.

share|improve this answer
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
show 1 more comment

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.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.