symmetric monoidal (∞,1)-category of spectra
In algebra, the Wedderburn-Artin theorem [Wedderburn 1908, Artin 1927] gives a clean characterization of those rings that are matrix algebras over division rings.
Concretely, the theorem says that any semisimple ring is a finite direct sum of matrix algebras over division rings. (Here a ring is called semisimple if each of its left or equivalently right modules is a finite direct sum of simple modules.)
(Wedderburn–Artin Theorem)
Every semisimple ring is isomorphic to a finite direct sum of matrix algebras over division rings. A semisimple ring is simple if and only if it is a matrix algebra over a division ring.
Beware: A simple ring may not be semisimple! A ring is simple iff it has no two-sided ideals, and in the absence of further hypotheses this does not imply that all of its left (or equivalently right) modules are direct sums of simple modules. For example, the Weyl algebra is simple but not semisimple, and not isomorphic to a matrix algebra over a division ring.
However, a simple ring that is left or right artinian is semisimple, so we have:
Every simple ring that is left or right artinian is isomorphic to a matrix algebra over a division algebra.
There is also a version of the Wedderburn–Artin theorem for associative algebras over fields:
(Wedderburn–Artin Theorem for Algebras over Fields)
Every semisimple algebra over a field is isomorphic to a finite direct sum of matrix algebras over division algebras over . A semisimple algebra over is simple if and only if it is a matrix algebra over a division algebra over .
There is also a version for endomorphism rigs in a semiadditive category:
(Wedderburn–Artin Theorem for Endomorphism rigs)
If is a semisimple object in a semiadditive category , then the endomorphism rig is a finite product of matrix rigs over division rigs.
There are many proofs of the Wedderburn–Artin theorem (Prop. ).
A common modern approach proceeds as follows.
Suppose is semisimple. Then by definition any right -module is a direct sum of simple -modules, and any finitely generated -module must be a finite direct sum of such. Thus, the right -module is isomorphic to a finite direct sum of simple -modules, which will be minimal right ideals of . Write this direct sum as
where the are mutually nonisomorphic simple right -modules, the th one appearing with multiplicity . Then we have for the endomorphisms
and we can identify with a ring of matrices
where the endomorphism ring of the right -module is a division ring by Schur's lemma because is simple. Since we conclude
(We use right modules because ; if we used left modules we would have , but the proof would still go through.)
A different style of proof can be found in Nicholson (1993). A key step in this approach is “Brauer’s Lemma”.
(Brauer’s Lemma)
Suppose is a ring and is a minimal left ideal with . Then for some with , and is a division ring.
Here a minimal left ideal is a nonzero left ideal for which the only smaller left ideal is .
For example, suppose is the ring of matrices over some division ring This has a minimal ideal consisting of matrices with just one nonzero column, say the th column. You can see . To write , we can take to be the matrix that’s zero everywhere except for a 1 in the th row and th column. Clearly , and consists of matrices that are zero except in the th row and th column, so is isomorphic to .
No deep techniques or preliminary lemmas are required to prove Brauer’s Lemma.
Proof of Brauer’s Lemma ()
Since , we must have for some . Of course . But is a left ideal contained in , so by minimality
Thus for some . Now, let
is a left ideal since for any we have
Note by definition, so by the minimality of we must have either or . But is in but not in , since , so we must have . In other words
To use this implication, recall that so so , and take above to be . We conclude that , so is idempotent:
Now we claim that . The reason is that is in the left ideal , so . Since is minimal this implies either or . But so .
Why is a division ring? Its unit is , of course. Suppose is nonzero. To show has an inverse, note that is a nonzero ideal contained in so by minimality . So, we must have for some . But this implies is the left inverse of in :
Finally, in a ring where every nonzero element has a left inverse, every element has a two-sided inverse, so it’s a division ring! To see this, say . It has a left inverse, say for short. To see that also the right inverse of note that must also have its own left inverse, say . Thus we have and , giving
Thus is the left inverse of . So is the right inverse of . ▮
The original articles:
Joseph H. M. Wedderburn: On hypercomplex numbers, Proc. London Math. Soc. s2-6 1 (1908) 77-118 [doi:10.1112/plms/s2-6.1.77]
Emil Artin: Zur Theorie der hyperkomplexen Zahlen, Abh. Math. Semin. Univ. Hambg. 5 (1927) 251–260 [doi:10.1007/BF02952526]
Review:
William K. Nicholson, A short proof of the Wedderburn–Artin theorem, New Zealand J. Math 22 (1993) 83-86 [pdf]
Tsiu-Kwen Lee, A short proof of the Wedderburn-Artin theorem, Communications in Algebra 45 7 (2017) [doi:10.1080/00927872.2016.1233242]
John Baez, The Wedderburn–Artin Theorem, n-Category Café, June 14, 2013 (web)
See also:
Last revised on September 2, 2024 at 14:34:56. See the history of this page for a list of all contributions to it.