nLab Wedderburn-Artin theorem




In algebra, the Wedderburn–Artin theorem 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 kk is isomorphic to a finite direct sum of matrix algebras over division algebras over kk. A semisimple algebra over kk is simple if and only if it is a matrix algebra over a division algebra over kk.

There is also a version for endomorphism rigs in a semiadditive category:


(Wedderburn–Artin Theorem for Endomorphism rigs)
If RR is a semisimple object in a semiadditive category A\mathsf{A}, then the endomorphism rig End(R)End(R) is a finite product of matrix rigs over division rigs.


There are many proofs of the Wedderburn–Artin theorem (Prop. ).

Common proof

A common modern approach proceeds as follows.

Suppose RR is semisimple. Then by definition any right RR-module is a direct sum of simple R R -modules, and any finitely generated RR-module must be a finite direct sum of such. Thus, the right RR-module R RR_R is isomorphic to a finite direct sum of simple RR-modules, which will be minimal right ideals of RR. Write this direct sum as

R R iI i m i, R_R \;\cong\; \bigoplus_i I_i^{\oplus m_i} \,,

where the I iI_i are mutually nonisomorphic simple right RR-modules, the iith one appearing with multiplicity m im_i. Then we have for the endomorphisms

End(R R) iEnd(I i m i) End(R_R) \;\cong\; \bigoplus_i End\big(I_i^{\oplus m_i}\big)

and we can identify End(I i m i)End\big(I_i^{\oplus m_i}\big) with a ring of matrices

End(I i m i)M m i(End(I i)), End\big(I_i^{\oplus m_i}\big) \;\cong\; M_{m_i}\big(End(I_i)\big) \,,

where the endomorphism ring End(I i)End(I_i) of the right RR-module I iI_i is a division ring by Schur's lemma because I iI_i is simple. Since REnd(R R)R \cong End(R_R) we conclude

R iM m i(End(I i)). R \;\cong\; \bigoplus_i M_{m_i}\big(End(I_i)\big) \,.

(We use right modules because REnd(R R)R \cong End(R_R); if we used left modules we would have REnd( RR) opR \cong End({}_R R)^{op}, but the proof would still go through.)

Nicholson’s proof

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 RR is a ring and KRK \subseteq R is a minimal left ideal with K 20K^2 \ne 0. Then K=ReK = R e for some eKe \in K with e 2=ee^2 = e, and eRee R e is a division ring.

Here a minimal left ideal is a nonzero left ideal for which the only smaller left ideal is {0}\{0\}.

For example, suppose RR is the ring of n×nn \times n matrices over some division ring DD This has a minimal ideal KK consisting of matrices with just one nonzero column, say the jjth column. You can see K 20K^2 \ne 0. To write K=ReK = R e, we can take ee to be the matrix that’s zero everywhere except for a 1 in the jjth row and jjth column. Clearly e 2=ee^2 = e, and eRee R e consists of matrices that are zero except in the jjth row and jjth column, so eRee R e is isomorphic to DD.

No deep techniques or preliminary lemmas are required to prove Brauer’s Lemma.

Proof of Brauer’s Lemma ()
Since 0K 20 \ne K^2, we must have Ku0K u \ne 0 for some uKu \in K. Of course u0u \ne 0. But KuK u is a left ideal contained in KK, so by minimality

Ku=K. K u = K.

Thus u=euu = e u for some eKe \in K. Now, let

L={aK:au=0} L = \{a \in K \colon a u = 0 \}

LL is a left ideal since for any rRr \in R we have

aLau=0rau=0raL. a \in L \; \implies \; a u = 0 \; \implies \; r a u = 0 \; \implies \; r a \in L \,.

Note LKL \subseteq K by definition, so by the minimality of KK we must have either L=0L = 0 or L=KL = K. But ee is in KK but not in LL, since eu=u0e u = u \ne 0, so we must have L=0L = 0. In other words

aKandau=0a=0. a \in K \; \text{and} \; a u = 0 \quad \implies \quad a = 0 \,.

To use this implication, recall that u=euu = e u so eu=e 2ue u = e^2 u so (ee 2)u=0(e - e^2)u = 0, and take aa above to be ee 2e - e^2. We conclude that ee 2=0e - e^2 = 0, so ee is idempotent:

e 2=e. e^2 = e \,.

Now we claim that K=ReK = R e. The reason is that ee is in the left ideal KK, so ReKR e \subseteq K. Since KK is minimal this implies either Re=0R e = 0 or Re=KR e = K. But e0e \ne 0 so Re0R e \ne 0.

Why is eRee R e a division ring? Its unit is ee, of course. Suppose beReb \in e R e is nonzero. To show bb has an inverse, note that RbR b is a nonzero ideal contained in Re=KR e = K so by minimality Rb=ReR b = R e. So, we must have e=rbe = r b for some rRr \in R. But this implies eree r e is the left inverse of bb in eRee R e:

(ere)b=er(eb)=erb=e 2=e. (e r e) b = e r (e b) = e r b = e^2 = e.

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 b0b \ne 0. It has a left inverse, say xx for short. To see that xx also the right inverse of bb note that xx must also have its own left inverse, say yy. Thus we have xb=1x b = 1 and yx=1y x = 1, giving

y=y(xb)=(yx)b=b. y = y (x b) = (y x) b = b .

Thus bb is the left inverse of xx. So xx is the right inverse of bb. ▮


  • 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 6, 2023 at 10:20:31. See the history of this page for a list of all contributions to it.