# Quick Question on a Proof of Artin-Wedderburn Theorem

Question [Edited]: [See below.] Are the isomorphisms in $(1)$ and $(2)$ (additive) group homomorphisms?

If I'm right, $\text{End}_R(M)$ is a ring, but $\text{Hom}_R\left(M_i^{n_i},M_j^{n_j}\right)$ is only a group (we can't compose two functions in it). However when coming back to the direct sum of $\text{End}_R\left(M_i^{n_i}\right)$ we get a ring again.

So, are we showing $\text{End}_R(M)\cong\displaystyle\bigoplus_{i=1}^k\text{End}_R\left(M_i^{n_i}\right)$ as rings by getting intermediate group isomorphisms?...

Also, where is the hypothesis "finite type" used in the proof? [Edit: This has been answered in the comments.]

Artin-Wedderburn Theorem: Let $M$ be a semisimple $R$-module of finite type. Then $$\text{End}_R(M)\cong\bigoplus_{i=1}^k M_{n_i\times n_i}(D_i)$$ for some division rings $D_i$.

Proof: $M=\displaystyle\bigoplus_{i=1}^k (M_i)^{n_i}$ where $M_i$ is simple. \begin{align} \text{End}_R(M)&=\text{End}_R\left(\bigoplus_{i=1}^k(M_i)^{n_i}\right)\\ &\cong\bigoplus_{1\leq i,j\leq k}\text{Hom}_R\left(M_i^{n_i},M_j^{n_j}\right)\tag{1}\\ &\cong\bigoplus_{i=1}^k\text{End}_R\left(M_i^{n_i}\right)\tag{2}\\ &\cong\bigoplus_{i=1}^kM_{n_i\times n_i}\left(\text{End}_R(M_i)\right)\tag{3}\\ \end{align} where $(2)$ is a consequence of Schur's Lemma and $(3)$ follows from a previous result.

$M_i$ simple $\stackrel{\text{Schur}}{\implies}$ $\text{End}_R(M_i)$ is a division ring.

All in all, $$\text{End}_R(M)\cong\bigoplus_{i=1}^kM_{n_i\times n_i}(D_i)$$ for some division rings $D_i$.$\blacksquare$

• I think since $M$ is a module of finite type, or as is common to say that as $M$ is a f.g $R$ module, you can write $M$ as a direct sum of finitely many simple submodules. So the very first step of the proof outlined above seems to use that. – SMG Nov 15 '14 at 2:32
• This is somewhat similar to something I asked long ago ( mathoverflow.net/questions/11576/apocryphal-maschke-theorem ). The intermediate isomorphisms (1) and (2) are a-priori just isomorphisms of additive groups, but if you look at the inverse isomorphism to their composition, you see that it just takes a list of endomorphisms of $M_i^{n_i}$ and combines them to an endomorphism of $\bigoplus_{i} M_i^{n_i}$; this is clearly a ring homomorphism, and thus an isomorphism because it is an inverse of a group isomorphism. – darij grinberg Nov 26 '14 at 17:33
• My comment was deleted since the answer I left it on was deleted. The isomorphism in (1) is akin to the block matrix decomposition of the endomorphism ring of a vector space given a direct sum decomposition of the vector space. – aes Nov 26 '14 at 17:45
• The individual summands in (1) don't have a ring structure, but the whole thing does: the multiplication is like matrix multiplication, as aes suggests above. When you make (1) into a ring this way, the first $\cong$ will be an iso of rings. (1)$\cong$(2) is then just the statement that the off-diagonal entries in all these "matrices" are zero, so the second iso is also a ring iso where (2) is a direct sum of rings. – Matthew Towers Nov 26 '14 at 18:34

## 1 Answer

It sounds a bit like by "are they group isomorphisms" you are really asking "are they only group isomorphisms and not ring isomorphisms?" Of course they are additive group isomorphisms, but they are actually more than that.

While the individual Hom groups aren't rings, the sum $\bigoplus_{1\leq i,j\leq k}\text{Hom}_R\left(M_i^{n_i},M_j^{n_j}\right)$ is a ring!

Perhaps the way to get comfortable with it is to imagine it as a "formal" ring of matrices with entries from the Hom groups. Then formal matrix multiplication composes them in a reasonable way such it is a ring. The matrix multiplication matches the composition in $\mathrm{End}_R(M)$ exactly through the map, so it is a ring homomorphism. This holds even if the $M_i$ lack their special properties (being pairwise nonisomorphic simple modules.)

In the passage from line (1) to (2), we simply eliminate a lot of superfluous terms in the direct sum and discover that the multiplication boils down to coordinatewise multiplication of a direct product of rings. That is what the pairwise nonisomorphic condition buys us.

So, the isomorphisms involved are ring isomorphism all the way through.

Here's another sort of example. Let $e$ be an idempotent in any ring with identity, and let $f=1-e$. Then one can show that $R\cong \begin{bmatrix}eRe&eRf\\fRe&fRf\end{bmatrix}$ as rings, where the multiplication on the right is formal matrix multiplication. Neither $eRf$ nor $fRe$ has to be a ring, but $eRe$ and $fRf$ are both rings.

You can look upon this as $R\cong\mathrm{End}_R(R)\cong \mathrm{End}_R(eR\oplus fR)=\mathrm{Hom}_R(eR\oplus fR,eR\oplus fR)\cong\begin{bmatrix}\mathrm{Hom}_R(eR,eR)&\mathrm{Hom}_R(fR,eR)\\\mathrm{Hom}_R(eR,fR)&\mathrm{Hom}_R(fR,fR)\end{bmatrix}$

The (group) isomorphism $\mathrm{Hom}_R(fR,eR)\cong eRf$ can be found in Lam's First course in noncommutative rings (Proposition 21.6.) It turns out to be a ring isomorphism if $e=f$.