How to prove that $$\operatorname{End}_R(V^{ \oplus n }) \cong M_n(D),$$ where $V$ is a simple left $R$-module and $D=\operatorname{End}_R(V)$.

This is part of the proof finally leads to prove that ring is simple and left semi-simple if and only if it is isomorphic to $M_n(D)$ for some division ring $D$, and what's already been proved is that $End_R(V \oplus W) \cong End_R(V) \oplus End_R(W)$ and its corollary.

  • $\begingroup$ Yes,it is just a title $\endgroup$
    – starry1990
    May 4, 2014 at 2:08
  • $\begingroup$ How would you do this if $R$ was a field and $V$ just a vector space? The idea is really very similar. Of course, I'm assuming that $V$ is simple here. $\endgroup$ May 4, 2014 at 7:14

1 Answer 1


Let $R=F$ be a field, and consider $V$ and $W$ with $F$ dimension each $1$. Then $\mathrm{End}_F(V)=\mathrm{End}_F(W)=F$, so their direct sum is $F\oplus F$. $W\oplus V$ is two dimensional, hence $\mathrm{End}_F(V\oplus W)$ is four dimensional. Thus you can see that $\mathrm{End}_R(V \oplus W)$ is not isomorphic to $\mathrm{End}_R(V) \oplus \mathrm{End}_R(W)$ in general.

The real picture is this, although it would need some explanation:

$\mathrm{End}_R(V\oplus W)\cong \begin{bmatrix}\mathrm{End}_R(V)&\mathrm{Hom}_R(W,V)\\\mathrm{Hom}_R(V,W)&\mathrm{End}_R(W)\end{bmatrix}$

You can picture these as working as matrix multiplication on the left side of pairs $\begin{bmatrix}v\\w\end{bmatrix}\in V\oplus W$. Of course when $V=W$, you just get

$\mathrm{End}_R(V\oplus V)\cong \begin{bmatrix}\mathrm{End}_R(V)&\mathrm{End}_R(V)\\\mathrm{End}_R(V)&\mathrm{End}_R(V)\end{bmatrix}=\begin{bmatrix}D&D\\D&D\end{bmatrix}$

So one way to fix the approach you've outlined is to prove the real version of the lemma you mentioned: $\mathrm{End}_R(\oplus_{i=1}^n V_i)\cong \oplus_{i=1}^n[\oplus_{j=1}^n\mathrm{Hom}_R(V_i,V_j)]$

Following the $2\times 2$ example above, you should be able to see the way.

  • $\begingroup$ Question: isn't the last isomorphism you mention one of abelian groups instead of algebras? $\endgroup$
    – Bib
    Jul 20, 2015 at 5:26
  • $\begingroup$ @bib nope: it is an algebra isomorphism (and a group isomorphism). The right hand side is not a direct sum of rings, yet it has a natural ring structure. $\endgroup$
    – rschwieb
    Jul 20, 2015 at 9:49
  • $\begingroup$ what is the multiplication? $\endgroup$
    – Bib
    Jul 20, 2015 at 20:32
  • $\begingroup$ @Bib formal matrix multiplication where the entries of the matrix come from the hom groups. $\endgroup$
    – rschwieb
    Jul 21, 2015 at 0:43

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