The Density Theorem

Let $M$ be an irreducible module and $\rho$ the corresponding representation. Define $\Delta=\textrm{End}_{R}M$. By Schur's lemma $\Delta$ is a division ring. We can consider $M$ as a left $\Delta$-module, for if $\delta\in\Delta,m\in M$ then define $\delta\cdot m=\delta(m)$. This gives $M$ a structure of a vector space over a division ring. We now consider $T=\textrm{End}_{\Delta}M$ the ring of transformations of $M$ as $\Delta$-module. In the same way, we can consider $M$ as a $T$-module. For $a\in R$, let $a_{M}:M\to M$ be $a_{M}(x)=ax$. For every $\delta\in\Delta$ we have

$$a_{M}(\delta\cdot x)=a\delta(x)=\delta(ax)=\delta\cdot a_{M}(x)$$

And this shows that $a_{M}\in\textrm{End}_{\Delta}M=T$. Note that $a_{M}=\rho(a)$ and therefore $\rho(R)\subset T$. It follows that if $N$ is a $T$-submodule of $M$ then $RT=\rho(R)T\subset T$ and therefore $N$ is a $R$-submodule of $M$.

Lemma 4.1   If $M$ is completely reducible $R$ module and $N\subset M$ then $N$ is also a $T$-submodule of $M$.

Proof. Let $M=N\oplus N'$ where $N'$ is another $R$-module. Define $\pi:M\to N$ to be the projection on $N$ which is induced from the above decomposition. Then $\pi\in\Delta$. Given any $t\in T$ we have:

$$tN=t(\pi M)=\pi(tM)\subset N$$

and therefore $N$ is a $T$-submodule. $\qedsymbol$

Theorem 4.2   Let $M$ be a completely reducible module and $x_{1},x_{2},\dots,x_{m}\in M$ and $t\in T$ as defined above. Then there is $a\in R$ such that $ax_{i}=tx_{i}$ for all $i$.

Proof. First we show this for $n=1$. Then, define $N=Rx_{1}$. By the previous lemma, $N$ is also a $T$-submodule and $N=Tx_{1}$. Then,

$$Tx_{1}=Rx_{1}$$

and it follows that there is $a\in R$ such that $ax_{1}=tx_{1}$. For $n>1$, consider the module $M^{(n)}$. It is clearly a completely reducible. It is easy to check that the mapping $(x_{1},\dots,x_{n})\to(tx_{1},\dots,tx_{n})$ is in the ring of endomorphisms of $M^{(n)}$ regarded as left $\textrm{End}_{R}M^{(n)}$ module. The above shows that there is some $a\in R$ such that $a(x_{1},\dots,x_{n})=(tx_{1},\dots,tx_{n})$ and the assertion follows. $\qedsymbol$