This is in relation to Lam's A first course in noncommutative rings Chpt 1, Sec 3, Thm 3.3 proof to demonstrate $M_n(D)$'s simple module is $D^n$ where $D$ is division ring.

Clearly $M_n(D)\to End_D(D^n_D)$ is isomorphism where $D^n$ is treated as right $D$ vector space but matrix multiplication is on the left. Furthermore, it is clear that any $v\in D^n-0$, $M_n(D)v=D^n$.

$\textbf{Q:}$ The book says "$M_n(D)$ can be identified with $End(V_D)$ where $V_D=D^n$ treated as right $D$ vector space. $V_D$ is faithful module and facts in linear algebra over division ring imply that it is simple $R-$module." What are the facts being used here? The methods I mentioned above is the alternative method mentioned in the book which is clear. It is basically showing there is no more invariant subspace of $D^n$ under $M_n(D)$ action and thus there is no smaller non-trivial submodule.

$\textbf{Q':}$ What is a non-trivial example of $M_n(D)$'s not faithful representation? One could imagine a ring homomorphism $M_n(D)\to M_n(D)/m$ with $m$ any maximal ideal of $M_n(D)$ but this quotient may not be representation of anything.

$\textbf{Q'':}$ The book says $End_D(_RD^n)\cong D$. Since $D^n$ is simple, by schur lemma one deduces $End_D(_R D^n)$ is $D$ itself but not some larger division ring. Why do I expect it being $D$? Clearly $D\to End_D(_R D^n)$ is division ring embedding.

  • $\begingroup$ @reuns $D^n$ is a simple left $R=M_n(D)$ module. And simple left modules of $M_n(D)$ are classified by its column vector spaces as $R$ left modules. $\endgroup$ – user45765 Jun 1 '19 at 0:53

1). I would recommend proving this lemma: an $R$ module $M$ is simple iff for every pair of nonzero elements $x,y\in M$, there exists $r\in R$ such that $xr=y$.

Let $v,w$ be any nonzero elements of $D^n$. By basic linear algebra, there exists a linear transformation mapping one to the other. This shows the module is simple.

2) $M_n(D)$ has exactly one nonfaithful representation: $\{0\}$. The annihilator if any other representation is a proper ideal, and this ring has only one proper ideal (the zero ideal.)

3) this is proven here, for example

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  • $\begingroup$ @reuns yes, the standard meaning of the phrase “an ideal” in ring theory is “a two-sided ideal. “ $\endgroup$ – rschwieb Jun 1 '19 at 17:05
  • $\begingroup$ Ok you meant for a two-sided ideal $I=RJR$ the kernel of the left-action of $R$ on $R/I$ is $I$. But for a left-sided ideal $I = RJ$, if $c$ is in the kernel then $RcR$ is also in the kernel, this is a two-sided ideal, which as you said for $R = M_n(D)$ must be $\{0\}$ (and $c=0$) or $R$ (and $R/I = \{0\},I = R$) $\endgroup$ – reuns Jun 1 '19 at 17:20
  • $\begingroup$ @reuns what I’m saying is a lot simpler than that. The kernel of action of $R$ on any module, left or right, is a two-sided ideal of $R$. $\endgroup$ – rschwieb Jun 1 '19 at 19:03

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