# matrix with coefficients of division ring's faithful representation and simple modules

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.

• @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. – 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.)
• 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 $R/I = \{0\},I = R$) – reuns Jun 1 '19 at 17:20
• @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 $R$. – rschwieb Jun 1 '19 at 19:03