# Example of a faithful irreducible module

Let $R$ be a non trivial simple ring. I am trying to show that there is a faithful irreducible left $R$-module.

Is the ring $R$ considered as a left module over itself such a module? I think it's faithful since the ring map $R \to R_\ell: r\mapsto (r_\ell:a\mapsto ra))$ has zero kernel but I am not sure that it's irreducible. How do you show that it has no proper left ideals?

Many thanks.

• I think $R$ needs to satisfy some chain condition for this to be true. – Geoff Robinson Mar 26 '14 at 0:05
• Sorry, my answer assumed commutative rings so I deleted it. I wasn't thinking about the non commmutative case. – Seth Mar 26 '14 at 0:07
• If $R$ is not a division ring, then (the left regular module) $R$ is not an irreducible module over itself. So for instance, if $R$ is the ring of $2\times 2$ matrices over a field, then $R$ is simple, so every module is faithful, but its simple modules all have dimension $2$ over the field, while $R$ itself has dimension $4$. – Jack Schmidt Mar 26 '14 at 3:53

Over a simple ring $R$, every nonzero unital module is faithful.
So just take any simple $R$ module $S$ and you have an irreducible faithful module.
• Can you please explain, why a simple module over $R$ must exist? Is $R$ assumed to be left-artinian? – Dune Mar 26 '14 at 10:02