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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.

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  • $\begingroup$ I think $R$ needs to satisfy some chain condition for this to be true. $\endgroup$
    – Geoff Robinson
    Mar 26, 2014 at 0:05
  • $\begingroup$ Sorry, my answer assumed commutative rings so I deleted it. I wasn't thinking about the non commmutative case. $\endgroup$
    – Seth
    Mar 26, 2014 at 0:07
  • $\begingroup$ 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 $4$. $\endgroup$
    – Jack Schmidt
    Mar 26, 2014 at 3:53

1 Answer 1

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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.

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  • $\begingroup$ Can you please explain, why a simple module over $R$ must exist? Is $R$ assumed to be left-artinian? $\endgroup$
    – Dune
    Mar 26, 2014 at 10:02
  • $\begingroup$ @dune Simple modules exist because maximal right (and left) ideals exist. The quotient of the ring by such a right ideal is simple. This is the case if R has identity, anyhow. (At least, I was going on the impression the user probably wanted identity) $\endgroup$
    – rschwieb
    Mar 26, 2014 at 10:13
  • $\begingroup$ Oh, right. I mixed up simple modules and simple left ideals. Thank you! $\endgroup$
    – Dune
    Mar 26, 2014 at 10:25

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