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$R$ is taken to be the ring of upper $3 \times 3$ matrices with entries in $\mathbb{R}$.

If I view $R$ as a module over itself, are any of its submodules free?

And how can I prove that its submodules are pojective $R$-modules?

Thanks!

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None of its proper submodules could be free. Think about it: $R$ is a $6$ $\Bbb R$ dimensional algebra. It could not contain even a single copy of itself properly, considering that a single copy would have dimension at least $6$.

It certainly contains projective submodules, though. For any idempotent $e$, $eR$ is going to be a summand of $R$, and hence a projective module.

Actually triangular matrix rings over fields are hereditary meaning that all their left and right ideals are projective.


The only proofs I'm aware of for this would be a little long to write out. I know a proof appears in section 2 of Lectures on Modules and Rings, and First Course in Noncommutative rings section 25, both books by T.Y. Lam.

The version I like is the one where you show that a ring is right hereditary iff its Jacobson radical is a projective right $R$ module.

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  • $\begingroup$ Thank you . great! but I still need to show why it's hereditary. Any hint? $\endgroup$ – Carla B. Jan 6 '14 at 14:13
  • $\begingroup$ I mean my definition of hereditary is that all left and right ideals are projective... So, I'm still facing the same problem :( $\endgroup$ – Carla B. Jan 6 '14 at 14:15
  • $\begingroup$ @CarlaB. You should really ask "how can I prove" rather than "can I know if" in the original question, then... I'll see what I can add. $\endgroup$ – rschwieb Jan 6 '14 at 14:15
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The algebra you're studying is the path algebra over the quiver $$ \underset{1}{\bullet}\xrightarrow{\alpha}\underset{2}{\bullet} \xrightarrow{\beta}\underset{3}{\bullet} $$ so it's hereditary as all path algebras. You find a proof of this result in a paper by C. M. Ringel, see section 3, page 3.

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