3
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This is a Wedderburn Theorem proof in Frank W. Anderson, Kent R. Fuller: Rings and Categories of Modules

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Please explain that: "Therefore $_RR$ has a composition series of length $n$". Exercise (11.5)

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  • $\begingroup$ For the first one: Every ring is a module over itself. $\endgroup$ – BIS HD Oct 26 '13 at 16:35
2
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(Maybe you didn't remember that $T^{(n)}$ is notation for $\oplus_{i=1}^{n}T$?)

Take the statement $_RR\cong T^{(n)}$ and look at that exercise. Since the sum is direct, the last line of (1) says that $c(_RR)=\sum_{i=1}^n c(_RT)$. But the composition length of $_RT$ is $1$, therefore $_RR$ has composition length $n$.

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  • $\begingroup$ Ok, thanks! But u can prove (1) to use it? $\endgroup$ – Rachel Oct 28 '13 at 15:13
  • $\begingroup$ @Rachel I can... if you want a solution for that question you might consider asking another question :) It's a bit too much for a comment, and not really relevant to the original question here. $\endgroup$ – rschwieb Oct 28 '13 at 17:58

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