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Structure of semisimple ring (Wedderburn-Artin) in Rings and Categories of Modules - Frank W. Anderson, Kent R. Fuller (auth.)

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Proof:

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Please explain that:

  1. "Now $_RR$ is direct sum off these traces". It's OK!
  2. "so (see 7) there is a finite set...". I read unit 7 and can't found that.
  3. $T_i$ is simple left ideal of the ring $Tr_R(T_i)$
  4. "Thus by 13.5 ..."

Thanks!

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So, turning back to 9.12, you see

9.12 Proposition. The socle of a left $R$-module $M$ is, as a left $R$-right $End(_RM)$-bimodule, a direct sum of its homogenous components.

In the discussion before this proposition, they give you the information to see that $Tr_R(T_i)$ is the homogenous component of $T_i$ in $M$. Explicitly, $Tr_R(T_i)=\sum \{_RS<_RR\mid _RS\cong _RT_i\}$.

For the case you are working on, $_RR=_RM$. Since $soc(_RR)=R$, Proposition 9.12 says for you "$_RR$ is a direct sum of $Tr_R(T_i)$".

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  • $\begingroup$ Since $R$ is semisimple ring $\Longrightarrow _RR$ is semisimple $\Longrightarrow _RR = \oplus _I (T_i), T_i$ is simple $\Longrightarrow Tr_R (_RR) = Tr_R \oplus _I (T_i)$ and Proposition 9.11 says: $_RR = \sum _I (Tr_R (T_i))$. Since $_RR$ is finite generated so $_RR = \oplus _{i=1} ^n (Tr_R (T_i))$. Right? $\endgroup$ – Rachel Nov 7 '13 at 14:15
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    $\begingroup$ Dear @Rachel : Well, 9.12 implies your last conclusion right away... are you trying to prove 9.12 from 9.11 now? Actually, I think A&F are laying on the terminology a little thick :) Many algebra books carry out this step this way: Show that the sum of mutually isomorphic left minimal ideals of a ring $R$ are an ideal of $R$ (called a homogeneous component for that isotype of ideal). Show that between different isotypes, the homogeneous components have intersection $\{0\}$, so their sum is direct. Finally, if $R$ is semisimple, the sum of all homogeneous components fills up $R$. $\endgroup$ – rschwieb Nov 7 '13 at 14:22
  • $\begingroup$ Thanks so much. But can u explain to me about my try above, is that ok? :) $\endgroup$ – Rachel Nov 7 '13 at 14:31
  • $\begingroup$ Dear @Rachel : Sure, I can give it a shot. The first line is certainly true, but I don't feel like the reasoning after "$T_i$ is simple $\implies$..." is very focused. Starting from your first line, think about grouping all the $T_i$ you wrote in $R\cong \oplus T_i$ by isotype. It turns out that the ones that are mutually isomorphic add up to an ideal. Instead of the $T_i$ then, you get $C_i=\oplus\{T_j\mid T_i\cong T_j\}$, and $\oplus C_i=R$. You're still working with the same direct sum, but now you collapsed groups of left ideals into ideals. $\endgroup$ – rschwieb Nov 7 '13 at 14:50
  • $\begingroup$ I mean, "$_RR= \oplus _I (T_i)$, where $T(i)$ is simple (semisimple module definition) $\Longrightarrow$ ..." $\endgroup$ – Rachel Nov 7 '13 at 15:03

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