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Reading through the proof of following theorem in 'Noncommutative Algebra by Farb and Dennis' :

If $R$ is a semisimple ring, then every simple $R$-module is isomorphic to a simple constituent of R.

The proof goes like this:

Suppose $R \cong \bigoplus_{i \in I} M_{i} $, and let $M$ be a simple R-module. Then, we have the maps :

$ \bigoplus_{i \in I} M_{i} \rightarrow R \rightarrow M $, where the second map is onto. Since, $M$ is simple, only one of the maps $ M_{i} \rightarrow M$ is non-zero, ........

I have problem with the bold part in the proof. The way I justified it was like this: Suppose $ i \neq j$, then $ M_{i} \ncong M_{j}$. So, if $ M_{i}\rightarrow M$ and $M_{j}\rightarrow M$ are non-zero, then $M_{i}, M_{j}$ and $M$ being simple, $M_{i} \cong M, M_{j} \cong M \implies M_{i} \cong M_{j}$. This is a contradiction.

Now, I am not quite satisfied with this argument, because 'if $ i \neq j$, then $ M_{i} \ncong M_{j}$' is true only if we combine together the simple summands of M, but then we will get $R \cong \bigoplus_{i \in I} M_{i}^{n_{i}} $, and $M_{i}^{n_{i}} $ are no longer simple. Please help me get this clear.

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  • $\begingroup$ I think it is better to say only one map upto isomorphism of R-modules is non-zero. I get that in any case, the argument after this statement remains same, and the proposition is proved, but the statement in bold seems false to me. $\endgroup$ – P-addict Sep 10 '19 at 6:56
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It seems already very clear to argue from first principles like this:

  1. There exists a homomorphism of right $R$ modules $\phi:R\to M$ that is onto.

  2. By the first isomorphism theorem, $R/\ker\phi\cong M$.

  3. Since $R$ is semisimple, $R=\ker\phi\oplus N$ for some submodule $N$ of $R$.

  4. By the second isomorphism theorem, $R/\ker\phi=(\ker\phi+N)/\ker\phi\cong N$.

Therefore $M\cong N$ where $N$ is a minimal right ideal of $R$.


If the argument given in the text is that among the injections of each $M_i$ into $R$ composed with the map $R\to M$, only one can be nonzero, then this is false.

For example, let $\phi: M_2(\mathbb R)\to \mathbb R^2$ simply extract the top row of the matrix, and express $M_2(\mathbb R)=\left\{\begin{bmatrix}a&b\\a&b\end{bmatrix}\right\}\oplus \left\{\begin{bmatrix}a&b\\0&0\end{bmatrix}\right\}$, the projection of both pieces is nonzero.

But if it is arguing that at least one is nonzero then this is fine. In your projection, you probably chose $1\mapsto m$ for some nonzero $m\in M$. Of course, $1=\sum m_i$ with each $m_i\in M_i$. Then $0\neq m=\sum\phi(m_i)$, and it cannot be that all the $\phi(m_i)$ are zero. Pick one that isn't zero, and you've proven that, for that particular $i$, the composed map from $M_i\to M$ is nonzero.

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    $\begingroup$ Thanks for your perfect answer! In the text, the proof goes like I have given, and yes, it says that only one of such maps can be nonzero. $\endgroup$ – P-addict Sep 10 '19 at 15:18
  • $\begingroup$ @P-addict No problem. Thanks for adding a quality post to the site! If only all users < 1000 rep took the time to ask as carefully as you did. $\endgroup$ – rschwieb Sep 10 '19 at 15:35

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