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

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.

• 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. – P-addict Sep 10 '19 at 6:56

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.

• 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. – P-addict Sep 10 '19 at 15:18
• @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. – rschwieb Sep 10 '19 at 15:35