# Simple $R$-module where $R$ is a semisimple ring. Possible small improvement of a proof.

Reading through the proof of the following theorem (in Introduction to Group Rings, by Milies and Sehgal)

Let $L$ be a minimal left ideal of a semisimple ring $R$ and let $M$ be a simple $R$-module. Then $LM \ne (0)$ if and only if $L \cong M$ as $R$-modules. In this case $LM = M$.

The proof of ($\Longleftarrow$) goes as follows

Assume $L \cong M$ as $R$-modules and let $f : L \to M$ be an isomorphism. As $R$ is semisimple there is an idempotent element $e$ such that $L = Re$ (by previous lemma). Set $m_0 = f(e)$. Since $f(re) = rf(e) = rm_0$ for all $r \in R$ it follows that $m_0 \ne 0$. Since $m_0 = f(e) = f(e^2) = em_0$ it follows that $LM \ne (0)$ as claimed.

Now, I'd like to argue that: $m_0 \ne 0$ simply because $e \ne 0$ and $f$ is injective. This seems simpler? Is there a reason why the authors have not done this? Is it perhaps just a matter of taste?

Regards

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The two ideas seem pretty much the same, IMO. – rschwieb May 16 '12 at 19:25