We would like to prove the following result.
Theorem. Let $R$ be a semisimple left Artinian ring. Then, every non-zero left ideal $J$ of $R$ can be written as a direct sum of
finitely many minimal ideals.
We can use the following results:
(1) Every non-zero left ideal of an Artinian Ring $R$, contains a minimal left ideal.
(2) If $I$ is a minimal left ideal of $R$, then either $I^2=\{ 0_R \}$ or there exists an idempotent element $e\in R$ such that $I=Re=Ie$.
(3) If $e\in R$ is an idempotent element of $R$, then $R=Re\oplus R(1_R-e)$.
In the proof in my notes, I find the first line confusing. It says that assume that the result fails, ie assume that $J$ is a left ideal, not a direct sum of minimal ideals but any left ideal $J'\subsetneqq J$ is a direct sum of minimal ideals.
So,
- Could you please explain why and how do we take this assumption?
- Could you please write down clearly the proof using the facts above?
Note that we say that $R$ is semisimple left Artinian, if it is left Artinian with nilradical $N(R)=\{0_R\}$ (where $N(R)$ is defined as the sum of all nilpotent ideals of $R$).
Thank you.
Best Answer
Consider
$$\mathcal{F}=\{I\subseteq R|\,I\mbox{ is a left ideal in }R\mbox{ and }I\mbox{ is not a direct sum of minimal left ideals of }R\}$$
Assume that $\mathcal{F}$ is nonempty (if it is empty, then we are done). Since $R$ is left artinian, there exists a minimal (with respect to inclusion) ideal $I\in \mathcal{F}$. Then $I$ is a left ideal and not a direct sum of minimal left ideals of $R$. Note that for every left ideal $J\nsubseteq I$ by minimality of $I$ we deduce that $J$ is a direct sum of minimal left ideals. This justifies the claim from your note.
Next consider $I$. Then there exists minimal left ideal $I_0\subseteq I$. We cannot have $I_0^2 = 0$ because $N(R) =0$. Thus $I_0 = Re$ is generated by an idempotent $e$. Next $R = Re\oplus R(1-e)$ as left $R$-module. We have $$I = I \cap (Re+R(1-e)) = I\cap (I_0 + R(1-e)) = I_0 + I\cap R(1-e)$$ and also $I_0 \cap (I\cap R(1-e)) = Re\cap (I\cap R(1-e)) = 0$. This implies that $$I = I_0\oplus (I\cap R(1-e))$$ as left $R$-modules. Then $J = (I\cap R(1-e))$ is a left ideal properly contained in $I$. Then $J$ is a direct sum of minimal ideals by assumptions. We derive that $I$ is a direct sum of minimal left ideals. This is contradiction. Thus $\mathcal{F}$ is empty.