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Suppose $M$ is a completely reducible left $R$-module for a ring $R$ and $I$ is an ideal of $R$. Then prove that the following are equivalent:

  1. $M=IM$

  2. If $I$ annihilates an element $x \in M$, then $x=0$

  3. For all $x\in M, x$ is contained in $Ix$

The implication $(3)\rightarrow(1)$ is simple. For $(2)\rightarrow(3)$, I took a projection from $Rx$ to $Ix$ ($Rx$ and $Ix$ are themselves completely reducible and hence this map is surjective), and tried showing it is 1-1, and hence a module isomorphism. For $(1)\rightarrow(2)$, I tried showing that the ideal $I$ cannot annihilate any non-zero $x\in IM$.

In both these cases, I have been stuck for hours with no progress. Any hints would be appreciated.

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up vote 1 down vote accepted

OK, let's look at $(1) \rightarrow (2)$ first. Suppose $Ix=0$ for some $x$ in our semisimple module $M=\oplus M_j,$ with the $M_j$ all simple. Then all the module homomorphisms $\phi_\alpha: M \rightarrow \alpha M, \alpha \in I$ have a kernel containing $Ix.$ Suppose $$x=\sum_{j=1}^n x_j, x_j \in M_{x_j}$$ Composing the factor inclusions with the action of each element of $I, M_j \rightarrow M \rightarrow \alpha M$, we get maps $\alpha M_j$ which have kernels for each $M_{x_j}$. Since $M_{x_j}$ is simple, this kernel must be the whole module; since $\alpha$ was arbitrary, what can we deduce about $IM_j$ and thus $IM$?

$(2) \rightarrow (3)$ will go similarly. First assume $x \neq 0$ is contained in one simple summand $M_x$ of $M$. Then if $Ix \neq 0, Ix=M_x,$ and $x \in Ix.$ More generally, $x$ could be a finite sum of elements $x_i$ of simple summands. But we've assumed $(2)$, so $Ix_i=M_{x_i}$ from the previous argument. Do you see how to wrap up from there?

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