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Suppose that $R$ is a ring with identity and $R^{\oplus n}$ is a free left $R$-module. Is every submodule of $R^{\oplus n}$ isomorphic to $A_1\oplus \ldots \oplus A_n$ where $A_i$ are left ideals of $R$?

Edit: It appears that this is false, and an example is given in a linked question. Can anyone explain why the example works?

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    Submodules of the form $\{(x,x) \in R^2 \mid x \in R \}$ are not what you're looking for, right? – 2017-02-23
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    @menag That submodule is isomorphic to $R$, which is a left ideal of $R$. I'm wondering if all submodules of this free module has to be externally isomorphic to some direct sum of left ideals of $R$. Since that one is isomorphic to $R$, it's an example. – 2017-02-23
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    Yes, that's why I'm asking. I'll think about it. – 2017-02-23
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    I had the feeling that vector fields on a sphere give a counter example. [Here](http://math.stackexchange.com/q/2166903/60589) is the question. Maybe it is correct? – 2017-03-01

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First recall that the ring in the linked answer is $R=\mathbb{Z}/4\mathbb{Z}[X]/(X^2)$.

An easy calculation shows that $R$ is local, with unique maximal ideal generated by $2$ and $X$, and has a unique simple ideal generated by $2X$.

The module in the linked answer is the submodule of $R^2$ generated by $(2,X)$. Since it is cyclic, it is a quotient of the free module $R$, and so has a unique maximal submodule. Therefore it is indecomposable, and so if it were a direct sum of ideals, it would have to be isomorphic to a single ideal.

But then it would have a unique simple submodule. However, it has two simple submodules, generated by $(2X,0)$ and $(0,2X)$ respectively.

A similar, and arguably clearer, example is given by $R=k[X,Y]/(X^2,Y^2)$ for a field $k$, with the submodule of $R^2$ generated by $(X,Y)$.