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Let $R$ be a ring with identity, $J$ its Jacobson radical. For any left $R$-module $M$, a submodule $N\leqslant M$ is called superfluous if for any submodule $K\leqslant M$, $K+N=M$ implies that $K=M$. It is well-known that $J$ is the largest superfluous submodule of $R$. Now my question is:

For any index set $I$, is $\text{rad}(R^{(I)})=J^{(I)}$ a superfluous submodule of $R^{(I)}$?

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    In the book "Rings and categories of modules" by Anderson and Fuller, Pro 5.20 saies it is hold if $I$ is finite. – 2017-01-05

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Let $k$ be a field, and $R=k[[t]]$ the ring of power series over $k$ in one variable. Then the Jacobson radical $J$ consists of the power series with zero constant term.

Take $I=\mathbb{N}$, and $K$ the submodule of $R^{(\mathbb{N})}$ generated by the elements $$(1,t,0,0,\dots), (0,1,t,0,\dots),(0,0,1,t,\dots),\dots.$$ Clearly $K+J^{(\mathbb{N})}=R^{(\mathbb{N})}$, but $K\neq R^{(\mathbb{N})}$ since the last non-zero coordinate of any element of $K$ has zero constant term. So $J^{(\mathbb{N})}$ is not superfluous.

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    Thank you, seems that your method works for every local PID. – 2017-01-05
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    @CensiLI Yes, or even more generally, any ring whose Jacobson radical contains an element that is not a right zero divisor. – 2017-01-05