Let $R$ be a ring, $M$ a left $R$-module and $N \subset M$ a submodule. We say $N$ is superfluous if, for all submodules $V \subset M$, if $V + N = M$ then $V = M$.
Let $P$ and $Q$ be left $R$-modules with $S \subset P$ and $T \subset Q$ superfluous. Is it true that $S \oplus T \subset P \oplus Q$ is superfluous?
I can get that it's true in a special case. Specifically when $J(R)$ is the Jacobsen radical of $R$, $P, Q$ are finitely generated and $S = J(R)P$, $T = J(R)Q$. $S$ and $T$ are superfluous by Nakayama and so is $S \oplus T = J(R)(P \oplus Q)$.
Let $\pi_1:P \oplus Q \to P$ be projection onto the first factor and $\pi_2:P \oplus Q \to Q$ be projection onto the second factor. For $V \subset P \oplus Q$ with $V + S \oplus T = P \oplus Q$ I can get that $\pi_1(V) = P$ since $$P = \pi_1(V + S \oplus T) = \pi_1(V) + S$$ and $S$ is superfluous. Similarly $\pi_2(V) = Q$. This however does not buy me that $P \oplus Q \subset V$, yet.
This is about all the progress I've made. Any hints/solutions would be greatly appreciated.
If $S$ and $T$ are both superfluous in $P$ then $S+T$ is superfluous in $P$. (If $S+T+V = P$, then $T+V=P$, so $V=P$).
Thus it suffices to prove that if $S$ is superfluous in $M$, and $M\subseteq N$, then $S$ is superfluous in $N$. For this, observe that if $S+V=N$, then if $m\in M$, there are $s\in S$ and $v\in V$ with $s+v=m$, so in fact $v\in V\cap M$. Thus $S+(V\cap M) = M$. Therefore $V\cap M = M$, so $S\subseteq V$, and therefore $S+V=V=N$.