$V\cdot\operatorname{Rad}(R)\subseteq\operatorname{Rad}(V)$ where equality holds if $V$ is free or projective. Question : Let $\operatorname{Rad}(R)$ denote the Jacobson Radical of $R.$ Show that $V\cdot\operatorname{Rad}(R)\subseteq\operatorname{Rad}(V)$ where equality holds if $V$ is free or projective. Show that equality fails when $R=\mathbb{Z}$ and $V=\mathbb{Q}_\mathbb{Z}$
My Thoughts : I feel like this is a straightforward proof, but I am just having some issues as to where to start dealing with a free and projective module. Would it be best to consider ideals, or would it be best to try and appeal to Nakayama's Lemma ?  I would be appreciative for any help or insight ! Thank you.
 A: First the counterexample. The radical of $\mathbb{Q}$ as a $\mathbb{Z}$-module is $\mathbb{Q}$, because this module has no maximal submodule.
Indeed a quotient of $\mathbb{Q}$ is a divisible abelian group, whereas no simple abelian group is divisible.
Now the inclusion. You just need to show that $\DeclareMathOperator{\Rad}{Rad}\Rad(R)$ annihilates all simple modules, which is easy: if $S$ is a simple (right) $R$-module, then for $x\in S$, $x\ne0$, you have that $I=\{r\in R:xr=0\}$ is a maximal right ideal. Since $\Rad(R)$ is the intersection of all maximal right ideals, you conclude that $S\Rad(R)=\{0\}$.
If $V$ is any right $R$-module and $L$ a maximal submodule thereof, then $(V/L)\Rad(R)=\{0\}$, which amounts to saying that $V\Rad(R)+L=L$, that is, $V\Rad(R)\subseteq L$. Therefore $V\Rad(R)\subseteq\Rad(V)$ (the intersection of all maximal submodules).
For a free module $V=R^{(\Lambda)}$ (a direct sum of copies of $R$), you can consider, for each $\lambda\in\Lambda$ and each maximal right ideal $I$ of $R$, the submodule $L_{\lambda,I}=\pi_\lambda^{-1}(I)$, where $\pi_\lambda\colon V\to R$ is the projection on the component. Then $V/L_{\lambda,I}$ is isomorphic to $R/I$ and so is simple, which implies $L_{\lambda,I}$ is maximal.
If you take the intersection of all these maximal submodules you clearly get $(\Rad(R))^{\Lambda}$ and therefore $\Rad(V)\subseteq(\Rad(R))^{\Lambda}=V\Rad(R)$. The previously proved inclusion shows equality.
For a projective module $V$, prove that $\Rad(M\oplus N)=\Rad(M)\oplus\Rad(N)$ and apply the equality for free modules to $V\oplus V'$ a free module.
