$\frac{\text{Radical}(M)}{K}\subseteq \text{Radical}(M/K)$

Question

I am trying to prove that for a $R$-module $M$, and any sub-module $K$, $\frac{\text{Radical}(M)}{K}\subseteq \text{Radical}(M/K)$

Proof: Let $x\in \frac{\text{Radical}(M)}{K}$, then $x=r+K$ where $r\in\text{Radical}(M)$. Since, $\text{Radical}(M)=\bigcap\{U\subset M; U~\text{is maximal in}~M\}$. Then, $x=r+K, r\in U; U$ is maximal in $M$. Thus, $r\in\text{Radical}(M/K)$

Is this a good proof?

Answer 1

The proof maybe becomes clearer when abstracting. I write $\operatorname{rad}$ for the radical.

For every $R$-linear map $f \colon M → N$, $f(\operatorname{rad} M) ⊆ \operatorname{rad} N$.

Hint. Let $U ⊆ N$ be any maximal module. Then $N / U$ is simple. Hence, $f^{-1}(U)$ is either all of $M$ or maximal itself because $M / f^{-1}(U)$ must be isomorphic to some submodule of $N / U$ by the first isomorphism theorem. What does this tell you about $\operatorname{rad} M$ and $f(\operatorname{rad} M)$?

Do you see how the claim you are trying to prove follows?