Proving $f(Rad(M)) \subseteq Rad(N)$ and other related properties about the radical of a module.

Question

I'm trying to prove the following facts about the radical of a module over a ring $R$. Recall that the radical of a module $M$ is the intersection of all maximal submodules of $M$, and we denote it as $Rad(M)$.

(1) If $f: M \to N$ is an $R-$ linear morphism, then $f(Rad(M)) \subseteq Rad(N)$ for all $R-$ modules $M$ and $N$,

(2) If $f:M \to N$ is an $R-$ linear epimorphism and $Ker(f) \subseteq Rad(M)$, then the equality in (1) holds. That is $f(Rad(M))= Rad(N)$,

(3) $Rad(\frac{M}{Rad(m)})=0$ for every $R-$ module $M$.

Let me you guys show you my attempt to prove the statements above. For $(1)$ I stumbled upon this well known proposition which states that for every module, $Rad(M)$ is equal to the sum of every small submodules of $M$. (A small submodule $S$ of $M$ is a module such that every time there is another submodules $K$ of $M$ such that $K+S=M$, it follows that $K=M$). Also it's not super difficult to prove that for every $R$ module morphism $f:M \to N$ we have that $f(S)$ is a small submodule of $N$ for every small submodule $S$ of $M$. Then since $Rad(M)= \sum_{\Delta} S_{\alpha}$ where $\lbrace S_{\alpha} \rbrace_{\Delta}$ is the family of all small submodules of $M$. Then:

$$f(Rad(M))= f(\sum_{\Delta} S_{\alpha})=\sum_{\Delta} f(S_{\alpha}) \subseteq Rad (N).$$

Im not sure if the contention $\sum_{\Delta} f(S_{\alpha}) \subseteq Rad (N)$. I mean $Rad(N)$ is the sum of all small submodules of $N$ and $f(S_{\alpha})$ is a small submodule in $N$ for every $\alpha \in \Delta$ so the sum of the small submodules of $\lbrace f(S_{\alpha})\rbrace_{\Delta}$ in $N$ should be contained in the sum of ALL small submodules of $N$, right ??

For (2) I was thinking about using the first isomorphism theorem so $N \cong \frac{M}{Ker(f)}$. Then $Rad(N) = Rad (\frac{M}{Ker(f)})$ and since $Ker(f) \subset Rad (M)$ then $Ker(f)$ behaves as maximal submodule obtaining $\frac{M}{Ker(f)}$ simple. Then $Rad (\frac{M}{Ker(f)})= \frac{M}{Ker(f)} = Rad (N)$ ??? But probably Im the wrong way to attain a succesful proof for this one.

For (3) I was thinking about using $(2)$,since we have a canonical epimorphism $\pi: M \to \frac{M}{Rad(M)}$ and $Ker(\pi) =Rad(M)$. So by $(2)$ we obtain $$ \pi (Rad(M))= Rad (\frac{M}{Rad(M)}).$$ But $\pi(Rad(M))=0$, so it follows that $Rad (\frac{M}{Rad(M)})=0$.

Are my proofs of $(1)$ and $(3)$ right? Is there something I can do to improve them? How can I end up proving $(2)$? Any help will be appreciated 🙂

Answer 1

Your proof of $(1)$ is correct, you just need to remember that the sum of a family of submodules is the “smallest” submodule containing each member of the family:

Given $\alpha \in \Delta$, $f[S_\alpha]$ is a small submodule of $N$, so $f[S_\alpha] \subseteq \operatorname{Rad}(N)$. But this means that $\operatorname{Rad}(N)$ is a submodule of $N$ containing each member of the family $\{f[S_\alpha] : \alpha \in \Delta\}$; hence its sum is “smaller” than $\operatorname{Rad}(N)$, i.e. $\sum_{\alpha \in \Delta} f[S_\alpha] \subseteq \operatorname{Rad}(N)$.

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For $(2)$, observe the following:

• If $K$ is a maximal submodule of $M$, then $f^{-1}[f[K]] = K$. To see this, note that $f^{-1}[f[K]] = K+\ker f$ (this works for any morphism) and since $\ker f \subseteq \operatorname{Rad}(M) \subseteq K$, it follows that $K+\ker f = K$.
• If $K$ is a maximal submodule of $M$, then $f[K]$ is a maximal submodule of $N$. First, $f[K]$ is proper: if $f[K]=N$, taking inverse images we have $K=M$, which is false. Second, if $L$ is any submodule of $N$ containing $f[K]$, then $K \subseteq f^{-1}[L]$, and so $f^{-1}[L]$ is $K$ or $M$, hence $L=f[f^{-1}[L]]$ is $f[K]$ or $f[M]=N$.

Now, if $y \in \operatorname{Rad}(N)$, take $x \in M$ with $f(x)=y$. We claim that $x \in \operatorname{Rad}(M)$. Indeed, given a maximal submodule $K$ of $M$, we know that $y \in f[K]$, which gives us that $x \in f^{-1}[f[K]] = K$.

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The proof of $(3)$ is fine like it is.