# Why Jacobson radical of a cyclic module over integral domain $R$ such that $\mathrm{Rad}(R)=(0)$ is zero? [closed]

Let $R$ be an integral domain with zero Jacobson radical and $M$ be a free $R$-module. Why Jacobson radical of $M$ is zero? Why $\mathrm{Rad}(Rm)=(0)$ where $m\in M$?

## closed as off-topic by user26857, C. Falcon, Namaste, Shailesh, JonMark PerryMay 14 '17 at 0:12

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• What is Rad(Rm) – user379195 May 13 '17 at 7:52
• That is Jacobson radical of $R$-module $Rm$ – ruholla kh May 13 '17 at 8:19
• – user26857 May 13 '17 at 20:56

## 2 Answers

Note that \begin{align}\mathrm{Rad}(M) & = \bigcap \{N: M/N\; \text{is a simple module} \} \\ & = \bigcap \{N: M/N\; \text{is a semi-simple module}\}. \end{align} And for a maximal ideal, $I_\text{max} \lhd R,$ we have that $M/I_\text{max}M$ is an $R/I_\text{max}$-vector space, and so $M/I_\text{max}M$ is a semi-simple module.

So $\mathrm{Rad}(M) \subseteq \bigcap \{IM : I\; \text{is a maximal ideal of}\; R\}$.

But as $M$ is free, if $\{m_1, ... ,m_n\}$ is a basis for $M$, and $i_1m_1 + \cdots + i_nm_n= j_1m_1 + \cdots +j_nm_n$ then we conclude that $i_k=j_k\; \forall k \in \{1, ..., n\}$.

Hence we see that \begin{align}\mathrm{Rad}(M) &\subseteq \bigcap \{IM : I\; \text{is a maximal ideal of}\; R\}\\ & \subseteq\mathrm{Rad}(R)M\\ & =\{0\}\ (\text{by hypothesis}). \end{align}

I don't think that I have used that $R$ is an integral domain here so maybe I have made a mistake but I can't see it... I hope this helps...

• Maybe integral domain is a extra condition. Please tell me for $R/i_{max}$-vector space $M/I_{max}$, Why $M/I_{max}M$ is a semi-simple module? – ruholla kh May 13 '17 at 18:43
• @ruhollakh this is because any vector space is a semi-simple module. It is the direct sum of the 1 dimensional subspaces being the spans of the basis elements. As these are 1 dimensional they are simple modules. – SEWillB May 13 '17 at 21:29

A free module is a direct sum of copies of $R$, say $$M=R^{(X)}=\{(r_x)_{x\in X}: \operatorname{supp}(r_x)_{x\in X}\text{ is finite}\}$$ where $\operatorname{supp}(r_x)_{x\in X}$ is the set $\{x\in X:r_x\ne0\}$.

For a maximal ideal $\mathfrak{m}$ of $R$, consider, for $y\in X$, $$\mathfrak{m}_y=\{(r_x)_{x\in X}\in M:r_y\in \mathfrak{m}\}$$ Then $\mathfrak{m}_y$ is a maximal submodule of $F$, as $$M/\mathfrak{m}_y\cong R/\mathfrak{m}$$ Also $$\bigcap_{\substack{\mathfrak{m}\text{ maximal}\\y\in X}}M_y=J^{(X)}$$ where $J=\operatorname{Rad}R$ is the Jacobson radical of $R$. Therefore $$\operatorname{Rad}M\subseteq J^{(X)}$$ In the present case $J=(0)$, so also $\operatorname{Rad}F$ is zero. (Actually, equality holds in general.)

If $m\in F$, then $Rm\cong R$, because a free module is torsion free (here we're using the fact that $R$ is a domain). Therefore $\operatorname{Rad}(Rm)=\{0\}$ by the isomorphism.