-1
$\begingroup$

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 $m\in M$?

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, C. Falcon, Namaste, Shailesh, JonMark Perry
If this question can be reworded to fit the rules in the help center, please edit the question.

0
$\begingroup$

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...

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – ruholla kh May 13 '17 at 18:43
  • $\begingroup$ @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. $\endgroup$ – SEWillB May 13 '17 at 21:29
0
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.