Find a simple $\mathbb{Z}[1/2]$-module.
What would be an example and how would we think about this type of problems?
First I looked at the $\mathbb{Z}$-module $\mathbb{Z}/p$ localized at the multiplicative set $S=\{1,2,2^2, 2^3,\cdots\}$ which is isomorphic to $\mathbb{Z}[1/2] \otimes \mathbb{Z}/p$ as a $\mathbb{Z}[1/2]$-module, but I am not sure what the submodule of a tensor product would look like...
Thank you.
Follow from Qiaochu's comment.
Given any non zero $m\in M$, $Rm = M$ since $M$ is simple, so we have the surjective $R$-module map $$f: R\rightarrow M$$ and $R/\ker(f) \cong M$ as $R$-module.
$\ker(f)$ must be a maximal ideal, or else it is contained in some other maximal ideal $I$, then we have a proper submodule of $M$ $$I/\ker(f) \subsetneq R/\ker(f) \cong M.$$
Since $\mathbb{Z}[1/2]$ is the $\mathbb{Z}$ localized at the multiplicative set $S=\{2^n: n=1,2, \cdots\}$, and there is a inclusion preserving bijection between maximal (also prime) ideals of $\mathbb{Z}$ with $I\cap S = \emptyset$ and $\mathbb{Z}[1/2]$ given by the map $I \mapsto I\mathbb{Z}[1/2]$. so $I = 3\mathbb{Z}[1/2]$ would be a maximal ideal and $R/I$ is a simple $\mathbb{Z}[1/2]$-module.