I'm reading number theory seminar notes from Stanford. I found the following exercise, posting the abridged version:
Let $R$ be a valuation ring with fraction field $K$. Let $\Gamma := \Gamma_R = K^{\times}/R^{\times}$ be the value group of $R$, and $v$ the valuation on $K$ given by reduction modulo $R^\times$. Consider the mapping $M_I:= I \mapsto v(I \backslash \{0\})$
We leave to the reader as an exercise to prove that this is an inclusion-preserving bijection from the set of ideals of $R$ onto the set of submonoids $M \subset Γ_{\le 1}$ such that $m \in M, \gamma \le m \Rightarrow γ \in M$
This exercise has defeated me: how should I think about proving it?
This is only the definitions, there is nothing to prove.
The word submonoid is useless, any subset $M$ of $v(R)$ satisfying $m \in M, \gamma\in v(R),\gamma \ge m \Rightarrow γ \in M$ represents the ideal $\{ a\in R,v(a)\in M\}$.
$v(R)$ is totally ordered by definition of "valuation ring", if $a,b\in R-0$ then either $a/b$ or $b/a$ is in $R$,
$v$ is the map $K^\times \to K^\times/R^\times$ extended with $v(0)=\infty$ and the order $v(b)\ge v(a)$ iff $b \in a R$.