Prime ideal in valuation ring Let  $R$   be  a  valuation  ring.  If  $ P = (a)$   is   a  prime  ideal  of  $R$ then  $P$  is  the   maximal  ideal.  I  know  that   the  maximal  ideal  is $\lbrace{ x \in R | v(x) > 0}\rbrace$.
 A: There is a characterization of principal ideals in a valuation domain $R$.

LEMMA: if $P=(a)$ is any principal ideal of a valuation domain $R$, then $P=\{ x \in R : v(x) \ge v(a)\}$.

I leave the proof to you as an exercize.
As the answer of your problem, let $M$ be the unique maximal ideal of $R$. Suppose by contradiction that $P$ is prime and $P \neq M$. Then there exists some $y \in M \setminus P$. Since $y \notin (a)$, by the lemma,
$$v(y)<v(a)$$
or, equivalently,
$$v(ay^{-1})=v(a)-v(y)>0$$
Which implies that $ay^{-1} \in R$. Now,
$$ay^{-1} \cdot y=a \in P$$
since $P$ is a prime ideal and $y \notin P$, necessarily $ay^{-1} \in P$. By the lemma,
$$v(a)-v(y)=v(ay^{-1}) \ge v(a)$$
or, equivalently,
$$v(y) \le 0$$
contradicting $y \in M = \{ x \in R : v(x) > 0\}$. We got a contradiction, thus necessarily $P=M$.
A: If $(a)$ is not maximal, then there exists $x$ such that $(a)\subsetneq (x)\subseteq M$ where $M$ is the maximal ideal. (The principal ideals of $R$ are linearly ordered, after all.)
Then $a=xr$, but $a$ is irreducible (since it is prime, since $(a)$ is a prime ideal), so $r$ is a unit, and $(a)=(x)$, a contradiction.
So, $(a)$ has to be maximal to begin with.
A: Partial answer is the image of the valuation is the integers.
Let $K$ be  the field of fractions of $R$ and $v$ the valuation.

*

*Let $x\in R$ suppose that there xists $y\in P$ such that $v(x)=v(y)$.
$y=(yx^{-1})x$, $v(y^{-1}x)=0$ implies that $yx^{-1}$ is in $R$, since $P$ is prime, $yx^{-1}\in P$ or $x\in P$, so $x\in P$.


*Let $x\in R$ with $v(x)>0$, there exists $p$ such that $v(x^p)=v(a^m)$, this implies that $x^p\in P$ and $x\in P$ since $P$ is a prime ideal.
