Prime ideal in valuation ring

Question

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 $\lbrace{ x \in R | v(x) > 0}\rbrace$.

Answer 1

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 $P=M$.

Answer 2

Partial answer is the image of the valuation is the integers.

Let $K$ be the field of fractions of $R$ and $v$ the valuation.

1. 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 $x\in P$.

2. 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.

Answer 3

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.