Ideals are totally ordered in valuation rings

Question

Show that for any valuation ring $V$ the ideals are totally ordered under inclusion

Definitions:

Valuation ring: For an integral domain $D$ and a field of fractions $K$, There is a totally ordered abelian group $Γ$ (called the value group) and a valuation $ν: K → Γ ∪ \{∞\}$ with $D = \{ x ∈ K | ν(x) ≥ 0 \}.$

Ideals: Consider a ring $R$, an ideal $I$ is a subgroup of $R$ as a group which absorbs elements from multiplication from $R$.

Total order: An order relation on a set following the following axioms:

1. $ a \leq a$, Reflexivity
2. If $ a \leq b$ and $ b \leq c$ then $ a \leq c$, Transitivity
3. If $ a \leq b$ and $b \leq a$ then $a=b$ (anti-symm)
4. If $ a \neq b$ then either $a \leq b $ or $b \leq a$ (Strongly connected)

Context:

I'm trying to proceed based on Bernard's hint here

Hint:

1. Prove first the principal ideal in $V$ are totally ordered by inclusion: for this, let $a,b\in V$. Show that if $Va\not\subset Vb$, then $Vb\subset Va$.
2. Deduce that, if $\mathfrak a$ and $\mathfrak b$ are two ideals in $V$, if $\mathfrak a\not\subset \mathfrak b$, then $\mathfrak b\subset \mathfrak a$ (take $a\in\mathfrak a$, $\;a\notin\mathfrak b$. Show that, for any $b\in\mathfrak b$, $b\in Va$).

Attempt

1. We have, $Va=\{ xa| x \in V \}$ and $Vb= \{yb| y \in V \}$ , if $V_a \not \subset Vb$ then there are some elements of form $yb$ in $Vb$ which are not in $V_a$. Not sure how to proceed after this.

2. Since $a$,$b$ are a subgroup of $R$ and both absorb elements from Ring, I believe it suffices to prove $b$ is a subgroup of $a$. This post tells me it is suffices to show that $b$ is just a subset of $a$. Not sure how to proceed after this.

There seems to be a related answer by Alex Mather's here showing equivalent definitions of Valuation but it doesn't help much.

P.S: I am only a beginner in field/ring theory , so this question is probably very noob level but thanks for checking it out!

Answer 1

For step 2, I think it is simpler framed as a proof by contradiction. That is, assume $\mathfrak a$ and $\mathfrak b$ are not comparable and then consider some $a \in (\mathfrak a \setminus \mathfrak b)$ and $b \in (\mathfrak b \setminus \mathfrak a)$. From these we can generate principal ideals $Va$ and $Vb$. Then looking at $Va \cap Vb$ should provide a contradiction.

For step 1, it is easiest to first prove an alternate characterization of valuation rings. In particular, for any $x \in K$, either $x \in V$ or $x^{-1} \in V$. This is true because $\nu(x^{-1}) = -\nu(x)$. Therefore, either $\nu(x) \geq 0$, in which case $x \in V$, or $\nu(x^{-1}) > 0$, in which case $x^{-1} \in V$.

From there, we take two ideals generated by nonzero $a$ and $b$. Notice from the above that either $ab^{-1} \in V$ or $ba^{-1} \in V$. Without loss of generality, assume the first is true. In particular, let $ab^{-1} = x$. Then $a = bx$, so $a$ is in the ideal generated by $b$.