Characterization of valuation rings

Question

There is the next characterization for valuation domains:

Let $D$ be an integral domain and $K$ its field of fractions. The following are equivalent:

1. For every nonzero $x$ in $K$, either $x$ in $D$ or $x^{−1}$ in $D$.

2. The ideals of $D$ are totally ordered by inclusion.

3. The principal ideals of $D$ are totally ordered by inclusion (i.e., the elements in D are totally ordered by divisibility.)

4. There is a totally ordered abelian group $Γ$ (called the value group) and a surjective group homomorphism (called the valuation) $v :$ $K^×$ $→ Γ$ with $D = \{ x ∈$ $K^×$ | $v(x)$ ≥ $0 \}$ $∪$ {0}.

The equivalence between 1, 2 and 3 are clear, and I understand that the fourth can be found in Krull's article (1936), but I don't speak german and I can't find a proof anywhere. Any link or hint will be much appreciated.

Answer 1

I think it's clear that 4 implies 1. On the other hand, to see 1 implies 4 I think it's easier if you allow the group $\Gamma$ to be written mutiplicatively (and then we should have $D=\{x\in K^\times\mid v(x)\le1\}$), then on your own figure out why a "multiplicative" valuation is equivalent to an "additive" one.

With this in mind, take $\Gamma=K^\times/D^\times$ with $aD^\times\le bD^\times\iff ab^{-1}\in D^\times$; then we have an obvious surjective group homomorphism $v:K^\times\to\Gamma$. Furthermore, you can check the order we have defined on $\Gamma$ is a partial order, and it is a total order if 1 is satisfied. It is also simple to check that $D=\{x\in K^\times\mid v(x)\leq 1\}\cup\{0\}$.