= Solution
The <valuation ring> and its <maximal ideal> are
$$
\mathcal O_K=\{x:v(x)\geq0\},
\qquad
\mathfrak m=\{x:v(x)>0\}.
$$
If $\mathcal O_K$ is <Noetherian ring>[Noetherian], its maximal ideal is finitely generated. Ideals in a valuation ring are totally ordered, so every finitely generated ideal is generated by one of its generators; write $\mathfrak m=(\pi)$. Then $v(\pi)$ is the smallest positive element of the <value group>. Subtracting integral multiples of this value shows that every value is an integral multiple of $v(\pi)$, so $v$ is discrete.
Now assume $K$ is complete and discretely valued. Parts i and ii, applied to $v$ and to the other discrete valuation $v'$, give
$$
\mathfrak m_v\subseteq S\subseteq\mathcal O_{v'}.
$$
If $v(a)=0$ and $\pi$ is a uniformizer for $v$, then $\pi a^r\in\mathfrak m_v$ for every $r\in\mathbb Z$. Hence
$$
v'(\pi)+r,v'(a)\geq0
$$
for every integer $r$, forcing $v'(a)=0$ and in particular $v'(a)\geq0$. Every nonzero $x$ has the form $x=\pi^m u$ with $v(u)=0$, so
$$
v'(x)=m,v'(\pi)=\frac{v'(\pi)}{v(\pi)}v(x).
$$
The proportionality constant is positive because $v'$ is nontrivial. Thus the valuations, and their associated absolute values, are equivalent.
Solved by gpt-5.6-sol high.
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