= Solution
The <valuation ring> is $\mathcal O=\{x\in K:|x|\le1\}$, and $\mathfrak m=\{x\in K:|x|<1\}$ is an ideal by the <ultrametric inequality>. Its complement inside $\mathcal O$ consists exactly of elements of <field absolute value> one, whose inverses also lie in $\mathcal O$. Every proper ideal must avoid units and hence lie in $\mathfrak m$. Therefore $\mathfrak m$ is the unique <maximal ideal>: \b[$\mathcal O$ is a <local ring>]. Also its fraction field is $K$, since every nonzero element outside the ring has its inverse inside it.
If $z\in K$ is integral over $\mathcal O$, it satisfies a monic equation $z^n+a_{n-1}z^{n-1}+\cdots+a_0=0$ with $|a_i|\le1$. Were $|z|>1$, the leading term would have strictly larger <field absolute value> than every other term, so the <ultrametric inequality> would prevent cancellation to zero. Thus $z\in\mathcal O$, proving \b[the <valuation ring> is integrally closed].
For the ideal criterion, assume the <valuation> is nontrivial and write $v=-\log|\cdot|$. If its <value group> is discrete, normalize it to $\mathbb Z$ and choose a <uniformizer> $\pi$ of <valuation> one. In any nonzero ideal $I$, the nonnegative integer valuations have a least value $m$. Choose $a\in I$ of that value. For every $b\in I$, $v(b/a)\ge0$, so $b\in a\mathcal O$. Hence $I=(a)=(\pi^m)$ and $\mathcal O$ is a <principal ideal domain>.
Conversely, if $\mathcal O$ is a <principal ideal domain>, its nonzero <maximal ideal> is $(\pi)$. For every element $x$ of positive <valuation>, $x\in\mathfrak m$ implies $x=\pi y$ with $y\in\mathcal O$, so $v(x)\ge v(\pi)>0$. Thus $\gamma=v(\pi)$ is the least positive value. For any value $t$, subtract $\lfloor t/\gamma\rfloor\gamma$. The remainder belongs to the <value group> and lies in $[0,\gamma)$, so is zero. The <value group> is therefore $\gamma\mathbb Z$, and the <valuation> is discrete. We have proved the <principal ideal criterion for a nontrivial rank-one valuation ring>.
The nontriviality qualification matters under the standard definition of <discrete valuation>, which requires <value group> isomorphic to $\mathbb Z$. With the trivial <field absolute value>, $\mathcal O=K$ is still a <principal ideal domain>, but the <value group> is zero and there is no <uniformizer>. Thus the literal unrestricted equivalence has a field exception. If “discrete” instead includes the trivial <value group> as a discrete subgroup of $\mathbb R$, that exceptional case satisfies the equivalence too.
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