= Solution
The <discrete valuation> identities $v(xy)=v(x)+v(y)$ and $v(x+y)\ge\min(v(x),v(y))$ make $R$ closed under products and sums; it contains $0,1$ and additive inverses. As a <subring> of the <field> $K$, it is an <integral domain>.
Its <units> are exactly the elements of valuation zero: $x^{-1}\in R$ if and only if $v(x)=0$. The remaining elements form the ideal $\mathfrak m=\{x:v(x)>0\}$, and every proper ideal is contained in it, since an ideal containing a unit is the whole ring. Hence $R$ is a <local ring> with this unique <maximal ideal>.
For a nonzero ideal $I$, the set of valuations of its nonzero elements is a nonempty subset of $\mathbb Z_{\ge0}$ and has a least value. Choose $x\in I$ attaining it. Every $y\in I$ satisfies $v(y/x)\ge0$, so $y/x\in R$ and $I=(x)$. The <zero ideal> is also <principal>. This proves that $R$ is a <discrete valuation ring>, in particular a <principal ideal domain>.
Choose a <uniformizer> $\pi$, with $v(\pi)=1$. For $z\in K$, some $n\ge0$ makes $\pi^nz\in R$, and $z=(\pi^nz)/\pi^n$ is a quotient of elements of $R$. Thus \b[$K$ is the <field of fractions> of $R$].
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