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In this localization of a ring, a nonzero fraction $a/b$ with odd $b$ is a unit exactly when $a$ is odd: its reciprocal then has odd denominator; if $a$ is even, an inverse would equate an even integer with an odd one after clearing denominators. Consequently every nonzero element has a unique formHere $r$ is the exponent of $2$ in the numerator of the reduced fraction, its nonnegative 2-adic valuation.

Every nonzero ideal has the form $dR$ by part (c). Write the nonzero integer $d=2^{r}v$ with $v$ odd; since $v$ is a unit, $dR=2^{r}R$. Hence the nonzero proper ideals are exactlyThe zero ideal and $R=2^{0}R$ complete the full list. These ideals are distinct: $2^r$ does not belong to $2^{r+1}R$, since $1/2\notin R$.

The reduction map $a/b\mapsto a\mathrm{mod}2$ is a well-defined surjective ring homomorphism $R\to {\mathbb{F}}_2$ with kernel of a ring homomorphism $2R$; odd denominators reduce to $1$. Thus $R/2R\cong {\mathbb{F}}_2$ is a field, so $2R$ is a prime ideal and $2$ is a prime element. Every nonzero nonunit is associated to $2^r$ with $r\ge 1$. If $r\ge 2$, the factorization $2^r=2\cdot 2^{r-1}$ is into two nonunits, so it is not an irreducible element and therefore not a prime element. We concludeThe failed step in the argument of part (b) is the guarantee that the product plus one is a nonunit and hence has an irreducible factor. Here every prime element is an even fraction, so a product of prime elements plus $1$ is an odd fraction, therefore a unit. Already $2+1=3$ has no prime divisor in this ring. In $k[t]$, positive degree prevented that failure.

For the zero ideal, choose $d=0$. Otherwise choose a nonzero $d\in I$ for which the Euclidean function $\delta (d)$ is minimal; the well-ordering principle guarantees such a choice. For any $a\in I$, perform Euclidean division:Because $I$ is an ideal, $r=a-qd\in I$. A nonzero remainder would contradict the minimal choice of $d$, so $r=0$ and $a\in Rd$. Thus $I\subseteq Rd$. Conversely, $d\in I$ implies $Rd\subseteq I$, provingEvery ideal is therefore a principal ideal; in particular, a Euclidean domain is a principal ideal domain.

For the zero ideal, choose $d=0$. Otherwise choose a nonzero $d\in I$ for which the Euclidean function $\delta (d)$ is minimal; the well-ordering principle guarantees such a choice. For any $a\in I$, perform Euclidean division:Because $I$ is an ideal, $r=a-qd\in I$. A nonzero remainder would contradict the minimal choice of $d$, so $r=0$ and $a\in Rd$. Thus $I\subseteq Rd$. Conversely, $d\in I$ implies $Rd\subseteq I$, provingEvery ideal is therefore a principal ideal; in particular, a Euclidean domain is a principal ideal domain.

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 ${\mathbb{Z}}_{\ge 0}$ and has a least value. Choose $x\in I$ attaining it. Every $y\in I$ satisfies $v(y/x)\ge 0$, 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 $n\ge 0$ makes ${\pi }^{n}z\in R$, and $z=({\pi }^{n}z)/{\pi }^n$ is a quotient of elements of $R$. Thus $K$ is the field of fractions of $R$.