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A ring is a Noetherian ring if it satisfies the ascending chain condition on ideals: every chaineventually becomes constant. Equivalently, every ideal has a finite generating set of an ideal. To see this equivalence, if an ideal were not finitely generated, one could repeatedly choose an element outside the ideal generated by the preceding choices, producing a strictly ascending chain. Conversely, the union of an ascending chain is an ideal; if it is generated by finitely many elements, all of them lie in one term of the chain, and every later term equals that term.

A ring is an Artinian ring if it satisfies the descending chain condition on ideals: every chaineventually becomes constant. These are stabilization conditions on all ideals, not merely on principal ideals.

Yes: $R[[x]]$ is Noetherian. We prove Noetherianity of a formal power series ring directly, using only the equivalence in part (a). Let $A=R[[x]]$, a formal power series ring, and fix an ideal $I\subseteq A$. For $n\ge 0$, let $C_n$ be the set of coefficients of $x^n$ in elements of $I\cap x^{n}A$. It is an ideal of $R$, because coefficient extraction is additive and respects multiplication by constants. Multiplication by $x$ givesSince $R$ is Noetherian, there is an $N$ with $C_n=C_N$ for all $n\ge N$. Each $C_n$, for $0\le n\le N$, has a finite generating set of an ideal $a_{n1},\dots ,a_{n{r}_n}$. Choose $f_{nj}\in I\cap x^{n}A$ whose coefficient of $x^n$ is $a_{nj}$. If $C_n=0$, use no generators at that index.

We claim that these finitely many $f_{nj}$ generate $I$ over $A$. Given $f\in I$, start with $g_0=f$ and recursively construct $g_n\in I\cap x^{n}A$. For $n<N$, express the coefficient of $x^n$ in $g_n$ as ${\sum }_j{c}_{nj}{a}_{nj}$, with $c_{nj}\in R$, and subtract ${\sum }_j{c}_{nj}{f}_{nj}$ to obtain $g_{n+1}\in I\cap x^{n+1}A$. For $n\ge N$, use $C_n=C_N$ and subtractto cancel that coefficient. Every step involves a finite sum and leaves a residual still in $I$.

For each $j$, define the formal power seriesThe residual after the $n$th cancellation lies in $x^{n+1}A$, so comparison of each coefficient gives the exact identityThe right side is a finite $A$-linear combination of the chosen generators. No assertion that arbitrary ideals are closed under limits is being used: the equality is verified coefficient by coefficient, and the multipliers $h_j$ belong to $A$. Thus every ideal of $A$ is finitely generated, proving

Distinct prime-power ideals are pairwise comaximal. For comaximal ideals, product equals intersection, so induction givesThe Chinese remainder theorem therefore gives the ring isomorphism

The ideal approximation theorem, proved by applying this Chinese-remainder map one prime power deeper, suppliesEquivalently, $v_{P_i}(\alpha )=m_i$ at every prime dividing $I$.

Since $(\alpha )\subseteq I$, the fractional idealis an integral ideal and $I{I}^{\prime }=(\alpha )$ is a principal ideal. At every $P_i\mid I$,
$v_{P_i}(I^{\prime })=v_{P_i}(\alpha )-m_i=0$, so $I+I^{\prime }={\mathcal{O}}_K$.

Choose $\beta \in I$ and $\gamma \in I^{\prime }$ with $\beta +\gamma =1$. Then
$\alpha ,\beta \in I$. For any $x\in I$,Thusso every ideal of ${\mathcal{O}}_K$ has a generating set of two elements.