 Solution

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