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Put $A=R[[X]]$, and let $J$ be any ideal of this formal power series ring. For $d\ge 0$ define an ideal of $R$ byMultiplication by $X$ shows $L_d\subseteq L_{d+1}$. Since $R$ is Noetherian, this chain stabilizes at some $d=N$, and each $L_d$ is finitely generated. For every $0\le d\le N$, choose finitely many series $f_{d,j}\in J\cap X^{d}A$ whose degree-$d$ coefficients generate $L_d$.

Given $f\in J$, cancel its degree-zero coefficient using the $f_{0,j}$, then its degree-one coefficient using the $f_{1,j}$, and so on. At a degree $d\ge N$, use $X^{d-N}{f}_{N,j}$. Each residual remains in $J$ and has zero coefficients below the next cancellation degree. For each of the finitely many chosen $f_{d,j}$, accumulate its multiplier into a formal power series $h_{d,j}$. The higher-degree increments of these multipliers converge coefficientwise. At every degree the cancellation procedure gives the exact coefficient identity, soThus the finitely many lifts generate $J$ as an actual ideal. No assumption that $J$ is closed has been used. This proves Noetherianity of formal power series rings:Iterating the argument proves the assertion for $R[[X_1,\dots ,X_s]]$, which is naturally the iterated formal power series ring.

For a surjective module map $f:M\to N$, define the adic completion of a module byTake $\hat{n}\in \hat{N}$ and choose representatives $n_r\in N$ for its components. Compatibility says $n_{r+1}-n_r\in I^{r}N$. Choose $m_1$ with $f(m_1)=n_1$. If $f(m_r)=n_r$, surjectivity gives $f(I^{r}M)=I^{r}N$, so choose ${\delta }_r\in I^{r}M$ with $f({\delta }_r)=n_{r+1}-n_r$ and put $m_{r+1}=m_r+{\delta }_r$. The classes of $m_r$ modulo $I^{r}M$ are compatible and map to the chosen components of $\hat{n}$. HenceThis surjectivity on adic completions does not require finite generation of $M$ or $N$. The compatibility of the lifts is the point; independent lifts of the finite-level classes would not suffice.

Since $R$ is Noetherian, write $I=(a_1,\dots ,a_s)$. Define a ring homomorphismModulo $I^r$, evaluating a series requires only its finitely many terms of total degree below $r$. These evaluations commute with the reduction maps, and define $\Phi$ without needing the original ring to be complete or separated.

To prove surjectivity, begin with $\hat{b}\in \hat{R}$. Choose $b_0\in R$ matching its component modulo $I$. After terms of total degree below $r$ have been chosen, their evaluation differs from $\hat{b}$ at level $R/I^{r+1}$ by a class in $I^r/I^{r+1}$. This class is an $R$-linear combination of the monomials $a^{\alpha }$ with $|\alpha |=r$. Choose the corresponding homogeneous degree-$r$ polynomial $p_r(X)$ as the next term. Then the series $b_0+{\sum }_{r\ge 1}{p}_r(X)$ evaluates to $\hat{b}$ at every level. This proves the power-series presentation of an adic completion, and thereforeA quotient of a Noetherian ring is Noetherian, since its ideals correspond to ideals containing the kernel. The cases $I=0$ and $I=R$ give $\hat{R}=R$ and $\hat{R}=0$, respectively, and satisfy the conclusion as well.

If $I=(a_1,\dots ,a_s)$, evaluation $X_i\mapsto a_i$ gives a ring map to the $I$-adic completion, with each series evaluated modulo $I^r$ using only terms of total degree below $r$. It is surjective: successively represent the residual class in $I^r/I^{r+1}$ by a homogeneous degree-$r$ polynomial in the $a_i$. If $R$ is a commutative Noetherian ring, Noetherianity of formal power series rings therefore makes its completion Noetherian.