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A Noetherian module satisfies the ascending chain condition on its submodules; equivalently every submodule is a finitely generated module. If the latter condition holds, the union of an ascending chain has a finite generating set lying in one member, so the chain stabilizes. Conversely, a submodule with no finite generating set allows successive choices outside the submodule generated so far, producing a strictly ascending chain. A Noetherian ring is a ring that is Noetherian as a module over itself, equivalently one whose ideals are all finitely generated.

The formal power series ring consists of sequences written ${\sum }_{i\ge 0}{a}_i{x}^i$, with coefficientwise addition and the Cauchy product:Every coefficient uses a finite sum; no analytic convergence is involved. Evaluation at zero gives $A[[x]]/(x)\cong A$. A quotient ring of a Noetherian ring is Noetherian, because its ascending chains of ideals lift to such chains upstairs. Thus one direction of Noetherianity of a formal power series ring is immediate.

For the other direction suppose $A$ is Noetherian, and let $I\subseteq A[[x]]$ be an ideal. Define the coefficient ideals of a formal power series idealEach is an ideal, and multiplication by $x$ gives $I_r\subseteq I_{r+1}$. Choose $N$ with $I_N=I_{N+1}=\cdots$. For $0\le r\le N$, choose generators $a_{r1},\dots ,a_{r{m}_r}$ for $I_r$, and choose series $g_{rj}\in I\cap x^{r}A[[x]]$ whose $x^r$ coefficients are those generators. Empty generating sets are allowed for zero ideals.

These finitely many series generate $I$. To prove this without assuming that $I$ is closed, start with $g\in I$ and cancel its coefficients successively. If the residual $h_r$ lies in $I\cap x^{r}A[[x]]$, its $x^r$ coefficient lies in $I_r$. Put $s=\min (r,N)$; since $I_r=I_s$ when $r\ge N$, write that coefficient as ${\sum }_j{c}_{rj}{a}_{sj}$ and subtract ${\sum }_j{c}_{rj}{x}^{r-s}{g}_{sj}$. The new residual belongs to $I\cap x^{r+1}A[[x]]$.

For $s<N$, the generators $g_{sj}$ are used only at the single step $r=s$. For $s=N$, their accumulated multipliers are the genuine formal power series ${\sum }_{r\ge N}{c}_{rj}{x}^{r-N}$. Thus the construction yields the exact coefficientwise identityEvery coefficient of the residual vanishes after finitely many steps. This is an ordinary finite generating expression in $A[[x]]$, not merely an approximation by elements of a smaller ideal. Therefore every ideal is finitely generated, provingFinally, for the surjective module endomorphism $f$, the kernels $\ker f\subseteq \ker f^2\subseteq \cdots$ stabilize because $M$ is a Noetherian module. Take $N$ with $\ker f^N=\ker f^{N+1}$. If $f(m)=0$, surjectivity of $f^N$ gives $u$ with $f^N(u)=m$. Then $f^{N+1}(u)=0$, so $u\in \ker f^N$ and $m=0$. Hence $f$ is injective as well as surjective; its inverse is automatically $A$-linear. This proves $f$ is an isomorphism, the statement that Noetherian modules are Hopfian.

A ring is Artinian when its ideals satisfy the descending chain condition: every chain $I_1\supseteq I_2\supseteq \cdots$ eventually becomes constant. It is Noetherian when its ideals satisfy the ascending chain condition, equivalently when every ideal is finitely generated. We first prove finite length of a commutative Artinian ring; this gives the stronger structural reason for its Noetherian property.

If $\mathfrak{p}$ is a prime ideal of an Artinian ring, the quotient $R/\mathfrak{p}$ is an Artinian integral domain. For a nonzero element $x$ of that domain, the chain $(x)\supseteq (x^2)\supseteq \cdots$ stabilizes. Thus $x^n=b{x}^{n+1}$ for some $b$, and cancellation gives $bx=1$. The quotient is a field, so every prime ideal is a maximal ideal.

There are only finitely many maximal ideals. Otherwise, choose distinct ones ${\mathfrak{m}}_1,{\mathfrak{m}}_2,\dots$. The intersections ${\mathfrak{m}}_1\cap \cdots \cap {\mathfrak{m}}_n$ form a strictly descending chain. To see strictness, for each $i\le n$ choose $a_i\in {\mathfrak{m}}_i\setminus {\mathfrak{m}}_{n+1}$; their product belongs to the first $n$ ideals but not to the next one, since ${\mathfrak{m}}_{n+1}$ is prime. This contradicts the descending chain condition.

Put $J={\bigcap }_i{\mathfrak{m}}_i$. This Jacobson radical is also the nilradical, since all primes are maximal. We need the stronger conclusion that $J$ is nilpotent, without assuming Noetherianity. Its powers stabilize, say $K=J^n=J^{n+1}=JK$. Suppose $K\ne 0$. By the descending chain condition, choose an ideal $L$ minimal subject to $KL\ne 0$. Some $x\in L$ has $Kx\ne 0$, so minimality gives $L=Rx$. Moreover $K(JL)=KL\ne 0$, and minimality gives $JL=L$. Hence $x=jx$ for some $j\in J$. But $1-j$ is a unit: it cannot lie in any maximal ideal, because $j$ lies in all of them. Thus $x=0$, a contradiction. Therefore $J^n=0$.

The Chinese remainder theorem gives $R/J\cong {\prod }_{i}R/{\mathfrak{m}}_i$, a finite product of fields. Each quotient $J^i/J^{i+1}$ is an Artinian module over this product, and each field component must be a finite-dimensional vector space; an infinite-dimensional vector space admits a strictly descending chain of subspaces. Consequently every layer has finite composition length. The finite filtrationshows that $R$ itself has finite composition length. A strict inclusion of submodules strictly increases length, so an ascending chain cannot continue indefinitely. Every commutative Artinian ring is therefore Noetherian. This is the Artinian rings are Noetherian result.

For the formal power series ring, let $B=R[[X]]$ with $R$ Noetherian, and let $H$ be any ideal of $B$. For $n\ge 0$ define a coefficient idealThese coefficient ideals of a formal power series ideal satisfy $A_n\subseteq A_{n+1}$, by multiplication by $X$. The ascending chain condition gives $A_n=A_N$ for all $n\ge N$. For each $0\le n\le N$, choose finitely many series $f_{n,j}\in H\cap X^{n}B$ whose coefficients at $X^n$ generate $A_n$.

We claim that these finitely many series generate $H$ as an ordinary ideal. Given $f\in H$, cancel its coefficient at $X^k$ successively. After coefficients below $k$ have vanished, its coefficient at $X^k$ lies in $A_k$. If $k\le N$, use an $R$-linear combination of the $f_{k,j}$. If $k>N$, use a combination of $X^{k-N}{f}_{N,j}$, since $A_k=A_N$. The remainder then belongs to $H\cap X^{k+1}B$.

Collect all the cancellations against each fixed generator. For $n<N$ its multiplier is a polynomial, while the multipliers of the $f_{N,j}$ are well-defined formal power series: at any fixed degree, only finitely many cancellation steps contribute. The remainder has every coefficient zero. ThusThis is a finite sum of ideal generators, rather than merely a topological closure assertion. Since $H$ was arbitrary,This coefficient-cancellation argument proves Noetherianity of a formal power series ring.

The corresponding Artinian assertion is false. For any nonzero ring $R$, the idealsin $R[[X]]$ are strictly decreasing, since $X^n$ has a nonzero coefficient in degree $n$ and no multiple of $X^{n+1}$ does. In particular, a field $k$ is Artinian, but $k[[X]]$ is not. The zero ring is the harmless exception.