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For an injective module endomorphism $f:M\to M$ of an Artinian module, the chain $M\supseteq fM\supseteq f^{2}M\supseteq \cdots$ stabilizes. If $f^{n}M=f^{n+1}M$, then every $x$ satisfies $f^{n}x=f^{n+1}y$. Injectivity of $f^n$ gives $x=fy$, proving surjectivity. No finite-length or Noetherian hypothesis is needed.

For a short exact sequence $0\to N\to M\to Q\to 0$, the middle module is Artinian exactly when both outer modules are Artinian. Submodules and quotients inherit the descending chain condition. Conversely, for a descending chain $M_i$, both $M_i\cap N$ and its image in $Q$ eventually stabilize. If $x\in M_i$, choose $y\in M_{i+1}$ with the same image in $Q$; then $x-y\in M_i\cap N=M_{i+1}\cap N$, giving $M_i=M_{i+1}$. Induction shows that finite direct sums of Artinian modules are Artinian.

A commutative Artinian ring has finite composition length as a module over itself. It has finitely many maximal ideals, and its Jacobson radical $J$ is nilpotent. The Chinese remainder theorem makes $R/J$ a finite product of fields; each Artinian module $J^i/J^{i+1}$ has finite-dimensional components over those fields. Adding their lengths along the finite radical filtration proves the assertion, and in particular the ascending chain condition.

Set the dimension of the zero module to $-1$. A nonzero module has dimension zero exactly when it is an Artinian module. Inductively, dimension at most $d$ means that in every descending chain the successive quotients eventually have dimension less than $d$. The least such ordinal is the Krull dimension; the ordinal definition also covers dimensions beyond finite integers. The left dimension of a ring is that of its left regular module. In noncommutative rings this is not defined simply as the length of a chain of two-sided prime ideals.

Let $R$ be a commutative Artinian ring. We prove finite composition length first, without assuming the conclusion that $R$ is Noetherian.

Every prime ideal is maximal. Indeed, an Artinian integral domain is a field: for $a\ne 0$, stabilization of $(a^m)$ gives $a^m=a^{m+1}b$, and cancellation gives $ab=1$. There are only finitely many maximal ideals. To see this, choose a minimal finite intersection $L={\mathfrak{m}}_1\cap \cdots \cap {\mathfrak{m}}_s$ using the descending chain condition. Minimality gives $L\subseteq \mathfrak{m}$ for any other maximal ideal $\mathfrak{m}$. Since ${\mathfrak{m}}_1\cdots {\mathfrak{m}}_s\subseteq L$, primeness forces one ${\mathfrak{m}}_i\subseteq \mathfrak{m}$, and maximality makes them equal. Thus the Jacobson radical $J$ is this finite intersection, and the Chinese remainder theorem gives

We also need nilpotence of $J$. Its powers stabilize; put $I=J^m$ at a stabilization stage, so $I^2=I$. If $I\ne 0$, choose an ideal $L$ minimal subject to $IL\ne 0$. Since $I(IL)=IL\ne 0$, minimality gives $IL=L$. Choose $a\in L$ with $Ia\ne 0$. Minimality applied to the subideal $Ra$ gives $L=Ra$. The equality $IL=L$ then yields $a=xa$ for some $x\in I\subseteq J$. But $1-x$ is a unit by the unit criterion for the Jacobson radical; hence $a=0$, a contradiction. Therefore $J^m=0$.

Each layer $J^i/J^{i+1}$ is an Artinian module over the finite product of fields $R/J$. Its components over those fields are finite-dimensional: an infinite-dimensional vector space has a strictly descending chain of subspaces. Each layer consequently has finite composition length. The finite filtrationproves that $R$ itself has finite composition length, and therefore satisfies the ascending chain condition on ideals. Every commutative Artinian ring is Noetherian. The zero ring satisfies the conclusion trivially.

The relevant Artin–Wedderburn theorem says that a semisimple ring which is a right Artinian ring has the formwhere the $D_i$ are division rings and the positive integers $n_i$ and division rings are unique up to permutation and isomorphism. The division rings need not be commutative or finite-dimensional over a field. Conversely every such finite product is semisimple and a right Artinian ring.

Here is a proof which also works if semisimplicity is initially expressed as $J(R)=0$. A nonzero right ideal contains a minimal nonzero right ideal $L$ by the descending chain condition; $L$ is a simple module. If $L^2=0$, then for every $l\in L$ and $r\in R$, $(lr{)}^2=0$, so $1-lr$ has inverse $1+lr$. The unit criterion for the Jacobson radical would give $L\subseteq J(R)=0$, a contradiction. Thus some $a\in L$ induces a nonzero map $L\to L$ by left multiplication. This map is an isomorphism, since $L$ is simple. Choose $e\in L$ with $ae=a$. Then $a(e^2-e)=0$ and injectivity on $L$ gives $e^2=e$. We obtain $L=eR$ and the splittingFor a right ideal $K$ containing $e$, the same argument splits $K=eR\oplus (K\cap (1-e)R)$. Repeatedly split a minimal right ideal out of the remaining summand. The remaining summands form a strictly descending chain of right ideals, so the process terminates. Thusfor pairwise nonisomorphic simple modules $S_i$.

A nonzero map between simple modules is an isomorphism: its kernel and image leave no alternatives. Thus $D_i={\mathrm{End}}_R(S_i)$ is a division ring, and ${\mathrm{Hom}}_R(S_i,S_j)=0$ for $i\ne j$. This is the needed Schur lemma. Taking endomorphisms of the direct sum gives ${\prod }_i{M}_{n_i}(D_i)$. Finally, left multiplication identifies $R$ with ${\mathrm{End}}_R(R_R)$: a right-module endomorphism sends $r$ to $f(1)r$, and composition corresponds to multiplication in $R$, not its opposite. This proves the displayed ring isomorphism.

For the converse, the right regular module of $M_n(D)$ is the direct sum of its $n$ row modules. Each row module is simple, since elementary matrices and nonzero scalar inverses send any nonzero row to any prescribed row. A finite product therefore has a finite semisimple regular module and is a right Artinian ring. The simple-module multiplicities and their endomorphism division rings recover the factors, establishing uniqueness.

Maximal right ideals of $R/J$ correspond to maximal right ideals of $R$, all of which contain $J$. Their intersection in the quotient is zero. Hence $R/J$ is a right Artinian ring with zero Jacobson radical, and part (a) makes it semisimple.

Part (b) gives the finite radical filtrationEach factor $J^i/J^{i+1}$ is annihilated by $J$, so it is a right module over $R/J$. Every module over a semisimple ring is semisimple: a free module is a direct sum of simple summands of copies of the regular module, and a quotient of a semisimple module is again semisimple, because the images of its simple summands are either zero or simple and span the quotient. A maximal independent family of these simple images spans: any simple image not contained in the sum intersects it trivially and could be added. Each factor is also Artinian, as a subquotient of $R_R$. A semisimple Artinian module has only finitely many simple summands, since an infinite direct sum would give a descending chain by successively deleting summands. Consequently every factor has finite module length.

Concatenating composition series of these finitely many factors makes $R_R$ a module of finite module length. Such a module satisfies both chain conditions, since every strict change consumes a positive amount of length. Therefore $R$ is right Noetherian. This is the Hopkins-Levitzki theorem, obtained from the two preceding results rather than invoked in their proofs.

The Jacobson radical is $J(R)={\bigcap }_{L}L$, where $L$ runs over the maximal right ideals. Every proper right ideal $I$ lies in such an $L$: the union of a chain of proper right ideals containing $I$ is a right ideal and remains proper because it omits $1$. Thus Zorn lemma applies.

The right-module form of Nakayama lemma says that a finitely generated module $M$ with $MJ=M$ is zero. More generally, if $N$ is a submodule and $M=N+MJ$, then $M=N$. To prove the first statement, suppose $M\ne 0$ and take a generating set $m_1,\dots ,m_s$ of minimal size. Since $MJ=M$, there are $a_i\in J$ withThe unit criterion for the Jacobson radical makes $1-a_s$ a unit. Hence $m_s$ is generated by the other $m_i$, a contradiction. Apply this result to the quotient module $M/N$ to obtain the general form.

For an injective module endomorphism $f$ of an Artinian module $M$, the descending chain condition givesfor some $n$. Given $x\in M$, write $f^n(x)=f^{n+1}(y)$; injectivity of $f^n$ gives $x=f(y)$. Thus $f$ is surjective, as asserted by Artinian modules are co-Hopfian.

Now $R$ is a semilocal ring and $\overline{V}=V/VJ$ is a finitely generated module over the right Artinian ring $\overline{R}=R/J$. A finite direct sum of copies of $\overline{R}$, and hence its quotient module $\overline{V}$, is an Artinian module. The module endomorphism $\alpha$ induces $\overline{\alpha }$ on $\overline{V}$, and the given inverse-image condition says precisely that $\ker \overline{\alpha }=0$. The preceding argument makes $\overline{\alpha }$ surjective, soThe quotient module $W=V/\alpha (V)$ is finitely generated and satisfies $WJ=W$. Applying Nakayama lemma yields $W=0$. Therefore $\alpha$ is surjective, by surjectivity from injectivity modulo the Jacobson radical.

For a finitely generated algebra over $k$, choose a finite-dimensional generating subspace $V$ containing one. Its Gelfand–Kirillov dimension isChanging generators bounds the two filtrations by constant rescalings of the index, so the value is independent of $V$.

For the noncommutative Krull dimension, set the zero module's dimension to $-1$. A nonzero module has dimension zero exactly when it is an Artinian module. Recursively, dimension at most $d$ means that in every descending chain of submodules the successive factors eventually have dimension less than $d$; use the ordinal version of this definition if necessary. The left Krull dimension of $R$ is that of its left regular module. For commutative Noetherian rings this agrees with the usual prime-chain Krull dimension, but prime-chain counting alone is not its definition for a noncommutative ring.

Write $U=U({\mathfrak{sl}}_2(\mathbb{C}))$ with $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. By the Poincaré-Birkhoff-Witt theorem, $f^i{h}^j{e}^k$ form a basis. In the total-degree filtration there are $\left(\genfrac{}{}{0ex}{}{n+3}{3}\right)$ such monomials of degree at most $n$, soThe commutators lower degree, giving $\mathrm{gr}U\cong \mathbb{C}[f,h,e]$, which also proves that $U$ is left and right Noetherian by finite leading-term cancellation.

To determine the module-theoretic Krull dimension, use the generalized Weyl algebra description and verify the dimension criterion, rather than substituting the three-dimensional graded ring. PutCommutation with $h$ is immediate; with $e$, the contributions are $4he-2(eh+he)-4e=0$, using $he-eh=2e$. The corresponding calculation with $f$ also vanishes. Thus $z$ is a Casimir element. The subalgebra $D=\mathbb{C}[h,z]$ is a polynomial ring in two variables: for a nonzero polynomial in $z$ with coefficients in $\mathbb{C}[h]$, its highest $z$ power contributes a nonzero PBW term with that many copies of both $f$ and $e$, and no lower power can cancel it.

Take $\sigma (h)=h-2$, $\sigma (z)=z$ and $t=(z-h^2-2h)/4$. The relations becomeThey identify $U$ with $D(\sigma ,t)$. Surjectivity follows from its generators; injectivity follows from PBW: the graded pieces $D$, $D{e}^r$ and $D{f}^r$ for $r>0$ are linearly independent, using the difference between the $e$ and $f$ exponents and then the highest paired exponent. Every PBW monomial lies in one of these pieces by replacing paired $f,e$ powers with polynomials in $h,z$.

We use the following general Krull dimension criterion for rank-one generalized Weyl algebras: for a commutative Noetherian base $D$ of dimension $d$, the generalized Weyl algebra has Krull dimension $d$ provided no height-$d$ maximal ideal is periodic under $\sigma$ and none contains infinitely many translates ${\sigma }^n(t)$. This is Theorem 5.3, specialized to the commutative base, in Krull dimension of Generalized Weyl Algebras with non-commutative coefficients.

Here every maximal ideal is ${\mathfrak{m}}_{\lambda ,\mu }=(h-\lambda ,z-\mu )$ and has height two. Its translates have $h$ coordinate $\lambda +2n$, so no nonzero power of $\sigma$ fixes it. Further,This is a nonzero quadratic polynomial in $n$, so it vanishes for at most two integers. Both exceptional mechanisms are excluded for every maximal ideal, and the base has dimension two. The criterion consequently givesThe orbit and root calculations are needed for this upper bound on the entire left-ideal lattice; merely exhibiting a length-two chain of prime ideals would only prove a lower bound.

An Artinian module is an $A$-module satisfying the descending chain condition on submodules: every chain $M_0\supseteq M_1\supseteq \cdots$ is eventually constant. An Artinian ring is a ring that is Artinian as a module over itself, so its descending chains of ideals stabilize.

We first prove the closure properties needed for finite generation. A submodule of an Artinian module is Artinian, because any descending chain in it is also a chain in the original module. A quotient module is Artinian, because inverse images turn a descending chain in the quotient into one upstairs. Conversely, consider a short exact sequencewith $N$ and $Q$ Artinian. For a descending chain $(M_i)$, the chains $M_i\cap N$ and $\pi (M_i)$ both stabilize. Choose $r$ beyond both stabilization indices. If $i\ge r$ and $x\in M_i$, equality of the images gives $y\in M_{i+1}$ with $\pi (y)=\pi (x)$. Thenso $x\in M_{i+1}$. Hence $M_i=M_{i+1}$. This proves the extension property of Artinian modules in a short exact sequence.

Apply this property inductively to $0\to A^{n-1}\to A^n\to A\to 0$. Every finite direct sum $A^n$ is Artinian when $A$ is an Artinian ring. If $m_1,\dots ,m_n$ generate $M$, the map $(a_i)\mapsto \sum a_i{m}_i$ is a surjection $A^n\twoheadrightarrow M$. The quotient property therefore provesThis argument also covers the zero module; it does not require a structure theorem for Artinian rings.

For a counterexample without finite generation, take any field $k$ and the moduleThe ring $k$ is Artinian, since its only ideals are $0$ and $k$. But $M_i⊋M_{i+1}$ for every $i$, since $e_i\notin M_{i+1}$. Thus an infinite-dimensional vector space over a field provides the requested non-Artinian module.

Finally, let $f:M\to M$ be an injective endomorphism with $M$ Artinian. Its image chain stabilizes, so $f^{r}M=f^{r+1}M$ for some $r$. Given $x\in M$, choose $y$ with $f^{r}x=f^{r+1}y=f^r(fy)$. Since $f^r$ is injective, $x=fy$. ThereforeThis is precisely the assertion that Artinian modules are co-Hopfian.

All rings below are nonzero and commutative with identity; the zero ring is trivially Artinian. A primary ideal is proper. An ideal $Q$ is primary precisely when every zero divisor of $R/Q$ is a nilpotent element. Its radical of an ideal $\mathfrak{p}=\sqrt{Q}$ is prime: if $ab\in \sqrt{Q}$ and $a\notin \sqrt{Q}$, apply the primary property to $(ab{)}^r\in Q$ to obtain a power of $b$ in $Q$. In a Noetherian ring, finite generation of $\mathfrak{p}$ gives ${\mathfrak{p}}^s\subseteq Q$ for some $s$. For generators $u_i$ with $u_i^{s_i}\in Q$, one can take $s=1+{\sum }_i(s_i-1)$.

We first prove existence of primary decomposition. The ascending chain condition implies that every proper ideal is a finite intersection of irreducible ideals. Otherwise choose a maximal counterexample $I$ under inclusion. It is not irreducible, so $I=J\cap K$ with both $J,K$ strictly larger; their finite irreducible decompositions give one for $I$, a contradiction.

An irreducible ideal in a Noetherian ring is primary. Pass to the quotient and suppose zero is irreducible. If $xy=0$ with $x\ne 0$, the ascending chain of annihilators of $y^j$ stabilizes, say at $j=r$. Any element of $(x)\cap (y^r)$ can be written $z=ax=b{y}^r$. Then $yz=0$, so $b\in \mathrm{Ann}(y^{r+1})=\mathrm{Ann}(y^r)$ and $z=0$. Irreducibility of zero implies $(y^r)=0$, since $(x)\ne 0$. Thus $y$ is nilpotent. This proves irreducible ideals are primary in Noetherian rings, and hence the Lasker–Noether theorem.

An intersection of finitely many primary ideals with the same radical $\mathfrak{p}$ is again primary: if $ab$ belongs to all of them and $b\notin \mathfrak{p}$, the primary property forces $a$ into every component. Combining such components and removing redundant ones yields a minimal primary decompositionThe radicals are uniquely determined, but the components need not all be unique. Here is a proof identifying the invariant radicals as the associated primes of a module $R/I$.

Every nonzero module over a Noetherian ring has a nonzero element with a prime annihilator: maximize the annihilator of a nonzero element using the ascending chain condition. If $abm=0$ and $bm\ne 0$, maximality gives $\mathrm{Ann}(bm)=\mathrm{Ann}(m)$ and hence $am=0$. This is the maximal annihilator of a module element is prime argument. For a $\mathfrak{p}$-primary quotient $R/Q$, the radical of the annihilator of every nonzero element is $\mathfrak{p}$: it is contained in $\mathfrak{p}$ by the primary property and contains a power of $\mathfrak{p}$ because ${\mathfrak{p}}^s\subseteq Q$. Thus the only possible associated prime is $\mathfrak{p}$, and it does occur.

The diagonal injection $R/I\hookrightarrow {⨁}_{i}R/Q_i$ shows that every associated prime of $R/I$ is one of the ${\mathfrak{p}}_i$. Indeed, for an element whose annihilator $P$ is prime, that annihilator is the intersection of the finitely many component annihilators, all containing $P$. Their product is contained in $P$, so primality forces one of them to equal $P$; its component element is nonzero and has associated prime ${\mathfrak{p}}_i$. Conversely, irredundancy gives $b\in {\bigcap }_{j\ne i}{Q}_j\setminus Q_i$. The nonzero cyclic submodule generated by $b+I$ embeds in $R/Q_i$. It has an associated prime, necessarily ${\mathfrak{p}}_i$, which is then an associated prime of $R/I$. We have proved the first uniqueness theorem for primary decomposition:Also the zero divisors on $R/I$ are exactly ${\bigcup }_i{\mathfrak{p}}_i$. For a scalar killing a nonzero element, extend its element annihilator to a maximal element annihilator containing it; the preceding argument gives an associated prime containing that scalar. The converse is immediate from the definition of an associated prime.

The minimal members of the set $\{{\mathfrak{p}}_i\}$ are the isolated primes of a primary decomposition. They are exactly the primes minimal over $I$: if a prime contains ${\bigcap }_i{\mathfrak{p}}_i$, it contains one ${\mathfrak{p}}_i$ by the product argument. A component belonging to an isolated prime is unique. Localize at ${\mathfrak{p}}_i$. Every other component becomes the whole ring, since its radical contains an element outside ${\mathfrak{p}}_i$ whose power lies in that component. A ${\mathfrak{p}}_i$-primary ideal contracts unchanged from this localization, because $sa\in Q_i$ with $s\notin {\mathfrak{p}}_i$ implies $a\in Q_i$. Hence the second uniqueness theorem for primary decomposition givesAn embedded primary component can vary. For example, in $k[x,y]$,Both second components are $(x,y)$-primary, since their quotients are dual-number algebras. For the second equality, reducing modulo $(x^2,y+x)$ makes $y=-x$; an element $xg$ vanishes exactly when $g$ has zero constant term, giving $(x^2,xy)$. The isolated component is $(x)$ and the embedded associated prime is $(x,y)$.

The relevance to Artinian rings is particularly sharp in Krull dimension zero. Suppose $R$ is Noetherian and all its primes are maximal. Apply a minimal primary decomposition to zero. Its radicals are distinct maximal ideals, so the components are pairwise comaximal ideals: each contains a power of its radical, and expanding $1=(a+b{)}^N$ for $a+b=1$ shows that powers of comaximal ideals remain comaximal. The Chinese remainder theorem gives the Artinian decomposition into local factorsEach factor has one maximal ideal, and that ideal is nilpotent. Its finite radical filtration has successive layers finitely generated over its residue field, hence of finite vector-space dimension. It follows that the factor, and therefore $R$, has a finite composition series of a module, so $R$ is Artinian. This proves the Noetherian dimension-zero criterion for an Artinian ring in this direction. Applied to $R/I$, it says that the quotient is Artinian exactly when all primes over $I$ are maximal. In that case no primary component is embedded, so all components are unique.

For completeness, the converse does not require initially assuming Noetherianity. In an Artinian ring, an Artinian domain is a field: stabilization of $(a^j)$ and cancellation gives an inverse to every nonzero $a$. Thus every prime is maximal. There are finitely many maximal ideals, since infinitely many distinct ones would give strictly descending finite intersections; comaximality guarantees strictness by the Chinese remainder theorem. Its nilradical $N$ is nilpotent. Indeed, its powers stabilize at $J=N^r$, with $J^2=J$. If $J\ne 0$, choose an ideal $C$ minimal among those satisfying $JC\ne 0$. Then $JC=C$, since $J(JC)=JC\ne 0$. Choose $x\in C$ with $Jx\ne 0$; minimality also gives $Rx=C$. Therefore $x=ax$ for some $a\in J\subseteq N$. But $1-a$ is a unit, contradicting $x\ne 0$. Hence $J=0$.

Now $R/N$ is a finite product of residue fields. Each $N^j/N^{j+1}$ is an Artinian module over that product, and therefore a finite direct sum of finite-dimensional vector spaces: an infinite-dimensional vector space would have a strictly descending chain of subspaces. The finite filtration by powers of $N$ makes $R$ a module of finite composition length, in particular Noetherian. Combining both directions yieldsThus primary decomposition separates the local pieces of a zero-dimensional ring, while their nilpotent maximal ideals record the multiplicities that the reduced set of primes alone does not detect.

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.

False. The module over a ring $\mathbb{Z}$ over itself is Noetherian, because its submodules are the principal ideals $n\mathbb{Z}$, but the descending chainshows that it is not an Artinian module.

A ring is right Artinian if its right ideals satisfy the descending chain condition, equivalently its right regular module is an Artinian module. Left and right chain conditions should be distinguished for a general ring.

For a finitely generated module $V$ on the right over a semilocal ring, the displayed condition makes the induced module endomorphism on $V/VJ$ injective. This quotient is an Artinian module, so Artinian modules are co-Hopfian makes that map surjective. The remaining quotient module $V/\alpha (V)$ equals its product with $J$, and Nakayama lemma makes it zero.