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Suppose $A\subseteq B$ are unital rings and $B$ is a finitely generated module over $A$ on the right. If $A$ is a right Noetherian ring, then $B_A$ is a Noetherian module. Every right ideal of $B$ is an $A$-submodule, hence has finitely many generators over $A$; these also generate it over $B$. Thus $B$ is a right Noetherian ring. The same argument works on the left under the corresponding hypotheses. Finite-index group ring extensions give an important example, using coset representatives as a finite module basis.

Every nonzero Noetherian module contains a uniform module: otherwise repeatedly split a nonuniform nonzero submodule into two disjoint nonzero submodules, retaining one and collecting the other, to build an infinite direct sum. Repeatedly adding a uniform submodule disjoint from the current finite sum must likewise stop. The final finite sum is an essential submodule, since any disjoint nonzero submodule would allow another addition.

For a multiplicative subset $S\subseteq R$ and an $R$-module $M$, the localization $S^{-1}M$ consists of fractions $m/s$, with $m/s=m^{\prime }/s^{\prime }$ when some $u\in S$ satisfies $u(s^{\prime }m-s{m}^{\prime })=0$. It is an $S^{-1}R$-module, and the canonical map is $m\mapsto m/1$. If $M$ is finitely generated over a Noetherian ring, then $S^{-1}M$ is a Noetherian module over $S^{-1}R$.

For a surjective module endomorphism of a Noetherian module, the chain $\ker f\subseteq \ker f^2\subseteq \cdots$ stabilizes at some $N$. If $f(x)=0$, choose $y$ with $f^N(y)=x$. Then $y\in \ker f^{N+1}=\ker f^N$, giving $x=0$. Thus the endomorphism is also injective and is an isomorphism.

For a short exact sequence $0\to N\to M\to M/N\to 0$, $M$ is a Noetherian module exactly when both the other terms are Noetherian modules. For the extension direction, an ascending chain $M_i$ has eventually constant intersections $M_i\cap N$ and images $(M_i+N)/N$. Equality of both forces equality of consecutive $M_i$: subtract an element with the same image and then use the common intersection.

The Hilbert basis theorem says that if $B$ is a commutative Noetherian ring, then $B[t]$ is Noetherian; iteration treats any finite number of variables. Here is a complete leading-coefficient proof. For an ideal $I\subseteq B[t]$, let $L_d$ be the set of degree-$d$ coefficients of its polynomials of degree at most $d$, including zero coefficients. These are ideals of $B$, and multiplication by $t$ gives $L_d\subseteq L_{d+1}$. The ascending chain condition makes the chain stationary from some $N$ onward. Choose finite generators for $L_0,\dots ,L_N$ and lift them to polynomials $f_{dj}\in I$ of degree at most $d$. If $f\in I$ has degree $m$, set $d=\min (m,N)$ and express its leading coefficient as ${\sum }_j{a}_{dj}{b}_j$, where $a_{dj}$ is the chosen degree-$d$ coefficient. Then $f-{\sum }_j{b}_j{t}^{m-d}{f}_{dj}$ has smaller degree. Induction proves that the finitely many lifts generate $I$. This proves the theorem, including coefficient rings with zero divisors.

Put $A=\mathbb{C}G$ and $L=\mathbb{C}[X,X^{-1}]$. The infinite dihedral group has normal forms $X^m$ and $X^{m}Y$. The relations move every $Y$ to the right and reduce its exponent modulo two; the action $X:n\mapsto n+1$, $Y:n\mapsto -n$ on $\mathbb{Z}$ shows that these forms are distinct. ThusThe Hilbert basis theorem makes $\mathbb{C}[X]$ Noetherian, and its localization $L$ is Noetherian: an ideal of a localization is generated by the images of the finite generators of its contraction. Since $A$ is free of rank two over $L$ on the left, it is a Noetherian module over $L$. Every left ideal of $A$ is an $L$-submodule, so all its ascending chains stabilize. Therefore $\mathbb{C}G$ is left Noetherian.

For the center of the infinite dihedral group algebra, write an element as $f(X)+g(X)Y$. Commuting it with $X$ gives $g(X)(X^{-1}-X)=0$ in the Laurent integral domain, forcing $g=0$. Commuting $f(X)$ with $Y$ then gives $f(X)=f(X^{-1})$. A symmetric Laurent polynomial is a linear combination of $1$ and $X^j+X^{-j}$, $j\ge 1$. With $z=X+X^{-1}$, the recursionexpresses $S_j=X^j+X^{-j}$ as a polynomial in $z$. Conversely every polynomial in $z$ is central. The highest Laurent exponent of a nonzero polynomial in $z$ shows that $z$ is transcendental. HenceThe identities $X^{-1}=z-X$ and $X^2-zX+1=0$ give $L=R+RX$. Independence follows by applying $X\mapsto X^{-1}$ to a relation $f(z)+Xg(z)=0$ and subtracting: $(X-X^{-1})g(z)=0$, so $g=f=0$. Consequentlyis a free $R$-module of rank four.

Let $M$ be a nonzero simple module on the left. For any $m\ne 0$, $M=Am$, so it is finite over $R$. Every central $r\notin P={\mathrm{Ann}}_R(M)$ acts bijectively: its kernel and image are $A$-submodules, and simplicity forces kernel zero and image $M$. To prove that $P$ is maximal, not merely prime, put $B=R/P$; then $M$ is a faithful finite $B$-module. For $0\ne b\in B$, $bM=M$. If $m_1,\dots ,m_d$ generate $M$, write $m_i=b{\sum }_j{c}_{ij}{m}_j$. The adjugate matrix identity for $I-bC$ shows that $\det (I-bC)$ annihilates every generator and hence $M$. Faithfulness makes this determinant zero in $B$. Its constant term is one, so it has form $1-bq$, proving $bq=1$. Thus $B$ is a field, and $P$ is a maximal ideal of $R$. This proves the general principle for simple modules over an algebra finite over its center in this instance.

Every maximal ideal of $\mathbb{C}[z]$ is $(z-\lambda )$ for some $\lambda \in \mathbb{C}$, by factorization of complex polynomials. The module $M=Am$ is therefore a quotient of the four-dimensional algebra $A/(z-\lambda )A$. In particular ${\dim }_{\mathbb{C}}M\le 4<\infty$, as required.

The actual simple dimensions are at most two. If $\lambda \ne \pm 2$, choose $u\ne \pm 1$ with $u+u^{-1}=\lambda$. The matrices $X=\mathrm{diag}(u,u^{-1})$ and $Y=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ give a representation of $A/(z-\lambda )A$ onto $M_2(\mathbb{C})$: polynomial functions of $X$ give the two diagonal matrix units, and multiplication by $Y$ gives the off-diagonal ones. Both algebras have dimension four, so this is an isomorphism. If $\lambda =\pm 2$, the element $X\mp 1$ has square zero and is conjugated to its negative by $Y$; its generated two-sided ideal is square zero. It annihilates every simple module, since its action on a simple module is either zero or the whole module, and the latter contradicts its square being zero. The remaining quotient is $\mathbb{C}[C_2]$, whose two simple modules have $Y=\pm 1$. Thus the generic simple modules have dimension two, and the four exceptional one-dimensional characters have $X=\pm 1$, $Y=\pm 1$.

Suppose first that $M$ is a Noetherian module. An ascending chain of submodules of $N$ is also a chain in $M$, so it stabilizes. A chain in the quotient module $M/N$ lifts to a chain in $M$, so it also stabilizes. Thus both $N$ and $M/N$ are Noetherian modules.

Conversely, let $M_1\subseteq M_2\subseteq \cdots$ be a chain of submodules of $M$. If $N$ and $M/N$ are Noetherian modules, both $M_i\cap N$ and $(M_i+N)/N$ stabilize, say for $i\ge i_0$. For $x\in M_{i+1}$, equality of the images gives $y\in M_i$ with $x-y\in N$. Then $x-y\in M_{i+1}\cap N=M_i\cap N$, and hence $x\in M_i$. This proves the extension direction of Noetherian modules in a short exact sequence.

The left regular module $R$ is Noetherian because $R$ is a left Noetherian ring. Applying the extension result inductively makes each finite direct sum $R^s$ Noetherian. Every finitely generated module on the left is a quotient module of such a direct sum, so it is Noetherian.

A poly-(cyclic or finite) group has a finite subnormal serieswhose factors are cyclic groups or finite groups. We prove that the group ring $R[G_i]$ is a left Noetherian ring by induction. Put $A=R[G_{i-1}]$. A finite factor makes $R[G_i]$ a finite free left $A$-module, using a coset transversal. Its left ideals are $A$-submodules, so finite-module extension preserves Noetherianity handles this case.

If the factor is infinite cyclic, choose a lift $t$ of a generator and let $\sigma (a)=ta{t}^{-1}$. The group ring is the skew Laurent polynomial ring $A[t,t^{-1};\sigma ]$. Here is the required left-sided skew Hilbert basis theorem. For a left ideal $L\subseteq A[t;\sigma ]$, defineEach $I_n$ is a left ideal of $A$. Left multiplication by $t$ gives $I_n\subseteq I_{n+1}$. The ascending chain condition makes these ideals constant for $n\ge n_0$; choose finitely many generators for each $I_n$, $0\le n\le n_0$, and lift them to polynomials $f_{nj}\in L$.

If $f\in L$ has degree $d\ge n_0$ and leading coefficient $a_d$, writeSubtracting ${\sum }_j{\sigma }^d(r_j)t^{d-n_0}{f}_{n_{0}j}$ cancels the leading coefficient. For $d<n_0$, use the lifts for $I_d$ instead. Induction on polynomial degree shows that the finitely many $f_{nj}$ generate $L$. Thus the skew polynomial ring of an automorphism is a left Noetherian ring.

Finally, for a left ideal $K$ of $A[t,t^{-1};\sigma ]$, its intersection with $A[t;\sigma ]$ has finitely many generators. Every $f\in K$ satisfies $t^{m}f\in A[t;\sigma ]$ for some $m\ge 0$, so those same generators generate $K$ after multiplication by $t^{-m}$. This completes the induction and proves that $R[G]$ is left Noetherian, the left-sided form of group rings of poly-(cyclic or finite) groups are Noetherian.

The prime radical of a noncommutative ring is $N(R)={\bigcap }_{P}P$, with $P$ ranging over all prime ideals of a noncommutative ring. Here primeness means that for two-sided ideals $A,B$, $AB\subseteq P$ implies $A\subseteq P$ or $B\subseteq P$. A minimal prime over an ideal $I$ is a prime ideal containing $I$ with no strictly smaller prime ideal containing $I$.

The ascending chain condition on left ideals also applies to two-sided ideals. We first prove by Noetherian induction that every proper $I$ contains a finite product of prime ideals, each containing $I$. If there were counterexamples, choose a maximal one. It is not prime, so there are strictly larger two-sided ideals $A,B$ with $AB\subseteq I$. Neither is a counterexample; products of prime ideals contained in $A$ and in $B$ then concatenate to a product contained in $I$, a contradiction.

Write $P_1\cdots P_t\subseteq I\subseteq P_j$. Every prime ideal $P$ over $I$ contains at least one $P_j$, by primeness. The inclusion-minimal members of the finite list $P_j$ are therefore precisely the minimal primes over an ideal $I$: any smaller prime over $I$ would contain a member of that same list. Consequently there are finitely many. Replacing each factor $P_j$ by a minimal prime contained in it giveswith each $Q_j$ minimal over $I$. Repeated factors may be necessary; in $k[\epsilon ]/({\epsilon }^m)$ the sole minimal prime $(\epsilon )$ requires its $m$th power to reach zero.

Apply this to $I=0$. Since $N(R)\subseteq Q_j$ for every $j$,Thus the prime radical is a nilpotent ideal, as in finite minimal primes and nilpotent prime radical.

Now assume $R$ is commutative. A prime ideal $P$ is an associated prime of a module $M$ when $P={\mathrm{Ann}}_R(m)$ for some $0\ne m\in M$. Every nonzero module over $R$ has an associated prime: the ascending chain condition gives a maximal annihilator of a nonzero element, and maximal annihilator of a module element is prime.

Since the given $M$ is a Noetherian module, repeatedly choosing such a cyclic submodule in the next quotient yields a finite prime filtrationThe process terminates by the ascending chain condition. In a short exact sequence $0\to U\to M\to W\to 0$, one has $\mathrm{Ass}(M)\subseteq \mathrm{Ass}(U)\cup \mathrm{Ass}(W)$. Indeed, if $P=\mathrm{Ann}(m)$ and $Rm\cap U=0$, the image of $m$ in $W$ still has annihilator $P$. Otherwise choose $0\ne am\in U$; then $a\notin P$, and primeness gives $\mathrm{Ann}(am)=P$. As $\mathrm{Ass}(R/P_i)=\{P_i\}$, induction along the prime filtration provesIn particular there are only finitely many associated primes of a module.

Finally let $P$ be a minimal prime ideal of $R$. Its localization at a prime ideal $R_P$ has exactly one prime, its maximal ideal $P{R}_P$. The preceding nilpotence result makes this maximal ideal nilpotent. Choose $0\ne y\in R_P$ annihilated by $P{R}_P$, for example a nonzero element in its last nonzero power, or $1$ if $P{R}_P=0$. Since elements outside the maximal ideal are units, ${\mathrm{Ann}}_{R_P}(y)=P{R}_P$.

Write $y=r/s$. The element $r/1$ is nonzero and also has annihilator $P{R}_P$. Generate $P$ by $p_1,\dots ,p_{\ell }$. For each $i$, choose $u_i\notin P$ with $u_i{p}_{i}r=0$, and put $u={\prod }_i{u}_i$. Then $m=ur$ remains nonzero after localization, while $Pm=0$. If $am=0$, localization forces $a/1\in P{R}_P$, and hence $a\in P$. Therefore ${\mathrm{Ann}}_R(m)=P$. This proves that minimal primes are associated primes.

Use unital rings, and distinguish left and right Noetherianity. A useful form of the noncommutative Hilbert basis theorem is the following: if a ring $B$ is generated by a subring $A$ and an element $x$, if $A+xA=A+Ax$, and if $A$ is a right Noetherian ring, then $B$ is a right Noetherian ring. The corresponding left-handed statement holds as well. In particular, a central polynomial ring over a Noetherian ring is Noetherian, and a skew polynomial ring of an automorphism preserves either side's Noetherianity.

Here is a proof of the general right-handed statement. Put $F_{-1}=0$ andThe equality follows inductively from $A+xA=A+Ax$. These groups form an exhaustive ascending filtration, stable under multiplication by $A$ on either side, with $F_j{F}_k\subseteq F_{j+k}$. For any $b\in A$ and $n\ge 0$, moving coefficients one step at a time also givesFix a right ideal $I$ of $B$ and define coefficient setsEach $L_n$ is a right ideal of $A$. Additive closure is immediate. If $a\in L_n$ and $b\in A$, use the preceding transport identity to obtainHere $(a{x}^n)c$ has this membership because $I$ is a right ideal and $F_{n-1}$ is stable under right multiplication by $A$. Also $L_n\subseteq L_{n+1}$, by multiplying $a{x}^n\in I+F_{n-1}$ on the right by $x$. The ascending chain condition in $A$ yields $N$ with $L_n=L_N$ for $n\ge N$.

For each $0\le k\le N$, choose finite right generators $a_{kj}$ of $L_k$ and lifts $f_{kj}\in I$ with $f_{kj}-a_{kj}{x}^k\in F_{k-1}$. These finitely many lifts generate $I$ as a right ideal of $B$. To prove this, let $f\in I\cap F_d$ and write $f=a{x}^d+v$, $v\in F_{d-1}$. Then $a\in L_d$. Put $k=\min (d,N)$ and write $a={\sum }_j{a}_{kj}{b}_j$. Transport each coefficient as $b_j{x}^d=x^d{c}_j+v_j$ with $v_j\in F_{d-1}$. ConsequentlyInduction on $d$, with $F_{-1}=0$, expresses $f$ in the right ideal generated by the lifts. Every right ideal is therefore finitely generated, proving the theorem. Passing to the opposite ring proves the left-handed version. This proof does not require unique coefficient normal forms.

A poly-(cyclic or finite) group has a finite subnormal serieswhose factors $G_i/G_{i-1}$ are cyclic or finite. Normality is required between consecutive terms. We show inductively that each group ring $\mathbb{Z}[G_i]$ is both a left Noetherian ring and a right Noetherian ring.

The starting ring $\mathbb{Z}$ is Noetherian, since every ideal is principal. The other module fact we use is that a finitely generated module over a Noetherian ring on the relevant side is a Noetherian module: it is a quotient of a finite direct sum of regular modules, and the ascending chain condition passes to finite sums and quotients. For a finite sum this follows by induction, comparing the intersections of a chain with one summand and its images in the quotient by that summand.

If $H◃K$ has finite index, choosing coset representatives shows that $\mathbb{Z}[K]$ is finite free as a left and as a right $\mathbb{Z}[H]$-module. By finite-module extension preserves Noetherianity, every one-sided ideal of $\mathbb{Z}[K]$ is finitely generated already as a $\mathbb{Z}[H]$-module, and those same generators generate it as a $\mathbb{Z}[K]$-ideal. Thus either chain condition passes through a finite factor.

If $K/H$ is infinite cyclic, choose $t\in K$ lifting its generator. Every element of $K$ has a unique form $h{t}^n$, and conjugation by $t$ gives an automorphism $\sigma$ of $\mathbb{Z}[H]$. HenceFor the polynomial subring $C=\mathbb{Z}[H][t;\sigma ]$, the condition $A+tA=A+At$ follows from invertibility of $\sigma$, so the proved noncommutative Hilbert basis theorem makes $C$ Noetherian on both sides. A right ideal $J$ of the skew Laurent polynomial ring is generated by $J\cap C$: for any $f\in J$, choose $m$ large enough that $f{t}^m\in C$, and recover $f=(f{t}^m)t^{-m}$. Finite generators of $J\cap C$ over $C$ therefore generate $J$ over the Laurent ring. Left multiplication by a large power of $t$ proves the left-handed statement. These are the Laurent-extension facts used in the induction.

Every factor of the subnormal series is covered by one of these two cases. ThereforeThe argument proves the more general result that group rings of poly-(cyclic or finite) groups are Noetherian over any coefficient ring satisfying both chain conditions.

Since $M$ is a finitely generated module, choose a surjection from a finite free module, $F_0\twoheadrightarrow M$. A finite free module over a left Noetherian ring is a Noetherian module, so the kernel $K_1$ is finitely generated. Choose a finite free module $F_1$ surjecting onto $K_1$, then compose its map with $K_1\hookrightarrow F_0$. Its kernel is again finitely generated. Repeating constructs an exact sequencewith all $F_i$ finite free modules, hence finitely generated projective modules. Thus the requested projective resolution exists, without asserting that it has finite length.

Assume now that $M\ne 0$ and ${\mathrm{pd}}_{R}M=n<\infty$. We may arrange a length-$n$ projective resolution with all terms finite projective. Here is why the preceding construction gives one. Set $K_0=M$, with short exact sequences $0\to K_{j+1}\to F_j\to K_j\to 0$. Applying the Ext functor, whose positive degrees vanish on a projective first argument, gives by repeated dimension shiftingfor every left module $N$. The final equality follows from the existence of any length-$n$ projective resolution. Vanishing of these degree-one groups means that every short exact sequence ending in $K_n$ splits: its extension class is zero. In particular a free surjection onto $K_n$ splits, so $K_n$ is a projective module. It is finitely generated by the Noetherian kernel construction. Use $P_n=K_n$ and $P_j=F_j$ for $j<n$ to obtainFor a left finite projective module $P$, put $P^{\ast }={\mathrm{Hom}}_R(P,R)$, a right module under $(fr)(p)=f(p)r$. It is a right finite projective module, and its natural evaluation mapis an isomorphism. These claims hold for a finite free module by coordinates, and pass to direct summands, proving them for every finite projective module without any right-Noetherian hypothesis.

Suppose $n\ge 1$ and ${\mathrm{Ext}}_R^n(M,R)=0$. Computing the Ext functor from this resolution shows that its last dual mapis surjective. Since $P_n^{\ast }$ is projective as a right module, this map has a right-module section $h:P_n^{\ast }\to P_{n-1}^{\ast }$, with $d_n^{\ast }h=1$. Dualizing gives $h^{\ast }{d}_n^{\ast \ast }=1$. Under the natural double-dual identifications, $d_n^{\ast \ast }$ is $d_n$, so this gives a left-module retraction of $d_n$.

Therefore $P_{n-1}=d_n(P_n)\oplus Q$ for a finite projective module $Q$. The next differential has kernel $d_n(P_n)$, so restricting it to $Q$ gives an exact resolution with $Q$ replacing $P_{n-1}$ and the initial $P_n$ removed. This has length $n-1$, contradicting the definition of projective dimension. For $n=0$, the nonzero $M$ is a finite projective module and a summand of some $R^d$. A nonzero vector in that summand has a nonzero coordinate, so one coordinate projection restricts to a nonzero map $M\to R$. Thus the degree-zero Hom functor is nonzero as well. We have provedThis is top Ext detects finite projective dimension over a left Noetherian ring. The nonzero-module convention matters when $n=0$: if one assigns projective dimension zero to the zero module, then ${\mathrm{Hom}}_R(0,R)=0$ is an exception. With the convention ${\mathrm{pd}}_{R}0=-\infty$, it is outside the asserted finite nonnegative case.

For an example of infinite projective dimension, take the dual numbers $R=k[\epsilon ]/({\epsilon }^2)$ over a field $k$, and $M=k=R/(\epsilon )$. The periodic resolution over dual numbers isIt is exact because both the kernel and image of multiplication by $\epsilon$ are the ideal $(\epsilon )$, and that is also the kernel of $R\to k$. Applying ${\mathrm{Hom}}_R(-,k)$ gives a copy of $k$ in every degree and zero differentials, since $\epsilon$ acts as zero on $k$. ConsequentlyA finite projective resolution would make these groups zero above its length, so

A uniform module is a nonzero module in which any two nonzero submodules intersect nontrivially. An essential extension $V\subseteq E$ means that $V$ is an essential submodule of $E$: every nonzero submodule of $E$ meets $V$. Since $R$ is a left Noetherian ring, the finitely generated $M$ is a Noetherian module.

Every nonzero Noetherian module contains a uniform submodule. Otherwise split a nonuniform submodule into two disjoint nonzero submodules, retain one for further splitting and set the other aside. If no uniform submodule is reached, the pieces set aside form an infinite direct sum, whose partial sums contradict the ascending chain condition. Now successively choose uniform submodules disjoint from the current direct sum. The process again must stop by Noetherianity, giving $V={⨁}_{i=1}^r{V}_i\subseteq M$. At termination no nonzero submodule is disjoint from $V$, soThis proves the finite uniform decomposition of a Noetherian module.

An injective hull $E(M)$ is an injective module containing $M$ essentially. It is equivalently a minimal injective extension; we use the general existence theorem for injective hulls. To prove uniqueness, let $E_1,E_2$ be two hulls of $M$. Injectivity of $E_2$ extends the identity on $M$ to $f:E_1\to E_2$. Its kernel is disjoint from $M$ and therefore zero by essentiality. Its image is injective, so its inclusion splits in $E_2$. The complementary summand is disjoint from $M$, hence zero. Thus $f$ is an isomorphism fixing $M$.

A finite direct sum of injective modules is injective, and a finite direct sum of essential inclusions is essential. For the latter assertion, start with a nonzero vector and process its coordinates in turn: if a coordinate lies outside its specified essential submodule, multiply by a ring element taking that coordinate nontrivially into that submodule. Already processed coordinates stay inside their submodules; the chosen coordinate keeps the vector nonzero. Essentiality is also transitive. Consequently both $E(M)$ and ${⨁}_{i}E(V_i)$ are injective hulls of $V$, so uniqueness givesThe essential extension of a uniform module is uniform, since intersecting any two nonzero submodules with the original uniform module already gives a nonzero common intersection.

Write $E=E(V_i)$ and $S={\mathrm{End}}_R(E)$. If $f\in S$ is injective, its image is isomorphic to the injective module $E$, hence is a direct summand. Uniformity leaves no nonzero complementary summand, so $f$ is an automorphism. Therefore the nonunits are exactly the noninjective endomorphisms. Let $J$ be their set. Two such maps have nonzero kernels, whose intersection is nonzero by uniformity; their sum kills this intersection. For a product, composition on the left preserves the original kernel. Composition on the right either has a noninjective right factor, or that factor is an automorphism and transports the original nonzero kernel. Thus $J$ is a proper two-sided ideal. Every element outside $J$ is a unit, so every nonzero residue class is invertible and every proper left or right ideal is contained in $J$. This proves the local endomorphism ring of a uniform injective module assertion:For the whole hull, let $E_i=E(V_i)$. Its endomorphism ring is the ring of finite block matriceswith multiplication given by summing compositions. The diagonal entries have the local rings just proved. Off-diagonal entries are homomorphism groups, not necessarily zero; in particular this ring is not generally the product of its diagonal rings. Repeated isomorphic summands give matrix blocks over their common local endomorphism ring.

For the left regular module of $A_1(\mathbb{C})$, the Bernstein filtration has graded ring $\mathbb{C}[x,\xi ]$, so finite leading-term cancellation proves that $A_1$ is left and right Noetherian and a domain. Let $Q$ be its classical division ring of fractions. It is an essential extension of $A_1$: a nonzero fraction $q=s^{-1}a$ satisfies $sq=a\ne 0$ in $A_1$.

To prove that $Q$ is injective, use the Baer criterion. For a homomorphism $f:I\to Q$ on a nonzero left ideal, fix $a\in I\setminus \{0\}$. For any nonzero $x\in I$, take common left multiples $sx=ta$ with $s,t\ne 0$. Then $sf(x)=tf(a)$, so in the division ringThus right multiplication by $a^{-1}f(a)$ extends $f$ from $I$ to $A_1$. The zero ideal is harmless. Hence $Q$ is injective and $E({}_{A_1}{A}_1)=Q$. For any $A_1$-linear endomorphism $F$ of $Q$, clearing the left denominator in $q$ gives $F(q)=qF(1)$. These are all right multiplications, and their composition reverses the multiplying elements. ThereforeThe opposite ring in the injective hull of an Ore domain computation is essential for left-module conventions.

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.

Choose a surjection $P_0=A^{r_0}\twoheadrightarrow M$. Since $A$ is a Noetherian ring, a finite free module is a Noetherian module, and its kernel $K_1$ is a finitely generated module. Choose a finite free surjection $P_1\twoheadrightarrow K_1$ and repeat. This constructs a resolution by finitely generated free moduleswhose terms are, in particular, finitely generated projective modules. It may be infinite; the request is finite generation of every term, not finite length.

Now suppose the nonzero finitely generated module $M$ has finite projective dimension $n$. For $n\ge 1$, its $n$th syzygy $K_n$ in this free resolution is projective. Indeed, dimension shifting along the successive kernel sequences givesfor every module $N$. The vanishing follows by computing the latter Ext functor from any length-$n$ projective resolution. Vanishing of ${\mathrm{Ext}}^1(K_n,-)$ makes every short exact sequence ending in $K_n$ split: apply the long exact sequence of Ext groups to a free surjection onto $K_n$ to lift its identity. Thus $K_n$ is a projective module. It is finitely generated by the preceding construction. Truncating giveswith every term finite projective.

Apply the Hom functor ${\mathrm{Hom}}_A(-,A)$ and write $P_i^{\ast }={\mathrm{Hom}}_A(P_i,A)$. At the top degree,If this group were zero, $d_n^{\ast }$ would be surjective. The dual module of a finite projective module is finite projective, so this surjection has a section $s:P_n^{\ast }\to P_{n-1}^{\ast }$. Finite projective modules identify with their double dual modules: the evaluation map is an isomorphism for a finite free module and passes to its direct summands. Dualizing $d_n^{\ast }s=\mathrm{id}$ therefore gives $s^{\ast }{d}_n={\mathrm{id}}_{P_n}$. Hence $d_n$ splits, and $P_{n-1}/d_n(P_n)$ is projective. Replacing the last two terms by this quotient shortens the projective resolution to length $n-1$, contradicting minimality. This proves top Ext detects finite projective dimension:For $n=0$, $M$ is a nonzero finite projective module and embeds as a summand of $A^r$. Some coordinate projection restricts nontrivially to $M$, proving ${\mathrm{Hom}}_A(M,A)\ne 0$ as required.

For infinite projective dimension, take the dual-number algebra $A=k[\epsilon ]/({\epsilon }^2)$ and $M=k=A/(\epsilon )$. Multiplication by $\epsilon$ has image and kernel both equal to $(\epsilon )$, giving the periodic resolution over dual numbersApplying ${\mathrm{Hom}}_A(-,k)$ makes every differential zero, so ${\mathrm{Ext}}_A^i(k,k)\cong k$ for every $i\ge 0$. A finite projective resolution would force all sufficiently high Ext functors to vanish. Thus ${\mathrm{pd}}_{k[\epsilon ]/({\epsilon }^2)}k=\infty$.

The finite-generation hypothesis is essential if the middle sentence's “a module” is interpreted as an unrestricted new module. A concrete vanishing top Ext without finite generation occurs for $A=k[[t]]$ and $K=A[1/t]=k((t))$. This Noetherian ring has the exact free presentationInjectivity of $d$ follows by examining the largest index of a finitely supported input, since $t$ is a non-zero-divisor. For exactness in the middle, use the relations to move a finite sum to its largest index $N$; its remaining coefficient is ${\sum }_{i=0}^N{a}_i{t}^{N-i}$, which is zero precisely when its image in $K$ is zero. The map onto $K$ is surjective because every Laurent series has a lower bound on its exponents.

Applying ${\mathrm{Hom}}_A(-,A)$ gives on products the map $(b_i)\mapsto (b_i-t{b}_{i+1})$. It is surjective: for any $(c_i)$, set $b_i={\sum }_{j\ge 0}{t}^j{c}_{i+j}$. These are well-defined formal power series, because each coefficient receives only finitely many contributions, and $b_i-t{b}_{i+1}=c_i$. Consequently ${\mathrm{Ext}}_A^1(K,A)=0$. But ${\mathrm{Hom}}_A(K,A)=0$: the image of every $t^{-i}$ must lie in ${\bigcap }_{r\ge 0}{t}^{r}A=0$, since it is divisible by every power of $t$. A nonzero projective module is a summand of a free module and therefore has a nonzero coordinate map to $A$. Thus $K$ is not projective, while its displayed resolution has length one. In particular,The nonvanishing conclusion proved above retains the finitely generated $M$ introduced at the start of the question.

Put $K=\mathrm{Frac}(A)$ and let $B$ be the integral closure of $A$ in $L$. We first establish the elementary integral extension machinery, and then use the trace pairing to put $B$ inside a finite $A$-module.

An integral element $x$ satisfies a monic polynomial over the base ring. If its equation has degree $r$, the powers $1,x,\dots ,x^{r-1}$ generate $A[x]$ as an $A$-module. More generally, adjoining finitely many integral elements gives a finitely generated module: reduce the exponent of each generator using its monic equation. Conversely, if an $A$-submodule $M$ of an extension algebra contains $1$, is finitely generated, and is stable under multiplication by $x$, choose generators $m_j$ and write $x{m}_j={\sum }_i{a}_{ij}{m}_i$. Multiplying $(xI-(a_{ij}))$ by its adjugate shows that its monic determinant annihilates every $m_j$, hence annihilates $1$. This finite-module criterion for integrality is the determinant trick.

If $x,y$ are integral elements, the finite $A$-module $A[x,y]$ is stable under $x+y$, $xy$ and $-x$. The criterion proves that these are integral elements. Thus the integral closure really is a subring. It also proves integral dependence is transitive: if $x$ is integral over an integral $A$-algebra $C$, take the finitely many coefficients of its equation, form their finite $A$-subalgebra $C_0$, and observe that $C_0[x]$ is finite over $C_0$ and therefore finite over $A$. The criterion applies to $x$. In particular, an algebra generated by finitely many integral elements is a finite integral extension, even though an arbitrary integral extension need not be finite.

Choose a $K$-basis $v_1,\dots ,v_d$ of $L$. Each basis element is algebraic over $K$. Ifchoose a nonzero $a_i\in A$ clearing all coefficient denominators. Then $e_i=a_i{v}_i$ satisfies a monic equation with coefficients $a_i^j{c}_{ij}\in A$. Hence all $e_i$ belong to $B$ and still form a $K$-basis. This is integral field basis by denominator clearing.

We next prove the trace of an integral element over a normal domain property. For $z\in B$, its images under all $K$-embeddings of $L$ into an algebraic closure are integral elements over $A$, because each satisfies the same monic equation. Their sum is integral by the subring property just proved. Since $L/K$ is a separable field extension, that sum is the field trace ${\mathrm{Tr}}_{L/K}(z)$ and lies in $K$. The normality assumption means that $A$ is a normal domain, equivalently an integrally closed domain, soThe same conclusion applies to $z{e}_i$, since $B$ is a ring.

The trace pairing of a finite separable field extension is nondegenerate. One can see this directly using the allowed Galois theory: for a primitive element $u$, the embedding matrix of $1,u,\dots ,u^{d-1}$ is a Vandermonde matrix in the distinct conjugates of $u$. Its determinant is nonzero, and the trace Gram matrix is its transpose times itself. Nondegeneracy is unchanged by a change of basis.

Let $M={\sum }_{i}A{e}_i$. Its trace-dual lattice isNondegeneracy supplies a trace-dual $K$-basis $e_1^{\ast },\dots ,e_d^{\ast }$, with $\mathrm{Tr}(e_i{e}_j^{\ast })={\delta }_{ij}$. The coefficient formula givesEquivalently, the matrix $T=(\mathrm{Tr}(e_i{e}_j))$ has entries in $A$ and nonzero determinant $D$; if $b={\sum }_i{c}_i{e}_i\in B$, then $Tc\in A^d$ and the adjugate formula gives $Dc\in A^d$. Thus $B\subseteq D^{-1}M$ as well. Both descriptions exhibit a finite free ambient $A$-module.

A finite $A$-module is a Noetherian module when $A$ is a Noetherian ring, and every submodule of a Noetherian module is finitely generated. Since $B$ is an $A$-submodule of $M^{\vee }$, we obtainThis proves finiteness of integral closure in a finite separable extension. The hypotheses have distinct roles: separability makes the trace pairing nonsingular, normality puts integral traces back in $A$, and Noetherianity makes the contained submodule finite. Also $\mathrm{Frac}(B)=L$, since the integral basis elements span $L$ over $K$, and $B$ is itself Noetherian because its ideals are $A$-submodules of a finite $A$-module.