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Use generators $x,D$ with $Dx-xD=1$. The ordered monomials $x^i{D}^j$ form a basis: construct the Weyl algebra as the differential Ore extension of $\mathbb{C}[x]$ with $Df=fD+f^{\prime }$, which has unique ordered polynomial expressions. Giving both generators degree one produces the Bernstein filtration, whose associated graded ring is $\mathbb{C}[x,\xi ]$. If $a,b\ne 0$, their nonzero leading symbols have nonzero product in this polynomial integral domain. Thus $ab$ has nonzero leading symbol and cannot vanish. Hence $A_1(\mathbb{C})$ has no nontrivial zero divisors.

Here is the left-handed Ore theorem. Let $S$ be a multiplicative set of regular elements of a ring, containing $1$ and excluding $0$. There is an injective Ore localization whose elements are left fractions $s^{-1}a$ if and only ifThis is the left Ore condition. For a noncommutative domain, taking $S=A\setminus \{0\}$ gives a division ring exactly when this condition holds. With zero-divisor denominators a further left reversibility condition is needed; the regular-denominator statement is the one used here.

Necessity follows by writing $a{s}^{-1}=t^{-1}b$ in the quotient and multiplying by $t$ and $s$, giving $ta=bs$ in the embedded original ring. For a construction proving sufficiency, use the left Ore fraction construction: pairs $(s,a)$ represent $s^{-1}a$, and $(s,a)$ and $(t,b)$ are equivalent whenfor suitable $u,v\in A$. Reflexivity and symmetry are immediate. For transitivity, combine $us=vt=w$ and $u^{\prime }t=v^{\prime }q=w^{\prime }$ by a common left multiple $hw=k{w}^{\prime }\in S$. Then $(hv-k{u}^{\prime })t=0$, so regularity gives $hv=k{u}^{\prime }$, which combines the numerator equalities and proves transitivity.

Addition uses the common denominator $w$, with numerator $ua+vb$. To multiply $s^{-1}a$ by $t^{-1}b$, choose $v\in S,c\in A$ with $va=ct$; setThe Ore condition supplies a common refinement whenever a representative or an Ore relation is changed, and regularity cancels the shared denominator, proving independence of choices. The same refinements verify the ring axioms; the formulas are simply addition and multiplication after clearing left denominators. The map $a\mapsto (1,a)$ is injective, since equivalence to zero would give $ua=0$ with $u\in S$. Each $s\in S$ becomes invertible. A homomorphism inverting $S$ extends uniquely by $(s,a)\mapsto f(s{)}^{-1}f(a)$, giving the universal property. For all nonzero denominators in a noncommutative domain, a nonzero fraction has inverse $(s^{-1}a{)}^{-1}=a^{-1}s$, so the quotient is a division ring. This is the requested proof sketch, with the fraction equivalence and multiplication made explicit.

It remains to check the left Ore condition for $A_1(\mathbb{C})$. Its graded polynomial ring is Noetherian by the Hilbert basis theorem. Lifting homogeneous generators of a graded left ideal and successively cancelling leading symbols proves that the filtered algebra is a left Noetherian ring: each subtraction lowers the nonnegative degree, so it terminates.

In any left Noetherian noncommutative domain, nonzero principal left ideals intersect. Otherwise $Ab\cap Aa=0$ with $a,b\ne 0$ would makean infinite direct sum of nonzero left ideals. To verify directness, reduce a finite relation modulo $Aa$ to make its first summand zero, then cancel $a$ on the right and repeat. Its increasing finite partial sums contradict the ascending chain condition. A nonzero element of $Aa\cap As$ therefore has the form $ta=bs\ne 0$, with $t\ne 0$, proving the required condition. Consequently $A_1(\mathbb{C})$ has a classical ring of left quotients, and it is a division ring.

First, a maximal left ideal $L$ is closed. Its norm closure is again a left ideal, since multiplication is continuous. If that closure were all of $A$, some $\ell \in L$ would satisfy $\Vert 1-\ell \Vert <1$ and would be invertible by the Neumann series. But a proper left ideal cannot contain an invertible element: ${\ell }^{-1}\ell =1$ would put $1$ in $L$. Maximality therefore forces $\overline{L}=L$.

Consequently $X=A/L$ is a nonzero quotient Banach space. It is a simple module for left multiplication by $A$, because the inverse image of any submodule under $A\to A/L$ is a left ideal containing $L$, and hence is either $L$ or $A$. Since $La\subseteq L$, right multiplication defines a well-defined bounded module endomorphismIndeed, changing $u$ by an element of $L$ changes $ua$ by an element of $L$, and taking the infimum over representatives proves the quotient norm bound. Right multiplication commutes with the left action by associativity.

Choose $\lambda$ in the nonempty spectrum of an element of $T$ in the Banach algebra $\mathcal{B}(X)$. Both the kernel and range of $T-\lambda I$ are submodules. If this module endomorphism were nonzero, simplicity would force its kernel to be zero and its range to be all of $X$. It would be a bounded bijection of Banach spaces, with bounded inverse by the bounded inverse theorem, contradicting the choice of $\lambda$. Thus $T=\lambda I$. Applying this identity to $1+L$ yields $a-\lambda 1\in L$. If also $a-\mu 1\in L$, then $(\lambda -\mu )1\in L$; since $1\notin L$, $\lambda =\mu$. ThereforeThis is the scalar right action on a maximal-left-ideal quotient; it does not assume that $L$ is a two-sided ideal or that $A/L$ is an algebra.

Now consider the center of an associative algebra $Z$. For each $u\in A$, the map $z\mapsto zu-uz$ is a continuous linear map, so its kernel is closed. Their intersection $Z$ is therefore closed. The identity belongs to $Z$, as do sums and scalar multiples of its elements. If $z,w\in Z$, then for every $u\in A$,Thus $zw\in Z$. Also $zw=wz$, since $z$ commutes with every element of $A$, including $w$. Hence the center is a closed commutative unital subalgebra, and is a Banach algebra in the inherited norm.

For every $z\in Z$, the inclusion $Lz=zL\subseteq L$ allows the first argument to apply. Let $\chi (z)$ be the unique scalar for which $z-\chi (z)1\in L$. Equivalently, the induced right multiplication on $A/L$ is $T_z=\chi (z)I$. Sincethe map $\chi :Z\to \mathbb{C}$ is a complex-linear unital character of an algebra. It is surjective because $\chi (\alpha 1)=\alpha$, and its kernel is precisely $L\cap Z$. ConsequentlyIndeed any ideal properly containing the kernel would have a nonzero image in the field $\mathbb{C}$ and thus would be all of $Z$. The algebra character is also continuous: $|\chi (z)|=\Vert \chi (z)I\Vert =\Vert T_z\Vert \le \Vert z\Vert$, although continuity is not needed for maximality.

Take rings and modules to be unital. Let $J_{\ell }$ denote the intersection of the maximal left ideals. For each nonzero vector $v$ of a simple module $M$ on the left, the map $R\to M$, $r\mapsto rv$, is onto and its kernel is a maximal left ideal. Consequently $J_{\ell }$ annihilates every simple module on the left. Conversely, annihilating $R/L$ forces an element to belong to the maximal left ideal $L$. Thuswhich shows that $J_{\ell }$ is a two-sided ideal.

For $j\in J_{\ell }$ and $r\in R$, the left ideal $R(1-rj)$ cannot be proper: a containing maximal left ideal would also contain $rj$ and hence $1$. Therefore $u(1-rj)=1$ for some $u\in R$. Here $u=1+urj$ also differs from $1$ by an element of $J_{\ell }$, so it too has a left inverse $v$. Multiplying the first equality by $v$ gives $1-rj=v$; hence $(1-rj)u=1$. We have proved that $1-rj$ is a unit, not just left invertible.

The identityis verified by multiplying on either side, using $(1-ab{)}^{-1}(1-ab)=(1-ab)(1-ab{)}^{-1}=1$. It implies that $1-jr$ is a unit as well. If $j$ were outside a maximal right ideal $I$, then $I+jR=R$, so $1-jr\in I$ for some $r$. A proper right ideal cannot contain a unit, giving a contradiction. Thus $J_{\ell }$ belongs to every maximal right ideal. Applying the same argument to the opposite ring proves the reverse inclusion. This proves the Jacobson radical is independent of handedness:For the zero ring the intersections are interpreted as the whole ring, so the equality remains valid.

If a two-sided ideal $N$ satisfies $N^m=0$, then for $x\in N$ and $r\in R$, $(rx{)}^m=0$. The finite geometric series ${\sum }_{i=0}^{m-1}(rx{)}^i$ inverts $1-rx$. The unit criterion for the Jacobson radical, also proved above by the maximal-ideal argument, gives $N\subseteq J(R)$. This assertion concerns a nilpotent ideal; arbitrary nilpotent elements of a noncommutative ring need not lie in its Jacobson radical.

For the formal power series ring over the p-adic integers, a series $f={\sum }_{i\ge 0}{a}_i{t}^i$ is a unit exactly when $a_0$ is a unit of ${\mathbb{Z}}_p$. Necessity follows by comparing constant coefficients. If $a_0$ is invertible, construct the inverse recursively by $b_0=a_0^{-1}$ and $b_m=-a_0^{-1}{\sum }_{i=1}^m{a}_i{b}_{m-i}$. Therefore the nonunits are exactly the series whose constant coefficient lies in $p{\mathbb{Z}}_p$. They form the unique maximal ideal $(p,t)$, and the quotient is ${\mathbb{F}}_p$. Hence

Give $U(\mathfrak{g})$ its degree filtration. The Poincaré-Birkhoff-Witt theorem identifies its associated graded ring with ${\mathrm{Sym}}_k\mathfrak{g}$, a polynomial ring in ${\dim }_k\mathfrak{g}$ variables. The Hilbert basis theorem makes this graded algebra Noetherian on both sides.

Here is the required transfer, including why a finite list of symbols gives actual generators. For a right ideal $I$ of a nonnegatively filtered ring $A$, its induced graded ideal $\mathrm{gr}I$ is a homogeneous right ideal of $\mathrm{gr}A$. Choose finite homogeneous generators and lift them to $u_1,\dots ,u_m\in I$. For $u\in I$ of degree $d$, express its leading symbol as a graded right linear combination of these symbols, using homogeneous coefficients of complementary degrees. Lift the coefficients to $A$ and subtract the corresponding ${\sum }_i{u}_i{a}_i$. The remainder has degree strictly below $d$. Induction, terminating because degrees are nonnegative, proves $I={\sum }_i{u}_{i}A$. The identical argument with coefficients on the left treats left ideals.

This ascending filtered-graded transfer of Noetherianity proves that $U(\mathfrak{g})$ is right and left Noetherian. A quotient ring inherits either property: pull an ascending ideal chain back to the original ring, or take images of ideal generators. Part (a) therefore gives $R$ right and left Noetherian. Alternatively, under the broader almost commutative algebra convention, its commutative finitely generated associated graded ring is directly a quotient of a finite-variable polynomial ring, and the same leading-symbol argument proves the conclusion.

Let $K=k(z)$, where $z$ is an indeterminate, and form the right Artinian triangular ring that is not left NoetherianMultiplication uses the natural $k$–$K$ bimodule structure on the upper-right entry. With the two diagonal idempotents $e_1,e_2$, the right regular module is $e_{1}T\oplus e_{2}T$. The off-diagonal submodule of $e_{1}T$ is one-dimensional over the bottom-right field $K$, hence simple. Its quotient is one-dimensional over the top-left field $k$, hence simple. Also $e_{2}T$ is a simple right module over $K$. Thus $T_T$ has a composition series of length three, and in particular $T$ is right Artinian.

On the left, multiplication of an off-diagonal matrix givesEvery $k$-subspace $U\subseteq K$ therefore yields a left ideal of off-diagonal matrices with entries in $U$. The subspaces $U_m={\mathrm{span}}_k\{1,z,\dots ,z^m\}$ are strictly increasing because $z$ is transcendental. Their associated left ideals violate the ascending chain condition. Hence $T$ is not left Noetherian, even though it is right Artinian and right Noetherian.

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.

Assume $R$ is a nonzero unital ring and all its modules are unital. Order the proper left ideals by inclusion. A chain has an upper bound given by its union, which is still a proper left ideal: if it contained $1$, one member of the chain would contain $1$ and be all of $R$. The family is nonempty because it contains $0$. The Zorn lemma gives a maximal proper left ideal $L$. The nonzero quotient $R/L$ has no proper nonzero submodules, because its submodules correspond to left ideals between $L$ and $R$. Thus $R/L$ is a simple left module. The nonzero-ring qualification is necessary: the zero ring has no nonzero unital simple module.

Define the Jacobson radical $J(R)$ as the intersection of all maximal left ideals. It is equivalently the intersection of the ideals ${\mathrm{Ann}}_R(S)$ over all simple modules, where $\mathrm{Ann}$ is the annihilator of a module. To verify this equivalence, for a nonzero vector $v$ in a simple module $S$, the surjection $R\to S$, $r\mapsto rv$, has a maximal left-ideal kernel. Thus $x\in J(R)$ implies $xv=0$ for every $v$ in every simple module. Conversely, if $x$ annihilates all simple modules, it annihilates $1+L$ in every $R/L$, so $x\in L$. An annihilator of a module is a two-sided ideal; this description therefore proves that $J(R)$ is a two-sided ideal.

For the forward implication of the unit criterion for the Jacobson radical, first let $y\in J(R)$. If $R(1-y)$ were a proper left ideal, it would lie in a maximal one $L$. Both $y$ and $1-y$ would belong to $L$, an impossibility. Hence $c(1-y)=1$ for some $c\in R$. A left inverse alone does not yet establish that $1-y$ is a unit in a general noncommutative ring. However, $c=1+cy$, with $cy\in J(R)$, so the same argument gives $dc=1$ for some $d$. Multiplying $c(1-y)=1$ by $d$ on the left gives $d=1-y$. Therefore $(1-y)c=dc=1$ as well. This proves that $1-y$ is a unit.

For $x\in J(R)$, its two-sided-ideal property gives $axb\in J(R)$ for all $a,b$, so $1-axb$ is a unit. Conversely, if $x\notin J(R)$, choose a maximal left ideal $L$ with $x\notin L$. Then $L+Rx=R$, so $l+ax=1$ for some $l\in L$, $a\in R$. Hence $1-ax=l\in L$ is not a unit: if it were invertible, multiplying it on the left by its inverse would put $1$ in $L$. Taking $b=1$ contradicts the given unit property. We have provedFinally let $R$ be commutative and $I$ an ideal. The set $S=1+I$ contains $1$, andso it is a multiplicative subset. In the localization of a module $I_S\subseteq R_S$, any element is $i/s$ with $i\in I$ and $s\in S$. For any $r/t\in R_S$,Both $ts$ and $ts-ri$ lie in $S$, since $ts\in 1+I$ and $ri\in I$. The fraction is consequently a unit in $R_S$. In a commutative ring, the product of any two coefficients is again a single coefficient, so the proved unit criterion for the Jacobson radical applies to all expressions $1-a(i/s)b$. ThusThis proves localization away from one plus an ideal lies in the Jacobson radical. If $I=R$, then $0\in S$, $R_S=0$, and the inclusion holds trivially.