Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 85 (Noetherian Algebras)

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

The variable $t$ is central. To prove the leading-coefficient proof for a central polynomial extension, let $I$ be any right ideal of $R[t]$ and setZero leading coefficients are allowed. Addition and right multiplication by constants show that $L_d$ is a right ideal. Multiplication by $t$ gives $L_d\subseteq L_{d+1}$. Since $R$ is a right Noetherian ring, this chain stabilizes, say at $L_N$, and each $L_d$ for $d\le N$ has finitely many generators $a_{dj}$. Choose $f_{dj}\in I$ of degree at most $d$ lifting these coefficients.

These finitely many polynomial lifts generate $I$ on the right. Indeed, for $f\in I$ of degree $m$, let $d=\min (m,N)$. Write its leading coefficient as ${\sum }_j{a}_{dj}{r}_j$. Thenbelongs to $I$ and has degree less than $m$, because $L_m=L_d$ when $m>N$. Induction on degree reduces $f$ to a right linear combination of the chosen lifts. This proves finite generation of every right ideal, and henceCentrality matters: it is what permits the displayed leading-coefficient cancellation without twisting the coefficients.

Use the degree-one convention implicit in this enveloping-algebra assertion: a degree-one almost commutative algebra has $F_{0}R=k$, a finite-dimensional $F_{1}R$ containing $1$ which generates $R$, and commutative associated graded ring. Taking the leading symbol of a commutator givesin particular $[F_{1}R,F_{1}R]\subseteq F_{1}R$. Thus $\mathfrak{g}=F_{1}R$, equipped with $[u,v]=uv-vu$, is a finite-dimensional Lie algebra. Bilinearity and antisymmetry are immediate, and expanding the three nested commutators proves the Jacobi identity.

Its inclusion into the associative algebra $R$ is a Lie algebra homomorphism. The universal enveloping algebra is $T(\mathfrak{g})$ modulo the relations $u\otimes v-v\otimes u-[u,v]$. Sending the tensor generators to their elements of $R$ kills these relations and gives an algebra homomorphism $U(\mathfrak{g})\to R$. It is surjective because $F_{1}R$ generates $R$. ThereforeThe vector $1\in F_{1}R$ is retained as a central Lie algebra generator and maps to the associative identity; its corresponding relation in the kernel is that generator minus $1_{U(\mathfrak{g})}$. One should not first replace $F_{1}R$ by $F_{1}R/k1$, since commutators of generators can be nonzero scalars.

Terminology is important here. Merely requiring a commutative finitely generated associated graded ring for an arbitrary weighted filtration is a broader definition of almost commutative algebra; it does not supply the finite-dimensional bracket-closed first step used in this argument. The degree-one assumptions make the requested enveloping presentation precise. The Noetherianity in part (b) also follows directly under the broader finite-generation convention.

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.

A right Ore localization at $S$ is a unital homomorphism $\iota :R\to Q$ such that every $\iota (s)$, $s\in S$, is a unit, every element of $Q$ is $\iota (r)\iota (s{)}^{-1}$, andEquivalently, the usual fraction construction has this zero criterion and is universal among homomorphisms that invert $S$. Injectivity is required only when the denominators are regular elements of a ring. If $0\in S$, every homomorphism inverting $S$ has zero target; this is the degenerate zero-ring localization. For a nonzero localization we require $0\notin S$.

The necessary and sufficient conditions are that $S$ be a right denominator set:

For necessity, write the element $\iota (s{)}^{-1}\iota (r)$ as $\iota (b)\iota (t{)}^{-1}$. Multiplying by $\iota (s)$ on the left and $\iota (t)$ on the right gives $\iota (rt-sb)=0$. The zero criterion supplies $u\in S$ with $(rt-sb)u=0$. Hence $r(tu)=s(bu)$, the right Ore condition in $R$, not merely an equality of images in $Q$. Also $sr=0$ makes $\iota (r)=0$ because $\iota (s)$ is invertible, and the zero criterion gives $rt=0$. This proves both necessities.

The sufficiency is the Ore theorem for denominator sets. Concretely, right fractions can be represented by $(r,s)$; a common-denominator equality has $su=tv\in S$ and $ru=r^{\prime }v$ for suitable $u,v\in R$. The two conditions make the induced fraction equivalence transitive and make addition and multiplication well defined; the zero fraction is exactly the kernel displayed above. The requested direction of the theorem was necessity, for which the complete argument has just been given.

For a right Noetherian domain, any two nonzero principal right ideals intersect. To see this without using the desired localization, suppose $bR\cap aR=0$ with $a,b\ne 0$. The sumis direct: reduce a finite relation modulo $aR$ to get its first summand zero, then cancel $a$ in the noncommutative domain and repeat. Every summand is nonzero, so the finite partial sums form a strictly ascending chain of right ideals, contradicting the ascending chain condition. A nonzero intersection $rR\cap sR$ thus provides $rt=sb\ne 0$, with $t\ne 0$. This is the right Ore condition for $S=R\setminus \{0\}$. Reversibility is automatic in a noncommutative domain, and the zero criterion makes the map injective. The localization is a division ring, since a nonzero $r{s}^{-1}$ has inverse $s{r}^{-1}$.

The Goldie theorem states that a semiprime ring has a semisimple Artinian classical right ring of quotients if and only if it has finite right uniform dimension and satisfies the ascending chain condition on right annihilators. A right Noetherian ring satisfies the annihilator condition because right annihilators are right ideals. Its right regular module has finite uniform dimension: an infinite direct sum of nonzero submodules would give a strictly ascending chain of finite partial sums. A prime ring is semiprime, so the theorem applies to a prime right Noetherian ring $R$.

To obtain a single matrix-ring factor, not just a finite product, prove that its quotient $Q$ is prime. A nonzero two-sided ideal $I\subseteq Q$ contains a nonzero fraction $r{s}^{-1}$ and hence the nonzero element $r=(r{s}^{-1})s\in I\cap R$. The same holds for any other nonzero ideal $K$. Since $R$ is prime, $(I\cap R)(K\cap R)\ne 0$, so $IK\ne 0$. Thus $Q$ is prime. The Artin–Wedderburn theorem gives a finite product of matrix rings over division rings; primeness excludes two different nonzero factors with zero product. This proves the prime classical quotient of a prime right Goldie ring:For the requested counterexample to Noetherianity, take $R=k[x_1,x_2,\dots ]$. It is a commutative integral domain and has its field of fractions as its classical right ring of quotients. But $(x_1)⊊(x_1,x_2)⊊\cdots$ is a strictly ascending ideal chain: evaluation at $x_1=\cdots =x_m=0$ shows $x_{m+1}$ is not in the preceding ideal. Therefore existence of a classical quotient does not imply right Noetherianity.

Use the descending chain condition on the finite intersections of maximal right ideals. It gives a minimal such intersection $K=M_1\cap \cdots \cap M_r$. Intersecting $K$ with any other maximal right ideal cannot decrease it, by minimality. Thus $K$ lies in every maximal right ideal, and $K=J(R)$. When $J(R)=0$, the diagonal mapis injective, and each summand on the right is a simple module.

For completeness, a submodule $N$ of a finite direct sum of simple modules is semisimple. Induct on the number of summands. For $S\oplus T$ with $S$ simple, $N\cap S$ is either $0$ or $S$. In the first case projection into $T$ identifies $N$ with a submodule of $T$, so induction applies. In the second case subtracting the $S$ coordinate shows $N=S\oplus (N\cap T)$, and induction again applies. This also shows finite module length. Apply it to the image of the diagonal map to obtainThis is the semisimplicity of a right Artinian ring with zero radical. It does not assume right Noetherianity in order to prove it.

The descending powers $J\supseteq J^2\supseteq \cdots$ stabilize because they are right ideals. Choose $m\ge 1$ with $J^m=J^{m+1}$ and put $I=J^m$. Then $I=J^{2m}=I^2$.

Suppose $I\ne 0$. Among the right ideals $K$ such that $KI\ne 0$, choose a minimal one; the set is nonempty since $RI=I\ne 0$. The right ideal $KI$ lies in $K$, and $(KI)I=K{I}^2=KI\ne 0$, so minimality forces $KI=K$. Choose $k\in K$ with $kI\ne 0$. Then $(kR)I=kI\ne 0$, and $kR\subseteq K$, so the same minimality gives $K=kR$.

Since $K=KI$ and $I\subseteq J$, its generator satisfies $k={\sum }_{\nu }k{r}_{\nu }{i}_{\nu }=kj$ for an element $j={\sum }_{\nu }{r}_{\nu }{i}_{\nu }\in J$. Thus $k(1-j)=0$. The unit criterion for the Jacobson radical makes $1-j$ invertible, implying $k=0$, contrary to $kI\ne 0$. Therefore $I=0$, andThe minimal-right-ideal proof of radical nilpotence is important here: one cannot apply the finite-generator form of Nakayama lemma directly to $J^m$ before proving it finitely generated. We made a suitable auxiliary right ideal cyclic before using the unit argument.

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.

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.

In a commutative ring, if $x^m=0$, then the principal ideal $(x)$ satisfies $(x{)}^m=0$. Thus a semiprime ring has no nonzero nilpotent elements. Conversely, if $I^m=0$ for an ideal $I$, then every $x\in I$ satisfies $x^m=0$; absence of nonzero nilpotent elements forces $I=0$. Hence commutative semiprimeness is exactly reducedness.

For a prime ideal $P$, let $S=R\setminus P$. Every ideal $I$ of $R_P=S^{-1}R$ is the extension of its contraction $I^c$: if $r/s\in I$, multiply by the unit $s/1$ to get $r/1\in I$, so $r\in I^c$. If $R$ is Noetherian, write $I^c=(r_1,\dots ,r_m)$; their images generate $I$. This proves that $R_P$ is Noetherian.

The extended ideal $P{R}_P$ is proper. Indeed $1=p/s$ would imply $u(s-p)=0$ for some $u\notin P$, and then $us=up\in P$, contradicting primeness and $u,s\notin P$. A fraction with numerator outside $P$ is invertible, with inverse $s/r$. A fraction with numerator in $P$ lies in the proper ideal $P{R}_P$ and cannot be invertible. Thus the nonunits are precisely $P{R}_P$, making it the unique maximal ideal. ThereforeTo prove reducedness is detected by prime localizations, let $x$ be a nilpotent element of $R$. Its image in every $R_Q$ is zero because these localizations are semiprime, hence reduced. If $x\ne 0$, its annihilator $\mathrm{Ann}(x)$ is a proper ideal; choose a maximal ideal $Q$ containing it. The zero criterion for localization says $x/1=0$ in $R_Q$ only if $sx=0$ for some $s\notin Q$, which would put $s$ in $\mathrm{Ann}(x)\subseteq Q$, a contradiction. Hence $x=0$, and $R$ is semiprime. This argument does not need Noetherianity.

Primeness is not detected in the same way. Take $R=k\times k$. Its two prime ideals are $P_1=0\times k$ and $P_2=k\times 0$, and its corresponding localizations are both isomorphic to the field $k$. To see these are all the prime ideals, the orthogonal idempotents $e_1,e_2$ have product zero, so a prime ideal contains one of them and is then the kernel of the other projection. Both localizations are prime rings, but $R$ is not prime: $e_1,e_2\ne 0$ and $e_1{e}_2=0$.

This part proves $(c)\Rightarrow (a)$. Write the Weyl algebra generators as $x_1,\dots ,x_n,D_1,\dots ,D_n$, with $[D_i,x_j]={\delta }_{ij}$ and the other generator commutators zero. The ordered monomials $x^{\alpha }{D}^{\beta }$ form a basis over any field: one may construct the algebra by successively adjoining the commuting derivations $D_i$ to $k[x_1,\dots ,x_n]$, with multiplication $D_{i}f=f{D}_i+\partial f/\partial x_i$, obtaining unique ordered forms. This also makes $A_n(k)$ infinite-dimensional for $n\ge 1$.

Assume $\mathrm{char}k=0$ and let $I$ be a nonzero two-sided ideal. Choose a nonzero element $a=\sum c_{\alpha \beta }{x}^{\alpha }{D}^{\beta }\in I$, and a nonzero term of maximal total degree, say $({\alpha }_0,{\beta }_0)$. The relations giveApply $(\mathrm{ad}D{)}^{{\alpha }_0}(\mathrm{ad}x{)}^{{\beta }_0}$ to $a$. Every lower-degree term vanishes. A term of the same degree survives only if all its exponents dominate $({\alpha }_0,{\beta }_0)$, which forces equality. Thus the resulting element of $I$ isThe factorials are nonzero in characteristic zero. This is a nonzero scalar, so $1\in I$ and $I=A_n(k)$. Therefore $A_n(k)$ is simple in characteristic zero. This scalar extraction by iterated Weyl commutators supplies the full simplicity proof.

This part proves $(a)\Rightarrow (b)$. If $M\ne 0$ were a finite-dimensional unital module, its action would give a homomorphismThe kernel is a two-sided ideal and is proper because the identity acts as the identity on $M$. A simple ring therefore makes the homomorphism injective. This is impossible: $A_n(k)$ has the infinitely many linearly independent ordered monomials established in part (a), whereas the endomorphism algebra has dimension $({\dim }_{k}M{)}^2$. Thus simplicity excludes every nonzero finite-dimensional module.

There is also a direct characteristic-zero check: the relation $[D_1,x_1]=1$ acts as a matrix identity. Taking its trace gives $0={\dim }_{k}M$ in $k$. In characteristic zero this forces $M=0$. This agrees with the Weyl algebra has no nonzero finite-dimensional modules criterion, but the injectivity argument proves the requested implication without presupposing characteristic zero.

For the contrapositive of $(b)\Rightarrow (c)$, suppose $\mathrm{char}k=p>0$. ConsiderLet $x_i$ act by multiplication and $D_i$ by formal differentiation. Differentiation preserves the defining ideal, since $D_j(x_i^{p}f)=x_i^p{D}_{j}f$ in characteristic $p$. The relations $[D_i,x_j]={\delta }_{ij}$ hold before quotienting and hence hold on $M$. Its basis consists of the monomials $x^{\alpha }$ with $0\le {\alpha }_i<p$, so ${\dim }_{k}M=p^n>0$. This is the explicit finite-dimensional module for the Weyl algebra in positive characteristic.

Therefore condition (b) fails in positive characteristic and implies characteristic zero. Combining this with the preceding two implications proves $(a)⟺(b)⟺(c)$. The same representation also has nonzero proper kernel, since the infinite-dimensional Weyl algebra cannot inject into a finite-dimensional endomorphism algebra, confirming its failure of simplicity in positive characteristic.

The Bernstein inequality for Weyl algebra modules requires characteristic zero: for a nonzero finitely generated left or right module $M$ over $A_n(k)$,Equivalently, its Bernstein growth dimension of a Weyl algebra module is at least $n$. Without the characteristic-zero hypothesis this is false, as the module in part (c) has dimension zero in the growth sense. The proof below works over any characteristic-zero field, not just an algebraically closed one.

For a left module, choose a nonzero finite-dimensional generating subspace $M_0$ and put $M_j=F_j{A}_n{M}_0$, where $F_j$ is the Bernstein filtration. Ordered monomials giveMultiplication defines the linear action mapWe prove this map injective by induction on $j$, which is the essential faithful finite-step action of a Weyl algebra argument. For $j=0$, an element is a scalar, and a scalar killing $M_0\ne 0$ is zero. Suppose $a\in F_j$ kills $M_j$. For any generator $z$ and $m\in M_{j-1}$, both $azm$ and $zam$ vanish because $zm\in M_j$ and $m\in M_j$. Thus $[a,z]\in F_{j-1}$ kills $M_{j-1}$, so induction gives $[a,z]=0$ for every generator. By the explicit commutator formulas in part (a), commuting with all $D_i$ forces all positive $x_i$ exponents of $a$ to vanish, and commuting with all $x_i$ forces all positive $D_i$ exponents to vanish. Characteristic zero is used to cancel the nonzero integer exponents. Hence $a$ is scalar and, because it kills $M_0$, is zero. This proves injectivity.

Taking dimensions yieldsThe left side is asymptotic to $j^{2n}/(2n)!$. Consequently ${\dim }_k{M}_{2j}\ge C{j}^n$ for all sufficiently large $j$ and some $C>0$. Taking logarithms and the limsup in the definition of Gelfand–Kirillov dimension of a module proves the claimed lower bound. For a good Bernstein filtration, the associated graded module is finite over $k[x_1,\dots ,x_n,{\xi }_1,\dots ,{\xi }_n]$; its cumulative Hilbert function is eventually polynomial, and the same comparison says its degree is at least $n$.

For right modules, use the opposite ring isomorphism $A_n(k)\cong A_n(k{)}^{\mathrm{op}}$ sending $x_i$ to $x_i$ and $D_i$ to $-D_i$. It respects the Bernstein filtration, so the identical estimate applies. The bound is sharp: the polynomial module $k[x_1,\dots ,x_n]$, generated by $1$, has $M_j$ equal to the polynomials of degree at most $j$, of dimension $\left(\genfrac{}{}{0ex}{}{j+n}{n}\right)$. Its Gelfand–Kirillov dimension of a module is exactly $n$.