Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 2 (Noetherian Algebras)

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.

Put $A=R[[X]]$, and let $J$ be any ideal of this formal power series ring. For $d\ge 0$ define an ideal of $R$ byMultiplication by $X$ shows $L_d\subseteq L_{d+1}$. Since $R$ is Noetherian, this chain stabilizes at some $d=N$, and each $L_d$ is finitely generated. For every $0\le d\le N$, choose finitely many series $f_{d,j}\in J\cap X^{d}A$ whose degree-$d$ coefficients generate $L_d$.

Given $f\in J$, cancel its degree-zero coefficient using the $f_{0,j}$, then its degree-one coefficient using the $f_{1,j}$, and so on. At a degree $d\ge N$, use $X^{d-N}{f}_{N,j}$. Each residual remains in $J$ and has zero coefficients below the next cancellation degree. For each of the finitely many chosen $f_{d,j}$, accumulate its multiplier into a formal power series $h_{d,j}$. The higher-degree increments of these multipliers converge coefficientwise. At every degree the cancellation procedure gives the exact coefficient identity, soThus the finitely many lifts generate $J$ as an actual ideal. No assumption that $J$ is closed has been used. This proves Noetherianity of formal power series rings:Iterating the argument proves the assertion for $R[[X_1,\dots ,X_s]]$, which is naturally the iterated formal power series ring.

For a surjective module map $f:M\to N$, define the adic completion of a module byTake $\hat{n}\in \hat{N}$ and choose representatives $n_r\in N$ for its components. Compatibility says $n_{r+1}-n_r\in I^{r}N$. Choose $m_1$ with $f(m_1)=n_1$. If $f(m_r)=n_r$, surjectivity gives $f(I^{r}M)=I^{r}N$, so choose ${\delta }_r\in I^{r}M$ with $f({\delta }_r)=n_{r+1}-n_r$ and put $m_{r+1}=m_r+{\delta }_r$. The classes of $m_r$ modulo $I^{r}M$ are compatible and map to the chosen components of $\hat{n}$. HenceThis surjectivity on adic completions does not require finite generation of $M$ or $N$. The compatibility of the lifts is the point; independent lifts of the finite-level classes would not suffice.

Since $R$ is Noetherian, write $I=(a_1,\dots ,a_s)$. Define a ring homomorphismModulo $I^r$, evaluating a series requires only its finitely many terms of total degree below $r$. These evaluations commute with the reduction maps, and define $\Phi$ without needing the original ring to be complete or separated.

To prove surjectivity, begin with $\hat{b}\in \hat{R}$. Choose $b_0\in R$ matching its component modulo $I$. After terms of total degree below $r$ have been chosen, their evaluation differs from $\hat{b}$ at level $R/I^{r+1}$ by a class in $I^r/I^{r+1}$. This class is an $R$-linear combination of the monomials $a^{\alpha }$ with $|\alpha |=r$. Choose the corresponding homogeneous degree-$r$ polynomial $p_r(X)$ as the next term. Then the series $b_0+{\sum }_{r\ge 1}{p}_r(X)$ evaluates to $\hat{b}$ at every level. This proves the power-series presentation of an adic completion, and thereforeA quotient of a Noetherian ring is Noetherian, since its ideals correspond to ideals containing the kernel. The cases $I=0$ and $I=R$ give $\hat{R}=R$ and $\hat{R}=0$, respectively, and satisfy the conclusion as well.

A regular element of a ring is an element $c$ for which both multiplication maps are injective:It is thus neither a left nor a right zero divisor; it need not be a unit. Here “ideals” means two-sided ideals. Put $A=A_n(\mathbb{C})$, and let $T_2(A)$ denote the specified upper triangular ring.

We first establish the additional facts about the Weyl algebra needed below. Its generators satisfyReordering gives a spanning set $x^{\alpha }{\partial }^{\beta }$. These monomials are independent. Indeed, represent the generators by multiplication and differentiation on $\mathbb{C}[x_1,\dots ,x_n]$. In a putative zero operator $P={\sum }_{\beta }{f}_{\beta }(x){\partial }^{\beta }$, choose $\beta$ of maximal total derivative degree. Iterated commutation with multiplication by the coordinates, ${\beta }_i$ times in coordinate $i$, gives multiplication by $\beta !f_{\beta }(x)$. All other derivative terms of that total degree disappear unless they have exactly the same multiindex, and lower-degree terms disappear. Thus $f_{\beta }=0$, and descending induction eliminates all coefficients. This proves the ordered monomial basis of a Weyl algebra and faithfulness of its polynomial action.

The Weyl algebra is a simple ring. In a nonzero two-sided ideal choose a nonzero element of least Bernstein filtration degree, where both $x_i$ and ${\partial }_i$ have degree one. Commuting it with any generator stays in the ideal and lowers that degree. Minimality therefore makes all these commutators zero. Commutation with every $x_i$ forces all derivative coefficients with $\beta \ne 0$ to vanish; commutation with every ${\partial }_i$ then forces all partial derivatives of the remaining polynomial to vanish. In characteristic zero the element is a nonzero scalar. The ideal consequently contains $1$. This proves simplicity of a Weyl algebra in characteristic zero. Its Jacobson radical is zero, since the radical is a proper two-sided ideal of a nonzero unital simple ring.

Use the two diagonal idempotents to isolate the entries of any ideal of $T_2(A)$. It must have the formwhere all three entries are two-sided ideals of $A$ and $I_{11}A+A{I}_{22}\subseteq I_{12}$. Conversely this condition makes the indicated set an ideal. Simplicity says that each $I_{ij}$ is $0$ or $A$, and either nonzero diagonal entry forces $I_{12}=A$. Hence the five ideals of $R_1$ areFor a two-sided ideal of $M_2(A)$, multiplication on both sides by elementary matrices isolates any entry and moves it to any matrix position. The entries obtained form one two-sided ideal $I$ of $A$, and the original ideal is $M_2(I)$. Thus the only ideals of $R_2$ are $0$ and $R_2$.

Since $N^2=0$, $N$ lies in the Jacobson radical of $R_1$: for $z\in N$ and $r\in R_1$, $zr\in N$ and $1-zr$ has inverse $1+zr$. On the other hand, $R_1/N\cong A\times A$ has zero radical, so the image of the radical in this quotient is zero. The elementary unit criterion for the Jacobson radical justifies both assertions. ThereforeThe second equality follows from simplicity of $R_2$.

We now construct the denominators used to describe regular matrices. A left Noetherian noncommutative domain satisfies the left Ore condition. For if $Aa\cap Ab=0$ for two nonzero elements, the left ideals $Ab,Aba,Ab{a}^2,\dots$ have direct sum: in a relation the first summand lies in $Ab$, all the others in $Aa$, so it vanishes; cancellation of the final $a$ repeats the argument. Their finite partial sums form a strictly ascending chain, contradicting left Noetherianity. Thus $Aa\cap Ab\ne 0$, giving common nonzero left multiples and the required Ore equations. The analogous argument with right ideals gives the right Ore condition under right Noetherianity. Both hypotheses for $A$ are allowed in the question.

The Ore localization therefore embeds $A$ in a division ring $Q$. Briefly, its elements are left fractions $a^{-1}b$, with nonzero $a$, and two fractions are identified after passing to a common left denominator. Common multiples supply addition and multiplication, while cancellation in the domain makes the embedding injective. A nonzero fraction has an inverse, obtained by interchanging its numerator and denominator in the appropriate Ore fraction calculation. The right fraction construction gives the same division ring, so finite lists have common left and common right denominators.

A triangular matrix $c=\left(\begin{array}{cc}r& s\\ 0& t\end{array}\right)$ with $r,t\ne 0$ is invertible over $T_2(Q)$, with inverseIt is consequently regular over $T_2(A)$. If $r=0$, then $c{E}_{11}=0$; if $t=0$, then $E_{22}c=0$. Hence

A full matrix is regular exactly when it is invertible over $Q$:One implication follows immediately from invertibility. Conversely, a singular matrix over a division ring has a nonzero right kernel column and a nonzero left kernel row, by row reduction. A common right denominator turns the column into a nonzero column over $A$, providing a right annihilator matrix over $A$. A common left denominator does the same for the row and gives a left annihilator. Thus a singular matrix is not regular. Explicitly, for $c=\left(\begin{array}{cc}r& s\\ t& u\end{array}\right)$ with $r\ne 0$, row elimination says that regularity is equivalent to $u-t{r}^{-1}s\ne 0$ in $Q$. If $r=0$, it is equivalent to both $s\ne 0$ and $t\ne 0$. A commutative determinant formula must not be substituted here.

Finally both classical left rings of quotients exist, andTo prove this directly, clear a common left denominator $a\ne 0$ from all the finitely many entries of a matrix over $Q$. It becomes $(a{I}_2{)}^{-1}b$ with $b$ in the original triangular or full ring and $a{I}_2$ regular. Every original regular element becomes a unit in the corresponding ring over $Q$, by the descriptions above. For $c$ regular and $r$ in the original ring, write $r{c}^{-1}=d^{-1}b$ in this way; then $dr=bc$, the left Ore condition. Since the denominators are regular, the additional annihilator condition for localization is automatic. This proves the matrix and triangular localization over an Ore domain assertions. In particular, the nonzero nilpotent radical of the triangular ring does not prevent its classical localization; that localization retains the triangular radical.

A precise version of the Hilbert-Serre theorem is as follows. Let $R={⨁}_{j\ge 0}{R}_j$ be a commutative graded ring, generated over an Artinian degree-zero ring $R_0$ by homogeneous elements $y_1,\dots ,y_s$ of positive degrees $d_1,\dots ,d_s$. Let $M$ be a finitely generated graded module, bounded below. Each component has finite module length over $R_0$. Its Poincare series of a graded module, or Hilbert series, isWhen $R_0$ is a field, these lengths are dimensions. The theorem assertsFor a nonnegatively graded module the numerator can be taken in $\mathbb{Z}[t]$. The finite-generation and finite-length hypotheses specify the meaning of the series; they are part of the theorem's statement.

We prove it by induction on $s$. Regard $M$ as a finite graded module over $S=R_0[Y_1,\dots ,Y_s]$, with $\mathrm{deg}{Y}_i=d_i$, acting through the specified generators of $R$. This polynomial ring is Noetherian by the Hilbert basis theorem and Question 1. When $s=0$, finitely many homogeneous generators over $R_0$ mean that only finitely many components are nonzero, so the series is a Laurent polynomial. For the induction step, put $y=Y_s$, $d=d_s$, $K=(0{:}_{M}y)$ and $C=M/yM$. Both $K$ and $C$ are finite graded modules; $y$ annihilates them, so they are modules over $R_0[Y_1,\dots ,Y_{s-1}]$. The finite generation of $K$ uses Noetherianity of $S$.

The exact sequence of graded modulesgives, by additivity of component lengths,Here the shift is defined so that $P_{M(-d)}=t^d{P}_M$. By induction both series on the right have denominator ${\prod }_{i<s}(1-t^{d_i})$. Dividing by $1-t^d$ proves the result. The annihilator term cannot be omitted unless multiplication by $y$ is injective.

In the standard grading $d_i=1$, the coefficients of $(1-t{)}^{-s}$ are $\left(\genfrac{}{}{0ex}{}{j+s-1}{s-1}\right)$. Multiplying by the finite numerator shows that the component dimensions agree with a polynomial for sufficiently large $j$. Cumulative dimensions have generating function $P_M(t)/(1-t)$ and are also eventually polynomial. This consequence supplies the dimension used for Weyl algebra modules.

Use the Bernstein filtration $B_m{A}_n$, in which both coordinate generators and differentiation generators have degree one. The ordered monomial basis of a Weyl algebra givesChoose a finite-dimensional generating subspace $M_0$ of a nonzero finitely generated left module $M$ and set $M_m=B_m{A}_n{M}_0$. This is a good filtration of a module: the symbols of the generators generate its associated graded module over the displayed polynomial ring. The Hilbert-Serre theorem makes ${\dim }_{\mathbb{C}}{M}_m$ an eventual polynomial. Define the Bernstein growth dimension of a Weyl algebra module byThe dimension is independent of the chosen finite generating subspace. Each of two finite generating sets is contained in a fixed filtration step for the other, giving $M_m\subseteq M_{m+c}^{\prime }$ and $M_m^{\prime }\subseteq M_{m+c^{\prime }}$. Such fixed shifts preserve the degree of growth. The same argument applies to any good Bernstein filtration. This dimension agrees with the Gelfand–Kirillov dimension of a module; one may set $d(0)=-\infty$.

Let the nonconstant element $D$ have Bernstein degree $d\ge 1$. Since ${\mathrm{gr}}_B{A}_n$ is a domain, the product of two nonzero symbols cannot vanish. HenceRight multiplication by $D$ is injective, using the domain property of $A_n$. Thus the quotient filtration on $M=A_n/A_{n}D$ has cumulative dimensionsThe leading terms of degree $2n$ cancel, and the leading surviving term isIt is nonzero, because $d\ge 1$. ThereforeEquivalently the Hilbert series of the associated graded quotient is $(1-t^d)/(1-t{)}^{2n}$. Here $d$ is Bernstein degree, not differential order: multiplication by $x_1$ has differential order zero but Bernstein degree one. Using the order filtration in this dimension argument would lose the finite-dimensional filtration pieces.

The base ring must be specified in the algebraic definition. For a commutative $k$-algebra $R$, work in ${\mathrm{End}}_k(R)$ and let $m_f$ mean multiplication by $f\in R$. Set $D_k^{-1}(R)=0$ and define recursivelyThese are the operators of order at most $i$. The order of a nonzero differential operator relative to a base ring is the least $i$ for which it belongs to $D_k^i(R)$; the zero operator can be assigned order $-\infty$. An operator of order zero commutes with all multiplications and is itself multiplication by $P(1)$, so $D_k^0(R)=R$. The union $D_k(R)={\bigcup }_{i\ge 0}{D}_k^i(R)$ is the ring of finite-order operators, with composition as multiplication. The identityand induction on the sum of orders give $D_k^p{D}_k^q\subseteq D_k^{p+q}$. Sums have order at most the maximum of their orders, proving that this union is a ring.

For an arbitrary commutative ring, the absolute version uses $k=\mathbb{Z}$, so the ambient endomorphisms are additive. For the polynomial ring in this question, the intended Weyl-algebra identification is the relative one with $k=\mathbb{C}$. Absolute operators would also include nonzero derivations of the coefficient field $\mathbb{C}$: a derivation on a rational function field in a transcendence basis over $\mathbb{Q}$ extends to its algebraic closure in characteristic zero, and hence to $\mathbb{C}$. Extending it coefficientwise to $\mathbb{C}[x]$ gives an order-one additive operator which is not complex-linear. Such an operator cannot belong to the complex Weyl algebra. This explains why the linearity convention matters.

Now take $R=\mathbb{C}[x_1,\dots ,x_n]$. Coordinate multiplication has order zero, and each partial derivative ${\partial }_i$ has order one because $[{\partial }_i,m_f]=m_{{\partial }_{i}f}$. They satisfy the defining Weyl algebra relations, giving an algebra homomorphism $A_n\to D_{\mathbb{C}}(R)$. Reordering spans $A_n$ by $x^{\alpha }{\partial }^{\beta }$. If an ordered expression acted as zero, commutation with the coordinates according to a maximal derivative multiindex would give multiplication by $\beta !f_{\beta }(x)=0$; removing the top coefficients successively proves that every coefficient is zero. Thus the homomorphism is injective, as in the ordered monomial basis of a Weyl algebra proof.

For surjectivity, let $P$ have order at most $m$, and put ${\delta }_{i}P=[P,m_{x_i}]$. The operations ${\delta }_i$ commute, since the coordinate multiplications commute. A product of more than $m$ of them kills $P$. Define polynomial coefficientsCommuting $P$ through $m_{x^{\alpha }}$ repeatedly gives the finite identityIt follows by induction on $|\alpha |$ from $P{m}_{x_i}=m_{x_i}P+{\delta }_{i}P$ and Pascal's identity. Terms with $|\beta |>m$ vanish. Sincefor $\beta \le \alpha$, we obtain $P(x^{\alpha })={\sum }_{|\beta |\le m}{a}_{\beta }(x){\partial }^{\beta }(x^{\alpha })$. The monomials are a complex basis of $R$, so complex linearity yields equality of the operators on all of $R$. HenceFor a nonzero ordered expression ${\sum }_{\beta }{a}_{\beta }(x){\partial }^{\beta }$, its order is exactly ${\max }_{a_{\beta }\ne 0}|\beta |$. The upper bound follows from composition; the lower bound follows by taking that many coordinate commutators to extract a nonzero multiplication operator. This describes the order filtration of a Weyl algebra:

The associated graded ring is thereforeThe symbol of ${\partial }_i$ is ${\xi }_i$. In a product of ordered operators, moving derivatives past coefficients introduces terms of lower differential order. Thus top-order symbols multiply as ordinary polynomials, and the ordered basis proves there are no further relations. In particular $[D^p,D^q]\subseteq D^{p+q-1}$, with $D^{-1}=0$: the top-order symbols of the two products agree and cancel. Unlike the Bernstein filtration, the order filtration's degree-zero part is the entire polynomial ring and its pieces are infinite-dimensional over $\mathbb{C}$.

Form the Rees ring of a filtered algebraThe central element $h$ is a non-zero-divisor. This ring is generated by $h$, the $x_i$, and $y_i=h{\partial }_i$, withIts quotient modulo $h$ is the associated graded ring just computed. Explicitly the degree-$m$ component of $h\mathcal{A}$ is $D^{m-1}{h}^m$, so the degree-$m$ quotient is $D^m/D^{m-1}$. The quotient modulo $h-1$ recovers $A_n$.

Since $\mathcal{A}/h\mathcal{A}$ is commutative, any two lifts $F,G\in \mathcal{A}$ of symbols have $[F,G]\in h\mathcal{A}$. DefineDivision by $h$ is unique because $h$ is a non-zero-divisor. If $F$ is replaced by $F+hU$, the quotient changes by $[U,G]$, which is zero modulo $h$; changing $G$ works the same way. Thus the bracket is well defined. Antisymmetry and bilinearity follow from the commutator. The identity $[F,GH]=[F,G]H+G[F,H]$ proves the Leibniz rule. For the Jacobi identity, lift each inner bracket by $[G,H]/h$; the cyclic sum of the outer expressions is the commutator Jacobi identity divided by $h^2$, hence zero. These verifications prove that the graded polynomial ring is a Poisson algebra. This is the Poisson bracket from a central deformation parameter construction.

The generators give $\{{\xi }_i,x_j\}={\delta }_{ij}$ and all coordinate-coordinate and symbol-symbol brackets zero. The Leibniz rule therefore determines the answer on every polynomial:The sign here follows the chosen convention $[P,Q]=PQ-QP$: in particular $\{{\xi }_i,x_j\}={\delta }_{ij}$, rather than $\{x_i,{\xi }_j\}={\delta }_{ij}$. For homogeneous symbols the bracket has order degree $p+q-1$, and can equivalently be written $\{{\sigma }_p(P),{\sigma }_q(Q)\}={\sigma }_{p+q-1}([P,Q])$.