Past exam of the mathematics course of the University of Cambridge / 2021 / iii

Suppose every submodule of $M$ is finitely generated. For an ascending chain $N_1\subseteq N_2\subseteq \cdots$, its union $N$ is a submodule. Finitely many generators of $N$ all lie in one $N_r$, so $N=N_r$ and the chain stabilizes. Thus the ascending chain condition holds.

If the ascending chain condition holds, any nonempty collection of submodules has a maximal member: otherwise, starting from one member and repeatedly choosing a strictly larger one constructs a nonstationary ascending chain.

Finally, assume the maximal condition. Among the finitely generated submodules of a given $N\subseteq M$, choose a maximal one $F$. If $F\ne N$, then $F+Rx$ is a larger finitely generated submodule for any $x\in N\setminus F$, a contradiction. Hence every $N$ is finitely generated. These are the three equivalent characterizations of a Noetherian module.

Let $n_1,\dots ,n_r$ generate $N$, and let the images of $m_1,\dots ,m_s$ generate $M/N$. For any $m\in M$, subtracting a suitable linear combination of the $m_j$ leaves an element of $N$, which is a combination of the $n_i$. Thus

If $M$ is Noetherian, every submodule of $N$ is a submodule of $M$, and submodules of $M/N$ correspond to submodules of $M$ containing $N$; hence both are Noetherian.

Conversely, suppose $N$ and $M/N$ are Noetherian. For any $P\subseteq M$, the intersection $P\cap N$ is finitely generated and the image $(P+N)/N$ is finitely generated. Lifting generators of the image and applying part i toshows that $P$ is finitely generated. Thus $M$ is Noetherian.

The exact sequenceand part ii show that a direct sum of two modules is Noetherian exactly when both summands are. Induction proves the assertion for every finite direct sum.

Each $R/I_j$ is Noetherian as an $R$-module because its $R$-submodules are precisely its ideals as a Noetherian ring. The diagonal maphas kernel ${\bigcap }_j{I}_j=0$. Hence $R$ is an $R$-submodule of a finite direct sum of Noetherian modules, so $R$ is a Noetherian $R$-module; equivalently, $R$ is a Noetherian ring.

DefineThis makes $M[X]$ an $R[X]$-module.

Let $N\subseteq M[X]$ be an $R[X]$-submodule. For $n\ge 0$, let $L_n\subseteq M$ consist of zero and the leading coefficients of elements of $N$ of degree $n$. Each $L_n$ is an $R$-submodule, and multiplication by $X$ givesBecause $M$ is Noetherian, this chain stabilizes at some $n_0$, and each $L_n$ for $n\le n_0$ is finitely generated. Choose finitely many polynomials $f_{nj}\in N$ of degree $n$ whose leading coefficients generate $L_n$.

For $f\in N$ of degree $d$, if $d\le n_0$, subtract an $R$-linear combination of the $f_{dj}$ to lower its degree. If $d>n_0$, use $L_d=L_{n_0}$ and subtract a combination of $X^{d-n_0}{f}_{n_{0}j}$. Induction on degree expresses $f$ in terms of the finite collection $\{f_{nj}\}$. Therefore every submodule $N$ is finitely generated andThis is the module form of the Hilbert basis theorem.

The localization of a ring $S^{-1}R$ consists of pairs $(r,s)\in R\times S$ modulofor some $u\in S$. Addition and multiplication are

The localization is the zero ring exactly when $1/1=0/1$, which means that some $u\in S$ is zero. Thus $0\in S$. This holds exactly when $S$ contains a nilpotent element: if $s^n=0$ and $s\in S$, multiplicative closure gives $0\in S$, while zero itself is nilpotent.

Let $e=(1,0)$. Since $e$ becomes invertible in $S^{-1}(A\times B)$ and $e(0,b)=0$, every $(0,b)$ becomes zero. The mapis therefore a well-defined ring isomorphism, with inverse $a\mapsto (a,0)/1$.

For a subring $\mathbb{Z}\subseteq R\subseteq \mathbb{Q}$, letIf $a/b\in R$ is in lowest terms, Bézout gives integers $u,v$ with $ua+vb=1$, soEvery prime divisor $p$ of $b$ then satisfies $1/p=(b/p)/b\in R$. Conversely, every rational whose denominator uses only primes in $\mathcal{P}$ lies in $R$. Hence all intermediate rings are exactlywhere $S$ is generated by an arbitrary set $\mathcal{P}$ of primes. For example, $1/3=1-2/3$, so $\mathbb{Z}[2/3]=\mathbb{Z}[1/3]=\{3^n:n\ge 0{\}}^{-1}\mathbb{Z}$.

For $x\in L$, choose a monic equation over $K$,Choose nonzero $a\in R$ with $a{c}_i\in R$ for every $i$. Multiplying by $a^n$ shows that $b=ax$ satisfieswhose coefficients lie in $R$. Thus $b\in \overline{R}$ and

Extension and contraction give the prime ideal correspondence for localizationExplicitly, $\mathfrak{p}$ maps to $S^{-1}\mathfrak{p}$, while $\mathfrak{q}\subseteq S^{-1}R$ maps to $\{r:r/1\in \mathfrak{q}\}$. Primality follows by clearing denominators, and the two operations are inverse because an ideal in a localization contains $r/s$ exactly when it contains $r/1$.

The weak Hilbert Nullstellensatz says that if a field is a finitely generated algebra over $k$, then it is a finite algebraic extension of $k$. Since every maximal ideal is prime, the nilradical $\mathcal{N}(R)$ is contained in the Jacobson radical $J(R)$.

Conversely, let $f$ be nonnilpotent. Then $R_f\ne 0$, and it is a finitely generated $k$-algebra, so it has a maximal ideal $\mathfrak{n}$. Its contraction $\mathfrak{p}$ to $R$ avoids $f$. Moreover,is a finitely generated $k$-domain inside a finite algebraic extension of $k$, hence is itself a field. Thus $\mathfrak{p}$ is maximal and does not contain $f$. Therefore $f\notin J(R)$, proving

The tensor product $M{\otimes }_{R}N$ is the free abelian group on symbols $(m,n)$ modulo additivity in each variable and the balancing relation $(rm,n)=(m,rn)$. It has the universal property that balanced bilinear maps $M\times N\to P$ correspond uniquely to homomorphisms $M{\otimes }_{R}N\to P$.

The mapis well defined and surjective. Its inverse sends $m+IM$ to $(1+I)\otimes m$; elements of $IM$ map to zero because $(1+I)\otimes am=(a+I)\otimes m=0$ for $a\in I$. Hence

Let $k=R/\mathfrak{m}$. Associativity and part i giveIf $M{\otimes }_{R}N=0$, the tensor product on the right is zero. Two nonzero vector spaces over a field have nonzero tensor product, so one factor vanishes. Nakayama lemma then gives $M=0$ or $N=0$.

An $R$-module $F$ is flat when $-{\otimes }_{R}F$ preserves injections, equivalently all finite exact sequences. Since $-{\otimes }_{R}R$ is naturally the identity functor, $R$ is flat. A free module is a direct sum of copies of $R$, and tensor products commute with direct sums, so every free module is flat.

As an $R$-module,is free. Therefore the polynomial algebra $R[X]$ is a flat $R$-algebra.

An element $x\in B$ is integral over $A$ when it satisfies a monic polynomial with coefficients in $A$. Equivalently, $A[x]$ is a finite $A$-module.

If $x,y$ are integral over $A$, then $A[x,y]$ is finite over $A$: it is generated by finitely many monomials $x^i{y}^j$. Multiplication by $x+y$, $x-y$, or $xy$ is an endomorphism of this finite module, so the determinant trick gives a monic annihilating polynomial. Hence the integral elements form a subring $C$ containing $A$.

The integral closure of $A$ in $B$ is this subring $C$. The ring $A$ is integrally closed in $B$ when $C=A$, and $B$ is integral over $A$ when $C=B$.

For $x\in A$, considerThis is monic, and every $\tau \in G$ permutes its factors, so all coefficients lie in $A^G$. Since $f_x(x)=0$, every $x\in A$ is integral over $A^G$. Thus

The Going-up theorem states: if $B$ is integral over $A$, ${\mathfrak{p}}_1\subseteq {\mathfrak{p}}_2$ are primes of $A$, and ${\mathfrak{q}}_1$ lies over ${\mathfrak{p}}_1$, then some prime ${\mathfrak{q}}_2\supseteq {\mathfrak{q}}_1$ lies over ${\mathfrak{p}}_2$.

Pass to $A/{\mathfrak{p}}_1\subseteq B/{\mathfrak{q}}_1$, which remains integral, and localize at the complement of ${\mathfrak{p}}_2/{\mathfrak{p}}_1$. The lying-over theorem supplies a prime of the localized upper ring over the maximal ideal of the localized lower ring. Contracting it to $B/{\mathfrak{q}}_1$, and then pulling it back to $B$, gives the required ${\mathfrak{q}}_2$.

If $x\in A$ is a unit in $B$, then $x^{-1}$ is integral over $A$:Multiplication by $x^{n-1}$ expresses $x^{-1}$ as an element of $A$, so $x$ is a unit in $A$.

Use the characterization $a\in J(A)$ exactly when $1-ra$ is a unit for every $r\in A$. If $a\in J(B)\cap A$, then $1-ra$ is a unit in $B$ and hence in $A$, proving $a\in J(A)$. Conversely, if $a\in J(A)$, every maximal ideal $\mathfrak{n}$ of $B$ contracts under the integral extension to a maximal ideal of $A$, which contains $a$. Thus every $\mathfrak{n}$ contains $a$, so $a\in J(B)$. Therefore

Put $S=A\setminus \mathfrak{p}$. Primes of $B_{\mathfrak{p}}=S^{-1}B$ correspond to primes $\mathfrak{q}$ of $B$ whose contractions are contained in $\mathfrak{p}$. By going up, each such $\mathfrak{q}$ is contained in a prime lying over $\mathfrak{p}$, and this prime is uniquely ${\mathfrak{p}}^{\prime }$. Hence ${\mathfrak{p}}^{\prime }{B}_{\mathfrak{p}}$ is the unique maximal ideal of $B_{\mathfrak{p}}$.

Localizing this already local ring at its unique maximal ideal changes nothing, soFinally, localization preserves integral extensions; therefore $B_{{\mathfrak{p}}^{\prime }}=B_{\mathfrak{p}}$ is integral over $A_{\mathfrak{p}}$.

The positive-degree part $A_{+}={⨁}_{n>0}{A}_n$ is an ideal and $A/A_{+}\simeq A_0$, so $A_0$ is Noetherian. Since $A$ is Noetherian, $A_{+}$ has finitely many homogeneous generators $x_1,\dots ,x_s$. Induction on degree shows that every positive-degree homogeneous element is a polynomial in the $x_i$ over $A_0$. Thus

For an additive length function $\lambda$ finite on the graded pieces, define the Poincare series of a graded moduleThe Hilbert-Serre theorem states thatfor a Laurent polynomial $f(t)\in \mathbb{Z}[t,t^{-1}]$.

Induct on $s$. For the last generator $x_s$ of degree $k_s$, multiplication gives an exact sequence whose kernel is the $x_s$-torsion and whose cokernel is $M/x_{s}M$. Additivity of $\lambda$ yieldsBoth modules on the right are finite graded modules over the algebra generated by $x_1,\dots ,x_{s-1}$. The induction hypothesis gives the asserted denominator. The case $s=0$ is a finite Laurent polynomial because $M$ is finitely generated over $A_0$.

When every $k_i=1$, cancel common factors to writewhere $d$ is the pole order at $t=1$. Sincethe coefficient of $t^n$ in $P(M,t)$ is, for all sufficiently large $n$, a fixed linear combination of shifted binomial polynomials. It is therefore a polynomial in $n$ of degree exactly $d-1$ unless $P=0$, in which case its degree is $-1$.

The Krull dimension $\dim R$ is the supremum of lengths $n$ of strict chainsof prime ideals. The transcendence degree ${\mathrm{trdeg}}_{k}L$ is the cardinality of a transcendence basis of $L/k$.

By Noether normalization lemma, there are algebraically independent $y_1,\dots ,y_d\in A$ such that $A$ is finite, hence integral, over $k[y_1,\dots ,y_d]$. Their fraction field has transcendence degree $d$, and $L$ is algebraic over it, so ${\mathrm{trdeg}}_{k}L=d$. Going up and incomparability show that an integral extension preserves Krull dimension, while a polynomial ring in $d$ variables over a field has dimension $d$. Hence

A chain of length $s$ in $R/\mathfrak{p}$ lifts to a chainin $R$. Since $R$ is a domain and $\mathfrak{p}\ne 0$, prepending $(0)⊊\mathfrak{p}$ gives a chain of length $s+1$. Therefore $s+1\le r$ and

Going up lifts every prime chain in $A$ to one in $B$, so $\dim B\ge \dim A$. Conversely, contracting a strict chain of primes of $B$ gives a chain in $A$, and the incomparability theorem for integral extensions ensures that no strict inclusion contracts to equality. Thus $\dim A\ge \dim B$, and

The height $\mathrm{ht}\mathfrak{p}$ is the supremum of lengths of strict chains of primes ending at $\mathfrak{p}$. The Krull principal ideal theorem says that in a Noetherian ring every prime minimal over a principal proper ideal has height at most one.

Induct on $n$. If $\mathfrak{p}$ is minimal over $I=(x_1,\dots ,x_n)$, choose a prime $\mathfrak{q}\subseteq \mathfrak{p}$ minimal over $(x_1,\dots ,x_{n-1})$ and localize appropriately. In $R/\mathfrak{q}$, the prime $\mathfrak{p}/\mathfrak{q}$ is minimal over the principal ideal generated by $x_n$, so its relative height is at most one. Induction gives $\mathrm{ht}\mathfrak{q}\le n-1$, hence

Write $\mathfrak{p}=(x)$. The principal ideal theorem and the hypothesis give $\mathrm{ht}\mathfrak{p}=1$. We use the standard principal prime in a Noetherian local ring lemma: a principal prime of positive height in a Noetherian local ring is generated by a nonzerodivisor and is the unique minimal prime above zero. The lemma follows by applying the associated-prime description of zero divisors and Nakayama's lemma to $\mathrm{ann}(x)$; if a nonzero annihilator or another minimal component existed, the principal prime would have height zero.

Here is the needed argument directly. For every $n\ge 1$,Indeed, if $y{x}^n=0$ with $y\notin \mathfrak{p}$, then $y$ is a unit in $R_{\mathfrak{p}}$, so $x^n=0$ there. The maximal ideal $\mathfrak{p}{R}_{\mathfrak{p}}=(x)$ would then be nilpotent, making $R_{\mathfrak{p}}$ zero-dimensional, contrary to $\mathrm{ht}\mathfrak{p}\ge 1$.

Now suppose $xy=0$. The displayed containment gives $y=x{y}_1$, then $x^2{y}_1=0$ gives $y_1=x{y}_2$, and inductivelyBecause $R$ is Noetherian, the ascending chain $\mathrm{ann}(x)\subseteq \mathrm{ann}(x^2)\subseteq \cdots$ stabilizes, say at $n=N$. Then $y_N\in \mathrm{ann}(x^{N+1})=\mathrm{ann}(x^N)$, and hence $y=x^N{y}_N=0$. Thus $x$ is a nonzerodivisor.

Let $N={\bigcap }_{n\ge 0}{x}^{n}R$. It is a finitely generated ideal. If $y=xz\in N$, then $xz\in x^{n+1}R$ for every $n$; cancellation of the nonzerodivisor $x$ gives $z\in x^{n}R$ for every $n$. Hence $N=xN$, and Nakayama lemma gives $N=0$ because $x$ lies in the maximal ideal.

Every nonzero element consequently has a finite $x$-adic order. If nonzero $a,b$ satisfied $ab=0$, write $a=x^r{a}^{\prime }$ and $b=x^s{b}^{\prime }$ with $a^{\prime },b^{\prime }\notin (x)$. Cancelling $x^{r+s}$ gives $a^{\prime }{b}^{\prime }=0$, contradicting primality of $(x)$. Equivalently, $(0)$ is prime, so

The formal power series ring $P=k[[X]]$ is Noetherian, so the finite product $R=P\times P$ is Noetherian. Its maximal ideals areso there are exactly two.

The idealis principal and prime because $R/\mathfrak{p}\simeq k$. The prime chainshows that it has height one, and no longer chain exists because $\dim P=1$. Yetwith both factors nonzero, so $R$ is not a domain. This shows why locality is essential in part i.

A stable $I$-filtration of $M$ is a descending sequence $M=M_0\supseteq M_1\supseteq \cdots$ with $I{M}_n\subseteq M_{n+1}$ for all $n$ and equality for all sufficiently large $n$. The Rees ring and associated Rees module areThe filtration condition makes multiplication by $I^r{t}^r$ send $M_n{t}^n$ into $M_{n+r}{t}^{n+r}$, so $M^{\ast }$ is a graded $R^{\ast }$-module.

If the filtration is stable from degree $\ell$, then $M^{\ast }$ is generated over $R^{\ast }$ by finite generating sets for $M_0,\dots ,M_{\ell }$. Conversely, let homogeneous elements of degrees at most $\ell$ generate $M^{\ast }$. In every degree $n>\ell$, each expression for an element of $M_n$ uses a positive-degree coefficient from $R^{\ast }$, so $M_n=I{M}_{n-1}$. Hence finite generation is equivalent to stability.

For $M^{\prime }\subseteq M$, takeThen $M^{\ast }$ is a graded submodule of the finite Rees module $⨁I^{n}M$. Since $R$ is Noetherian and $I$ is finitely generated, $R^{\ast }$ is Noetherian; hence $M^{\ast }$ is finite and the filtration is stable. Therefore, for some $\ell$ and all $n\ge \ell$,This is the Artin-Rees lemma.

Let $N={\bigcap }_{n\ge 0}{I}^{n}M$. Apply Artin--Rees to $N\subseteq M$. Since $N\subseteq I^{n}M$ for every $n$, stability givesfor all sufficiently large $n$, and in particular $N=IN$. The module $N$ is finitely generated because $M$ is Noetherian. The determinant trick applied to a finite generating set of $N=IN$ produces $a\in I$ withTaking $x=-a\in I$ yieldswhich is Krull intersection theorem.

The space $\mathrm{End}(V)$ becomes a Lie algebra under the commutatorA Lie subalgebra $L$ is abelian when $[L,L]=0$; nilpotent when its Lower central series of a Lie algebra ${\gamma }_{r+1}=[L,{\gamma }_r]$ reaches zero; and soluble when its derived series of a Lie algebra $L^{(r+1)}=[L^{(r)},L^{(r)}]$ reaches zero.

If $H⊊L$ and $L$ is nilpotent, choose the least $r$ with ${\gamma }_r(L)\subseteq H$. Then ${\gamma }_{r-1}(L)⊈H$, whileThus the normalizer of a Lie subalgebra strictly contains $H$. If $H$ is maximal proper, its normalizer must be all of $L$, so $H$ is an ideal. Solubility is insufficient: in the two-dimensional affine Lie algebra $L=⟨x,y⟩$ with $[x,y]=y$, the maximal subalgebra $\mathbb{C}x$ is not an ideal.

A derivation of a Lie algebra is a linear map $D$ satisfyingIt is inner when $D=\mathrm{ad}z$ for some $z\in L$.

Every nonzero finite-dimensional nilpotent Lie algebra has an outer derivation of a nilpotent Lie algebra. Choose a codimension-one maximal subalgebra $K$; it is an ideal by the result above, and write $L=K\oplus \mathbb{C}z$. The centralizer $C_L(K)$ is nonzero because it contains $Z(L)$. Let $n$ be largest such thatand choose $z_0\in C_L(K)\setminus {\gamma }_{n+1}(L)$. DefineBecause $K$ is an ideal and $z_0$ centralizes $K$, the derivation identity holds on $K\times K$ and on $z\times K$, hence everywhere. If $D=\mathrm{ad}x$, then $D(K)=0$ would put $x$ in $C_L(K)\subseteq {\gamma }_n(L)$, socontrary to the choice of $z_0$. Thus $D$ is outer.

The analogous assertion fails for soluble algebras. In the affine example, a derivation hasand equals $\mathrm{ad}(dx-by)$. Thus every derivation is inner although $L$ is nonzero and soluble.

A Lie algebra is semisimple when its soluble radical is zero. Its Killing form isThe radical of this invariant symmetric form is an ideal. The solvability result behind the Cartan criterion for semisimplicity applied to that ideal shows that it is soluble; semisimplicity therefore makes it zero. Hence $B_L$ is nondegenerate.

An abelian subalgebra $H$ is a Cartan subalgebra when its elements are semisimple and it is maximal toral, equivalently when $C_L(H)=H$. An arbitrary abelian subalgebra need not lie in one: in ${\mathfrak{sl}}_2$, the line spanned by the nilpotent matrix $e$ is abelian, whereas every element of a Cartan subalgebra is semisimple.

Let $H=C_L(h_0)$ for a regular $h_0\in H$, as allowed. Generalized eigenspaces of $\mathrm{ad}{h}_0$ giveIf $x\in H$ is orthogonal to $H$, invariance givesThus $x$ is orthogonal to all of $L$, and nondegeneracy gives $x=0$. Therefore $B_L{|}_H$ is nondegenerate.

The commuting semisimple maps $\{\mathrm{ad}h:h\in H\}$ can be simultaneously diagonalized. ConsequentlyHere $L_0=C_L(H)=H$, the nonzero weights $\Phi \subseteq H^{\ast }$ are the roots, and the Jacobi identity gives $[L_{\alpha },L_{\beta }]\subseteq L_{\alpha +\beta }$.

A finite-dimensional Lie algebra representation is a homomorphism $\rho :L\to \mathfrak{gl}(V)$. It is irreducible when $V$ has no invariant subspaces other than $0$ and $V$.

The algebra ${\mathfrak{sl}}_2$ has basis $e,f,h$ withFor every $n\ge 0$, its $(n+1)$-dimensional irreducible module $V(n)$ has basis $v_0,\dots ,v_n$ and actionwith out-of-range vectors zero. Any nonzero invariant subspace contains a weight vector; repeated application of $e$ reaches $v_0$, and repeated application of $f$ then generates the whole module, proving irreducibility. The adjoint module of ${\mathfrak{sl}}_2$ is $V(2)$, so every ideal is an invariant subspace and ${\mathfrak{sl}}_2$ is simple.

For a root $\alpha$, nondegeneracy of the Killing pairing between $L_{\alpha }$ and $L_{-\alpha }$ allows choices $e_{\alpha },f_{\alpha }$ withAfter rescaling, $(e_{\alpha },f_{\alpha },h_{\alpha })$ obey the ${\mathfrak{sl}}_2$ relations, giving a copy of ${\mathfrak{sl}}_2$ in $U_{\alpha }$.

Restrict the adjoint representation of $L$ to this copy. Finite-dimensional ${\mathfrak{sl}}_2$ theory shows that the $\alpha$-string through zero has one nontrivial summand $V(2)$, with weight spaces $L_{-\alpha }$, $\mathbb{C}{h}_{\alpha }$, and $L_{\alpha }$. Any additional vector in $L_{\alpha }$ would generate another weight-two summand and another independent zero-weight coroot, contradicting nondegeneracy of the root--coroot pairing on $H$. Hence

For $\beta \notin \mathbb{Z}\alpha$, the spaceis stable under $e_{\alpha },f_{\alpha },h_{\alpha }$. Adjacent raising and lowering maps are nonzero until the endpoints, and every root space is one-dimensional, so $W$ is a simple ${\mathfrak{sl}}_2$-module of highest weight $p+q$. Its lowest and highest $h_{\alpha }$-weights giveand therefore $\beta (h_{\alpha })=q-p$. Since $[L_{\alpha },L_{-\alpha }]=\mathbb{C}{h}_{\alpha }$, every $x$ in this bracket line satisfies