Past exam of the mathematics course of the University of Cambridge / 2023 / iii

The assertion is false. The integers $\mathbb{Z}$ form a Noetherian ring, since every ideal is principal, but this ring is not an Artinian ring: the strictly descending chainnever stabilizes.

The assertion is true. This is the Artinian commutative ring is Noetherian theorem. One proof uses the nilpotent nilradical $N$ of an Artinian ring $A$. The quotient $A/N$ is a finite product of fields. Each quotient $N^j/N^{j+1}$ is an Artinian module over the semisimple ring $A/N$, hence has finite length and is Noetherian. The finite filtrationthen makes $A$ a Noetherian module over itself, which is exactly the ascending chain condition on its ideals.

The assertion is false. The module over a ring $\mathbb{Z}$ over itself is Noetherian, because its submodules are the principal ideals $n\mathbb{Z}$, but the descending chainshows that it is not an Artinian module.

The assertion is false. For a prime number $p$, the Prüfer p-group $C_{p^{\infty }}$ is an Artinian $\mathbb{Z}$-module: every proper subgroup is a finite cyclic group, so no infinite strictly descending chain of subgroups exists. It is not Noetherian because its cyclic subgroups form the strict ascending chain

Let $\mathfrak{m}$ be the maximal ideal and $k=A/\mathfrak{m}$ the finite residue field. The maximal ideal of an Artinian local ring is nilpotent, so ${\mathfrak{m}}^r=0$ for some $r$. Every quotientis both an Artinian $A$-module and a vector space over $k$. An Artinian vector space is finite-dimensional, hence each quotient is a finite set. The finite filtrationtherefore proves that the underlying set of $A$ is finite. This is the Finiteness criterion for an Artinian local ring.

Suppose for a contradiction that $m>n$. Compose the given injective module homomorphism with the standard injection $B^n\hookrightarrow B^m$ that appends $m-n$ zero coordinates. This gives an injective endomorphism $T$ of the finite free module $B^m$ whose matrix has a zero final row.

Its characteristic polynomial has zero constant term, so the Cayley-Hamilton theorem givesInjectivity lets us cancel $T$. Repeating this argument eventually gives the identity endomorphism equal to zero. That would imply $B^m=0$, contrary to $B\ne 0$. Hence $m\le n$. This proves the rank inequality for an injection of finite free modules.

A ring extension $A\subseteq B$ is an integral extension when every $b\in B$ is an integral element over $A$: there is a monic polynomialwith all $a_i\in A$.

A ring extension $A\subseteq B$ is finite when $B$ is a finitely generated module over $A$. Every finite extension is integral by the determinant trick.

The prime ideal correspondence for localization identifies primes of $B_{\mathfrak{p}}$ with primes $\mathfrak{q}$ of $B$ disjoint from $A\setminus \mathfrak{p}$. Passing to the quotient by $\mathfrak{p}{B}_{\mathfrak{p}}$ retains exactly those containing $\mathfrak{p}{B}_{\mathfrak{p}}$. Thus the image is the fiber of the map on spectra:

The claim fails for a general extension. If $k$ is an infinite field, all the infinitely many maximal ideals $(x-a)$ of $k[x]$ contract to $(0)$ in $k$.

It still fails for an integral extension. Take a finite field $k={\mathbb{F}}_q$ andEvery $b\in B$ satisfies the monic equation $b^q-b=0$, so $B$ is integral over the diagonal copy of $k$. The coordinate kernels are infinitely many distinct maximal ideals, all lying over $(0)$.

The claim is true for a module-finite ring extension. The primes above $\mathfrak{p}$ correspond to the primes of the fiber ringThis is a finite-dimensional algebra over the residue field $\kappa (\mathfrak{p})$, hence an Artinian ring, and an Artinian ring has only finitely many prime ideals.

Form the finite-dimensional $k$-algebraBecause a finite extension is integral, the Lying-over theorem supplies a prime of $B$ above $\ker f$; after localization and extension of the residue field to $k$, this shows $C\ne 0$. Therefore $C$ has at least one maximal ideal.

As an Artinian ring, $C$ has only finitely many maximal ideals. For each such ideal $\mathfrak{n}$, the quotient $C/\mathfrak{n}$ is a finite field extension of the algebraically closed field $k$, so it equals $k$. Consequently the quotient maps $C\to k$ are in bijection with the required extensions $g:B\to k$. The set of extensions is therefore finite and nonempty.

Zariski lemma says that a field which is a finitely generated algebra over a field $k$ is a finite algebraic extension of $k$. The Strong Hilbert Nullstellensatz says that, for an ideal $I\subseteq k[T_1,\dots ,T_n]$ over an algebraically closed field,

Let the unique point of $V(\mathfrak{a})$ be $x=(x_1,\dots ,x_n)$ and letThe Nullstellensatz gives $\sqrt{\mathfrak{a}}=\mathfrak{m}$, so $\mathfrak{a}\subseteq \mathfrak{m}$. Each of the finitely many generators $u_i=T_i-x_i$ of $\mathfrak{m}$ has some power $u_i^{e_i}\in \mathfrak{a}$. Ifthen every monomial of total degree $r$ in the $u_i$ is divisible by one of the $u_i^{e_i}$. HenceThus the assertion is true; algebraically, the quotient defines a punctual scheme supported at $x$.

Choose finite generating sets $I=(g_1,\dots ,g_s)$ and $J=(f_1,\dots ,f_t)$, using the Hilbert basis theorem. The assumed inclusion says that each $f_j$ vanishes on $V_{\mathbb{C}}(I)$. By the Strong Hilbert Nullstellensatz, for every $j$ there is an exponent $N_j$ such that

The coefficients in an expression $f_j^{N_j}={\sum }_i{q}_{ij}{g}_i$ solve a finite system of linear equations with rational coefficients. Since it has a complex solution, Gaussian elimination gives a rational solution. Clearing the finitely many denominators produces a nonzero integer $D$ such thatfor every $j$.

For any prime $p\nmid D$, reduce these identities modulo $p$. At a common zero of ${\pi }_p(I)$ in the algebraic closure ${\overline{\mathbb{F}}}_p$, they give ${\pi }_p(f_j{)}^{N_j}=0$, hence ${\pi }_p(f_j)=0$, for all $j$. Thereforefor every prime except the finitely many divisors of $D$. This is the spreading out of an affine zero-set inclusion.

An $A$-module $M$ is a flat module when the tensor functor $-{\otimes }_{A}M$ preserves injections, equivalently when it is exact.

Suppose first that $M$ is flat. For any nonzero $a\in A$, tensor the injection $A\stackrel{a}{\to }A$ with $M$. The resulting map $M\stackrel{a}{\to }M$ is injective, so $am=0$ implies $m=0$. Thus $M$ is a torsion-free module.

Conversely, suppose $M$ is torsion-free over the principal ideal domain $A$. Every finitely generated submodule of $M$ is a finitely generated torsion-free module over a PID, hence a finite free module and therefore flat. The module $M$ is the filtered colimit of these submodules. Tensor products commute with filtered colimits, and filtered colimits of modules preserve exact sequences, so $M$ is flat. This proves that a torsion-free module over a principal ideal domain is flat.

Suppose $xy\in J$ and $x\notin J$. Then the colon ideal $(J:x)$ properly contains $J$ because it contains $y$. If also $y\notin J$, then $J+(x)$ properly contains $J$. The maximality of $J$ in the family makes both larger ideals lie outside the family. Applying the stated closure condition with $I=J$ and $a=x$ would imply $J$ lies outside the family, a contradiction. Hence $x\in J$ or $y\in J$, and $J$ is a prime ideal. This argument is the Prime ideal principle for an Oka family.

Let $\mathcal{F}$ be the proper nonprincipal ideals. A chain in $\mathcal{F}$ has a nonprincipal union: if its union were $(a)$, then $a$ would belong to one member of the chain, forcing that member to equal the union and be principal. The union is also proper. Thus Zorn lemma gives a maximal member whenever $\mathcal{F}$ is nonempty.

The family satisfies the condition from part (a). Indeed, supposeWrite $a=bd$ and $b=i+ra$, where $i\in I$. Every $x=bs\in I$ has $as=dx\in I$, so $s\in (c)$ and $I\subseteq (bc)$. Conversely $ac\in I$ because $c\in (I:a)$, while $ic\in I$ because $i\in I$. Thus $bc=ic+rac\in I$, proving $(bc)\subseteq I$. Hence $I=(bc)$ is principal.

If a nonprincipal ideal existed, part (a) would therefore produce a nonprincipal prime ideal, contrary to the hypothesis. Every ideal is principal, so the integral domain is a principal ideal domain.

Suppose first that $A$ is a unique factorization domain, and let $\mathfrak{p}$ be a minimal nonzero prime ideal. Choose $0\ne a\in \mathfrak{p}$ and factor it into irreducibles. In a UFD each irreducible is a prime element, so one factor $\pi$ belongs to $\mathfrak{p}$. The nonzero prime ideal $(\pi )\subseteq \mathfrak{p}$ must equal $\mathfrak{p}$ by minimality.

Conversely, the ascending chain condition in the Noetherian domain implies that every nonzero nonunit factors into irreducibles. Let $\pi$ be irreducible and choose a prime $\mathfrak{p}$ minimal over $(\pi )$. The Krull principal ideal theorem gives $\mathrm{ht}\mathfrak{p}\le 1$. Since $A$ is a domain and $\mathfrak{p}\ne (0)$, it is a minimal nonzero prime and hence is principal, say $\mathfrak{p}=(q)$. The divisibility $q\mid \pi$ and irreducibility of $\pi$ force $q$ to be associate to $\pi$, so $(\pi )=\mathfrak{p}$ is prime. Thus every irreducible is prime, proving that $A$ is a UFD. This is the Minimal-prime criterion for a Noetherian unique factorization domain.

Let $A$ be a principal ideal domain that is not a field. A PID is Noetherian and is a unique factorization domain. Every nonzero prime ideal is generated by a prime element. Ifthen $a\mid p$, so primality of $p$ makes $a$ a unit or an associate of $p$. The proper alternative is $(a)=(p)$, and every nonzero prime is therefore maximal. Since $A$ has a nonzero prime ideal, its Krull dimension is exactly one.

Conversely, let $A$ be a Noetherian UFD of Krull dimension one. Every nonzero prime is minimal among nonzero primes, and the forward argument in part (c) makes it principal. The zero ideal is principal as well, so part (b) shows that $A$ is a PID. A field has Krull dimension zero, hence $A$ is not a field.

This is the Artin--Tate lemma. Choose $A$-algebra generators $x_1,\dots ,x_r$ of $C$ and $B$-module generators $e_1,\dots ,e_s$ of $C$. Writewith coefficients $a_{ij},b_{ij\ell }\in B$, and let $B_0$ be the $A$-subalgebra of $B$ generated by these finitely many coefficients.

The module ${\sum }_j{B}_0{e}_j$ contains the $x_i$, is closed under multiplication, and contains $A$ after including an expression for $1$ among the chosen coefficients. It therefore equals $C$. Thus $C$ is a finite $B_0$-module. The ring $B_0$ is Noetherian by the Hilbert basis theorem, and $B\subseteq C$ is a $B_0$-submodule, so $B$ is a finite $B_0$-module. It follows that $B$ is a finitely generated $A$-algebra.

Let $K$ be a field finitely generated as a $\mathbb{Z}$-algebra. If $K$ has positive characteristic $p$, it is a finitely generated algebra over ${\mathbb{F}}_p$; Zariski lemma makes it a finite algebraic extension of the finite field ${\mathbb{F}}_p$, so $K$ is finite.

Suppose instead that $K$ has characteristic zero. Then $K$ is a finitely generated $\mathbb{Q}$-algebra and Zariski lemma makes it a number field. For algebra generators ${\alpha }_1,\dots ,{\alpha }_r$, choose a nonzero integer $N$ such that every ${\alpha }_i$ is integral over $\mathbb{Z}[1/N]$. The whole algebra $K=\mathbb{Z}[{\alpha }_1,\dots ,{\alpha }_r]$ would then be integral over $\mathbb{Z}[1/N]$. But for a prime $q\nmid N$, the element $1/q\in K$ is not integral over the integrally closed domain $\mathbb{Z}[1/N]$, a contradiction. Hence every field finitely generated over the integers is finite, and in particular no infinite field has that property.

The Poincare series of a graded module isIt is the generating function of the Hilbert function $n\mapsto {\dim }_k{A}_n$.

A Hilbert polynomial of $A$ is a polynomial $h_A(X)\in \mathbb{Q}[X]$ such thatfor every sufficiently large integer $n$. Eventual equality makes this polynomial unique.

Take $A=k[x]$ with $\mathrm{deg}x=2$. This is a Noetherian graded algebra with $A_0=k$, butNo polynomial can agree eventually with these alternating values, so $A$ has no Hilbert polynomial.

A sufficient condition is that $A$ be a standard graded algebra: it is generated as a $k$-algebra by finitely many elements of degree one. The Hilbert-Serre theorem then makes $P_A(t)$ a rational function whose denominator, after cancellation, is a power of $1-t$. When the eventual Hilbert polynomial is nonzero,where the right side uses the order of the pole at $t=1$.

The degree-$n$ component isDimensions therefore satisfy the Cauchy product rule, giving

For an affine algebraic group $G$, its Lie algebra is the tangent space at the identity,Equivalently, it is the space of left-invariant derivations of the coordinate ring $\mathbb{C}[G]$. The Lie bracket is the commutator of derivations,It is antisymmetric because $[Y,X]=-[X,Y]$. Associativity of composition givesafter all six triple products cancel in pairs, proving the Jacobi identity. This construction is the Lie algebra of an affine algebraic group.

The distinct affine algebraic groupsboth have the Special linear Lie algebra ${\mathfrak{sl}}_2(\mathbb{C})$. Quotienting a Lie group by a discrete central subgroup does not change its tangent Lie algebra.

The finite-dimensional irreducible rational representations of $S{L}_2(\mathbb{C})$ areof dimension $n+1$. The central element $-I$ acts on $V_n$ by $(-1{)}^n$. Therefore precisely the even-indexed representationsdescend to irreducible representations of $PG{L}_2(\mathbb{C})$. This is the Descent of an irreducible SL2 representation to PGL2.

A bilinear form $B$ on a Lie algebra is invariant whenequivalently $B([x,y],z)+B(y,[x,z])=0$.

Choose a basis $(x_i)$ of $\mathfrak{g}$ and its $B$-dual basis $(x^i)$. The Casimir element iswhere $U(\mathfrak{g})$ is the universal enveloping algebra. The tensor ${\sum }_i{x}_i\otimes x^i$ corresponds under $B:\mathfrak{g}\cong {\mathfrak{g}}^{\ast }$ to the identity endomorphism, so invariance of $B$ makes it fixed by the diagonal adjoint action. Applying multiplication $U(\mathfrak{g})\otimes U(\mathfrak{g})\to U(\mathfrak{g})$ givesfor every $x\in \mathfrak{g}$. Thus $\Omega$ lies in the center of an associative algebra of $U(\mathfrak{g})$.

Use the standard basisof ${\mathfrak{sl}}_2$. For the invariant trace form $B(x,y)=\mathrm{tr}(xy)$, the dual basis is $(f,e,h/2)$, soOn a highest-weight vector $v$ in the $(n+1)$-dimensional irreducible module, $ev=0$ and $hv=nv$. Since $efv=[e,f]v=nv$ and $fev=0$,Centrality and Schur lemma make this the eigenvalue on the whole module. Thus $\lambda =n(n+2)/2$, the Casimir eigenvalue for sl2. If the form is instead the Killing form, which is four times the trace form on ${\mathfrak{sl}}_2$, the corresponding Casimir and eigenvalue are divided by four.

No. Let $\mathfrak{g}={\mathfrak{sl}}_2$, let $\mathfrak{s}=\mathbb{C}e$ be the one-dimensional subalgebra generated by the standard raising operator, and let $L={\mathbb{C}}^2$ be the irreducible defining representation. On restriction to $\mathfrak{s}$, the element $e$ acts by a nonzero Nilpotent Jordan block. A direct sum of irreducible representations of the one-dimensional abelian Lie algebra would make $e$ diagonalizable, so this restriction is not completely reducible. The Complete reducibility of semisimple Lie algebra representations applies when the restricting algebra is semisimple, which $\mathfrak{s}$ is not.

The Cartan solvability criterion states that a finite-dimensional complex Lie algebra $\mathfrak{g}$ is solvable exactly whenwhere $\kappa$ is the Killing form. In the matrix form of the criterion, a Lie subalgebra $\mathfrak{g}\subseteq \mathfrak{gl}(V)$ is solvable iffor all $x\in [\mathfrak{g},\mathfrak{g}]$ and $y\in \mathfrak{g}$.

A torus $\mathfrak{t}$ in a Lie algebra is an abelian subalgebra whose elements act semisimply in the Adjoint representation. Its weight-space decomposition is

Invariance of the nondegenerate Trace form of a Lie algebra representation givesNondegeneracy therefore forces $\beta =-\alpha$. The standard sl2 subalgebra associated with a root argument gives one-dimensional opposite root spaces with vectors $e\in {\mathfrak{g}}_{\alpha }$, $f\in {\mathfrak{g}}_{-\alpha }$ and $h_{\alpha }\in \mathfrak{t}$ satisfyingTheir span is a copy of ${\mathfrak{sl}}_2$.

SetThen $\dim \mathfrak{h}=\dim \mathfrak{t}-1$, $\mathfrak{t}=\mathbb{C}{h}_{\alpha }\oplus \mathfrak{h}$, and every element of $\mathfrak{h}$ commutes with $e$, $f$, and $\mathfrak{t}$. Hence $\mathfrak{h}$ is abelian andas a direct sum of commuting Lie algebras. This is the two-root decomposition with a nondegenerate trace form.

Let $r=\dim \mathfrak{h}$ and choose a dual basis of weights ${\epsilon }_1,\dots ,{\epsilon }_r\in {\mathfrak{h}}^{\ast }$. TakeThe ${\mathfrak{sl}}_2$ factor acts in its defining representation on ${\mathbb{C}}^2$ and trivially on the other summands; $\mathfrak{h}$ acts trivially on ${\mathbb{C}}^2$ and by the displayed characters on the one-dimensional summands. The resulting trace form is the nondegenerate trace form on ${\mathfrak{sl}}_2$, ison $\mathfrak{h}$, and has zero cross terms. It is therefore nondegenerate on ${\mathfrak{sl}}_2\oplus \mathfrak{h}$.

Write an element of the diagonal torus asand let ${\epsilon }_i$ extract $t_i$. The root-space decomposition iswith one-dimensional root spaces. Thus the $D_n$ root system is

The upper-triangular choice givesA compatible simple system isThe highest root and Weyl vector areThe fundamental weights are

Using the paper's letters, the root lattice and weight lattice are respectivelyTheir quotient is

The Dynkin diagram is the $D_n$ diagram: a chain ${\alpha }_1-\cdots -{\alpha }_{n-2}$ whose last node is joined to both ${\alpha }_{n-1}$ and ${\alpha }_n$. The Extended Dynkin diagram adds ${\alpha }_0=-\theta$ joined to ${\alpha }_2$. For $D_4$, the central node ${\alpha }_2$ consequently has the four leaves ${\alpha }_0,{\alpha }_1,{\alpha }_3,{\alpha }_4$.

Let $V={\mathbb{C}}^4$. The Special linear Lie algebra $\mathfrak{sl}(V)$ acts on ${\Lambda }^{2}V$. The wedge productis a nondegenerate symmetric bilinear form, and the action preserves it because $\mathfrak{sl}(V)$ acts trivially on ${\Lambda }^{4}V$. This gives an injective homomorphismBoth Lie algebras have dimension $15$, so the map is an isomorphism. This realizes the Isomorphism between so6 and sl4.

For every root, the Weyl reflection isFor $\alpha ={\epsilon }_i-{\epsilon }_j$, it swaps the $i$th and $j$th coordinates. For $\alpha ={\epsilon }_i+{\epsilon }_j$, it sendsand fixes all other coordinates. The Weyl group of $D_n$ is therefore the group of signed permutations with an even number of sign changes,

For a dominant integral weight $\lambda$, the Weyl character formula isHere $W$ is the Weyl group, $\ell (w)$ its Coxeter length, $\rho$ the Weyl vector, and $\mathrm{ch}$ the formal character of a weight module. Taking the limit at the identity gives the Weyl dimension formula

Choose a short simple root ${\alpha }_1$ and a long simple root ${\alpha }_2$. The G2 root system has positive rootsThe two hexagons formed by the short and long roots give the usual twelve-root diagram. The fundamental weights areThe second is the highest root, so the irreducible module $L({\omega }_2)$ is the Adjoint representation of a Lie algebra.

The seven weights of $L({\omega }_1)$ are zero and the six short roots, each with multiplicity one. Its crystal, with arrows denoting the lowering operators ${\tilde{f}}_i$, is the chain

Applying the Weyl dimension formula to $\lambda =n_1{\omega }_1+n_2{\omega }_2$ gives

For type $B_n$, the spinor representation has highest weightand its weights are the $2^n$ sign vectorsEach weight has multiplicity one. Along the simple root ${\alpha }_j={\epsilon }_j-{\epsilon }_{j+1}$, the Kashiwara operator ${\tilde{e}}_j$ can raise a weight exactly when $(i_j,i_{j+1})=(-1,+1)$, when it replaces that pair by $(+1,-1)$. For the short root ${\alpha }_n={\epsilon }_n$, ${\tilde{e}}_n$ replaces a final $-1$ by $+1$. This proves the stated crystal by the root-string property of a crystal.