Past exam of the mathematics course of the University of Cambridge / 2022 / iii

An ideal of definition of a Noetherian local ring $(A,\mathfrak{m})$ is an $\mathfrak{m}$-primary ideal $I$; equivalently, $\sqrt{I}=\mathfrak{m}$, or some power ${\mathfrak{m}}^r$ is contained in $I$.

If $M$ is a finitely generated module of dimension $d$, the Hilbert–Samuel functionagrees for all sufficiently large $n$ with a polynomial of degree $d$. Its leading term iswhere $e_I(M)$ is the Hilbert–Samuel multiplicity.

Assume first that $f\ne 0$ and letbe the least total degree of a nonzero homogeneous part of $f$. The associated graded ring iswhere $f_d$ is the initial homogeneous form. Multiplication by the nonzero polynomial $f_d$ is injective in $k[x,y]$, so the degree-$j$ component has dimensionSumming the components of degrees below $n$ givesThus the Hilbert polynomial isIts leading coefficient is the order of vanishing, or multiplicity, of the plane curve $f=0$ at the origin. If $f=0$, no relation is imposed and $\chi =n(n+1)/2$ for every $n$.

A monomial of weighted degree $n$ is $x^{n-2j}{y}^j$ with $0\le j\le \lfloor n/2\rfloor$. ThereforeThis is a quasipolynomial of period two, not eventually one polynomial: it equals $n/2+1$ for even $n$ and $(n+1)/2$ for odd $n$. The usual eventual-polynomial theorem for a graded algebra assumes a standard graded algebra, generated in degree one. Here the generator $y$ has degree two, so there is no contradiction.

The ring of p-adic integers is the inverse limitEquivalently, each element has a unique convergent expansion ${\sum }_{j\ge 0}{b}_j{p}^j$ with digits $0\le b_j<p$. It is a complete discrete valuation ring with maximal ideal $(p)$ and residue field ${\mathbb{F}}_p$.

Addition preserves the condition becauseFor multiplication, write $c_n={\sum }_{i+j=n}{a}_i{b}_j$. Given $q$, choose $N$ so that $a_i,b_i\in p^q{\mathbb{Z}}_p$ for $i\ge N$. If $n\ge 2N$, every pair $i+j=n$ has $i\ge N$ or $j\ge N$, so every summand lies in $p^q{\mathbb{Z}}_p$. Hence $\mathrm{ord}(c_n)\to \infty$, proving that ${\mathbb{Z}}_p⟨T⟩$ is a subring of the formal power series ring ${\mathbb{Z}}_p[[T]]$.

The $(p)$-adic completion isA compatible system of polynomials determines coefficients $a_n\in {\mathbb{Z}}_p$. For each $q$, its reduction has finite degree, so all but finitely many $a_n$ lie in $p^q{\mathbb{Z}}_p$. This is exactly $\mathrm{ord}(a_n)\to \infty$. Conversely, every such restricted series reduces modulo $p^q$ to a polynomial and hence defines a compatible system. Therefore

An element lies in ${\mathfrak{p}}^{\ast }$ exactly when all its homogeneous components lie in $\mathfrak{p}$. Suppose $ab\in {\mathfrak{p}}^{\ast }$ but $a,b\notin {\mathfrak{p}}^{\ast }$. Choose the least-degree components $a_i\notin \mathfrak{p}$ and $b_j\notin \mathfrak{p}$. In the degree $i+j$ component of $ab$, every term other than $a_i{b}_j$ contains a lower component of $a$ or $b$ and hence lies in $\mathfrak{p}$. Since the whole component lies in $\mathfrak{p}$, it follows that $a_i{b}_j\in \mathfrak{p}$, contradicting primality. Thus

First, $\sqrt{{\mathfrak{q}}^{\ast }}=\mathfrak{p}$. Indeed, each homogeneous element of $\mathfrak{p}=\sqrt{\mathfrak{q}}$ has a power in $\mathfrak{q}$, and that power is homogeneous and hence lies in ${\mathfrak{q}}^{\ast }$; finite homogeneous generators of the Noetherian ideal $\mathfrak{p}$ give the assertion for every element.

Now suppose $ab\in {\mathfrak{q}}^{\ast }$ and $a\notin \mathfrak{p}$. Choose the least homogeneous component $a_i\notin \mathfrak{p}$. If $b\notin {\mathfrak{q}}^{\ast }$, choose the least component $b_j\notin \mathfrak{q}$. The degree $i+j$ component of $ab$ differs from $a_i{b}_j$ by terms in $\mathfrak{q}$. Since it lies in $\mathfrak{q}$, we get $a_i{b}_j\in \mathfrak{q}$. The $\mathfrak{p}$-primary property and $a_i\notin \mathfrak{p}$ imply $b_j\in \mathfrak{q}$, a contradiction. Hence $b\in {\mathfrak{q}}^{\ast }$, proving that

Pass to the graded domain $R=S/{\mathfrak{p}}^{\ast }$ and let $\overline{\mathfrak{p}}=\mathfrak{p}/{\mathfrak{p}}^{\ast }$. This is a nonzero prime containing no nonzero homogeneous element. Localize at the multiplicative set $U$ of all nonzero homogeneous elements. Every nonzero homogeneous element of $U^{-1}R$ is a unit; its nonzero graded pieces are one-dimensional over the degree-zero field, so after reindexing degrees this localization is a Laurent polynomial ring $K[t,t^{-1}]$. The extended prime $U^{-1}\overline{\mathfrak{p}}$ is therefore a nonzero prime of height one.

Any prime strictly between $0$ and $\overline{\mathfrak{p}}$ would remain a nonzero prime strictly below it after localization, impossible in $K[t,t^{-1}]$. Contracting back proves that no prime lies strictly between ${\mathfrak{p}}^{\ast }$ and $\mathfrak{p}$. The graded height theorem, equivalently the same localization argument applied to saturated chains, then gives

Choose a finite presentation $F_1\to F_0\to M\to 0$ with $F_0,F_1$ finite free. Applying ${\mathrm{Hom}}_A(-,A)$ givesLet $N=F_0^{\ast }$ and let $P$ be the image in $F_1^{\ast }$. Thenis exact, $N$ is finite free, and $P$ is torsion-free because it is a submodule of the free module $F_1^{\ast }$ over the domain $A$.

The reflexive-module second-syzygy criterion says that the kernel of a map from a finite free module to a torsion-free module over a Noetherian domain is reflexive. Applying it to this sequence shows that $M^{\ast }$ is reflexive. Concretely, after localizing at the fraction field, every functional on $(M^{\ast }{)}^{\ast }$ represented generically by an element of $N$ has no denominator: torsion-freeness of $P$ forces its image to vanish already over $A$. Thus the natural evaluation map $M^{\ast }\to M^{\ast \ast \ast }$ is an isomorphism.

Put $g=y^2-zw$. Every generator of $I$ is divisible by $x^2$ and belongs to $(y,z{)}^2$, while the displayed decomposition in the next part will show thatIts two minimal primes, and hence its isolated associated primes, areThe corresponding isolated primary components are

ConsiderIt is $(x,y^2-zw)$-primary. A direct ideal-intersection calculation givesThis decomposition is irredundant and its radicals areThereforeIn particular, ${\mathfrak{p}}_3=(x,y^2-zw)$ is the embedded associated prime and the decomposition also confirms $\sqrt{I}=(x)\cap (y,z)$.

Restriction of scalars makes every $S^{-1}A$-module an $A$-module, and multiplication by $s\in S$ is invertible with inverse multiplication by $1/s$.

Conversely, suppose each multiplication map $s_M:m\mapsto sm$ is an automorphism of the $A$-module $M$. DefineThe universal property of localization shows that this is well-defined and gives the unique $S^{-1}A$-module structure extending the $A$-action. These constructions are inverse.

Under this structure the natural map $M\to S^{-1}M$ has inverseso it is an isomorphism.

Localization of the inclusion $N\hookrightarrow M$ gives an injection $S^{-1}N\hookrightarrow S^{-1}M$, whose image lies in $N^{\prime }$. Conversely, take $m/s\in N^{\prime }$. Since $s/1$ is a unit and $N^{\prime }$ is an $S^{-1}A$-submodule,By the definition of the inverse image, $m\in N$, and hence $m/s$ lies in the image of $S^{-1}N$. Therefore

Write $D=\mathrm{ad}y$. The generalized eigenspace decomposition of the linear map $D$ isfor any sufficiently large $N$. Because $D$ is a derivation, the generalized-eigenspace bracket lemma givesConsequently $L_{0,y}$ is a Lie subalgebra.

The set $I(K)=\{x\in L:[x,K]\subseteq K\}$ is the normalizer of a Lie subalgebra $K$. Certainly $L_{0,y}\subseteq I(L_{0,y})$. Conversely, if $x\in I(L_{0,y})$, then $y\in L_{0,y}$ givesOn the direct sum of the nonzero generalized eigenspaces, $D$ is invertible. Hence the nonzero-eigenvalue component of $x$ vanishes, and

Now let $K$ be a Lie subalgebra containing $L_{0,y}$. Since $y\in K$, the subspace $K$ is $D$-invariant. The generalized zero eigenspace of the induced map on $L/K$ is the image of $L_{0,y}$, hence is zero. If $x\in I(K)$, then $Dx=[y,x]\in K$, so $x+K$ lies in that zero eigenspace. Thus $x\in K$ and

A Nilpotent Lie algebra is one whose lower central serieseventually reaches zero. Suppose $L$ is nilpotent and $K⊊L$. Choose the least $r\ge 2$ for which ${\gamma }_r(L)\subseteq K$. Then ${\gamma }_{r-1}(L)⊈K$, and anysatisfies $[x,K]\subseteq [L,{\gamma }_{r-1}(L)]={\gamma }_r(L)\subseteq K$. Therefore $x\in I(K)\setminus K$, proving the normalizer condition for a nilpotent Lie algebra

It remains to prove the converse needed here. The Engel lemma states that if a finite-dimensional Lie algebra of linear maps consists of nilpotent maps, then the maps have a common nonzero vector in their kernels. To prove it, induct on the dimension of the algebra. For a maximal proper subalgebra $H$, induction applied to the action of $H$ on $L/H$ produces $x\notin H$ with $[H,x]\subseteq H$. Thus $H$ is an ideal of codimension one. Induction also gives a nonzero common kernelThe ideal property makes $W$ invariant under $L$; a nilpotent representative of a basis of $L/H$ has a nonzero kernel on $W$, yielding a vector killed by all of $L$.

Apply the lemma to the Adjoint representation. It produces a nonzero element of the Center of a Lie algebra. Induction on $\dim L$, followed by passage to the quotient by this center, proves Engel theorem: if every $\mathrm{ad}y$ is nilpotent, then $L$ is nilpotent. The hypothesis $L_{0,y}=L$ says exactly that every $\mathrm{ad}y$ is nilpotent, so

A finite root system in a real Euclidean vector space $E$ is a finite spanning set $\Phi \subset E\setminus \{0\}$ such that, for every $\alpha \in \Phi$, the root reflectionpreserves $\Phi$, and the Cartan integer $2(\beta ,\alpha )/(\alpha ,\alpha )$ is an integer for all $\alpha ,\beta \in \Phi$. It is reduced when the only scalar multiples of $\alpha$ in $\Phi$ are $\alpha$ and $-\alpha$. Its Weyl group is the subgroup of the orthogonal group generated by the reflections $s_{\alpha }$. A base of a root system $\Delta$ is a basis of $E$ such that every root is an integer combination of elements of $\Delta$ whose nonzero coefficients all have the same sign.

The coroot of $\alpha$ isLet $C$ be the Weyl chamber determined by $\Delta$:The roots $\alpha$ and ${\alpha }^{\vee }$ are positive scalar multiples, so their reflecting hyperplanes and their positive half-spaces are identical. The same chamber $C$ therefore defines positivity in the coroot system ${\Phi }^{\vee }$. Its walls correspond exactly to the rays ${\mathbb{R}}_{>0}{\alpha }^{\vee }$ for $\alpha \in \Delta$. Hence its simple roots arewhich is therefore a base of ${\Phi }^{\vee }$.

For an arbitrary finite-dimensional complex Lie algebra $L$, a Cartan subalgebra is a nilpotent Lie subalgebra $H$ equal to its own normalizer. When $L$ is semisimple, this is equivalently a maximal abelian subalgebra consisting of elements that act semisimply in the Adjoint representation.

Choose such an $H$. Simultaneous diagonalization gives the root-space decompositionThe nonzero weights $\alpha \in H^{\ast }$ are the roots. The restriction of the Killing form $B$ to $H$ is a nondegenerate bilinear form, so each $\alpha$ corresponds to a unique $t_{\alpha }\in H$ with $\alpha (h)=B(t_{\alpha },h)$. On the real span of these $t_{\alpha }$, the restriction of $B$ supplies a positive-definite inner product after choosing the standard real form. The sl2 subalgebra associated with a root givesand shows that these reflections preserve the finite set $\Phi$. Thus the roots form a finite reduced crystallographic root system, whose Weyl group is generated by these reflections.

A finite-dimensional Lie algebra $L$ is semisimple when its solvable radical is zero, equivalently when it has no nonzero solvable ideals.

We prove the Weyl complete reducibility theorem. Induct on the dimension of a finite-dimensional $L$-module $V$. It is enough first to split a submodule $W$ of codimension one. The one-dimensional quotient is trivial because a semisimple Lie algebra is perfect. By induction, $W$ is a direct sum of irreducible modules. A Casimir element formed using the Killing form commutes with the $L$-action, acts as zero on every trivial summand, and acts by a nonzero scalar on every nontrivial irreducible summand. Its image is therefore the sum $W_1$ of the nontrivial summands, while its kernel contains the trivial summands $W_0$ and maps onto $V/W$. ThusInside $\ker \Omega$, choose a lift $v$ of a basis of $V/W$. For $x\in L$, $xv\in W_0$, and $L$ acts trivially on $W_0$. Hence $[x,y]v=0$ for all $x,y\in L$. Since $L=[L,L]$, actually $xv=0$ for every $x$, so $\mathbb{C}v$ is the required invariant complement.

This codimension-one case implies the general case. For an arbitrary submodule $W\subseteq V$, letwith the natural Hom representation. The maps vanishing on $W$ form an $L$-submodule $X_0$ of codimension one. Splitting $X_0$ supplies an $L$-equivariant $f$ with $f{|}_W=I_W$. Thenso every invariant subspace has an invariant complement and every finite-dimensional representation is completely reducible.

For the requested example, embed ${\mathfrak{sl}}_2$ as the upper-left $2\times 2$ block in the Special linear Lie algebra ${\mathfrak{sl}}_3$. Under the restricted Adjoint representation,Here $V_n$ is the irreducible ${\mathfrak{sl}}_2$-module of highest weight $n$. The first summand is the three-dimensional adjoint module, the second is trivial, and the last two are the two-dimensional defining module and its dual, which are isomorphic. This explicit direct sum demonstrates complete reducibility.

Because $M$ is a finite-dimensional semisimple module, it has an isotypic decompositionwhere the $S_i$ are pairwise nonisomorphic simple right $R$-modules. By Schur lemma,is a division ring, while ${\mathrm{Hom}}_R(S_i,S_j)=0$ for $i\ne j$. Consequently every endomorphism preserves the isotypic summands and is a matrix of entries from $D_i$ on each one. ThereforeIf the ground field is algebraically closed and the $S_i$ are finite-dimensional over it, Schur lemma gives $D_i=k$.

A vertex $i$ of a quiver is a sink when no arrow starts at $i$. A representation of a quiver assigns a vector space $V_j$ to every vertex and a linear map $V_a:V_j\to V_{\ell }$ to every arrow $a:j\to \ell$. A morphism $f:V\to W$ is a family of linear maps $f_j:V_j\to W_j$ such that $f_{\ell }{V}_a=W_a{f}_j$ for every arrow. The quiver has finite representation type when it has only finitely many isomorphism classes of indecomposable finite-dimensional representations.

At a sink $i$, the Bernstein–Gelfand–Ponomarev reflection functor replacesand reverses the arrows ending at $i$; the new arrow maps are the kernel inclusion followed by the coordinate projections. Every representation is a direct sum of copies of the simple representation $S_i$ and a representation for which the displayed incoming map is surjective. On the latter representations, reflection at the resulting source, using the corresponding cokernel, is inverse up to natural isomorphism. Thus reflection gives a bijection between indecomposable representations other than $S_i$ on the two sides. Adding the one omitted simple representation on each side proves that reversing all arrows into a sink preserves finite representation type.

For the four-arrow star $Q_1$, take $V_1=\cdots =V_4=k$, $V_5=k^2$, and let the four arrows have imagesAn endomorphism must preserve the first two lines, so its map on $V_5$ is diagonal. Preserving the third forces its diagonal entries to agree, and then all vertex maps are multiplication by that same scalar. The endomorphism ring is therefore $k$, so this representation is a brick module and hence indecomposable. Any isomorphism between parameters preserves the first three labelled lines; the induced projective linear transformation is therefore the identity, and the fourth line gives $\lambda =\mu$. Since $k$ is infinite, this is an infinite family of pairwise nonisomorphic indecomposables. Hence $Q_1$ is not of finite representation type. Repeatedly applying the reflection result to sinks or, dually, to sources shows that every orientation obtained by reversing some of its arrows also has infinite representation type.

Because the generating sets are disjoint, the free product has the presentationThis is immediate from the universal property of a group presentation: a map out of this presented group is exactly a pair of homomorphisms from $G$ and $H$ to the target group.

A reduced word is either the empty word or a productin which every syllable $u_i$ is a nonidentity element of $G$ or $H$, and consecutive syllables belong to different factors.

Let $\mathcal{W}$ be the set of reduced words. Each $g\in G$ acts on the right of $\mathcal{W}$: if the last syllable lies in $H$, append $g$; if it lies in $G$, multiply it by $g$ and delete it when the product is the identity. Define the action of each $h\in H$ analogously. These rules give genuine actions of the two factors by permutations of $\mathcal{W}$, hence an action of the free group $F(X\bigsqcup Y)$. Every relation in $R\bigsqcup S$ acts trivially, so the action factors through the displayed presentation of $G\ast H$.

The element represented by a reduced word $w$ sends the empty word to $w$. Therefore two reduced words representing the same element induce the same permutation and have the same value on the empty word. They must be identical. This proves the normal form theorem for a free product.

Let $w$ be a nonempty reduced word. If its first syllable belongs to $G$, choose any nonidentity $h\in H$. The reduced form of $hw$ begins with an $H$-syllable, whereas that of $wh$ begins with the original $G$-syllable; reduction at the right end cannot change the first syllable. Hence $hw\ne wh$. The same argument with a nonidentity $g\in G$ applies when $w$ begins in $H$. No nonidentity element is therefore central, and

Choose distinct nonidentity elements $g_1,g_2\in G$ and $h_1,h_2\in H$, and putExpand a freely reduced word in $u^{\pm 1},v^{\pm 1}$. At a boundary where two syllables from one factor meet, their product is one of $g_1^{-1}{g}_2$, $g_2^{-1}{g}_1$, $h_1{h}_2^{-1}$, or $h_2{h}_1^{-1}$, all nonidentity by the choices above. Every other boundary already alternates between the factors. Thus the expansion reduces to a nonempty reduced word in $G\ast H$ and cannot represent the identity. The homomorphism from the rank-two free group sending its free generators to $u,v$ is injective, so

The Integer Heisenberg groupis generated by the matrices with $(a,b,c)=(1,0,0)$ and $(0,1,0)$. Their group commutator is the nonidentity central matrix with $(a,b,c)=(0,0,1)$. Thus it is nonabelian and nilpotent of class two, while being finitely generated.

The countable direct sumis abelian and therefore nilpotent of class one. It is not finitely generated, whereas every polycyclic group is finitely generated. Hence it is nilpotent but not polycyclic.

The lamplighter groupis generated by one lamp switch and one translation. It is metabelian, hence solvable. Its base subgroup ${⨁}_{\mathbb{Z}}{C}_2$ is not finitely generated. Every subgroup of a polycyclic group is finitely generated, so the lamplighter group is not polycyclic.

Let $N$ be the subgroup whose upper-left block is $I_2$. It is a finitely generated nilpotent subgroup of the integer upper unitriangular group $U{T}_4(\mathbb{Z})$ and is normal in $L$. The block-diagonal matrixgenerates an infinite cyclic quotient, soFinitely generated nilpotent groups are polycyclic, and an extension of polycyclic groups is polycyclic. Hence $L$ is polycyclic.

Inside $L$, retain only $t$ and the entries in positions $(1,3)$ and $(2,3)$. They form a subgroupThe characteristic polynomial of $A$ is ${\lambda }^2-5\lambda +1$, so its eigenvalues areOne has modulus greater than one, and the resulting semidirect product has exponential growth. Every finitely generated virtually nilpotent group has polynomial growth, as does each of its finitely generated subgroups. Therefore $L$ cannot be virtually nilpotent.

The empty word is the unique vertex of degree three in the underlying tree; every other vertex has degree four. Every graph automorphism therefore fixes the empty word and permutes its three neighbours. Those neighbours are precisely $\{0,1,2\}$, so this set is invariant.

The automorphism $a$ cyclically permutes the first letter and leaves the remaining suffix unchanged, so $a^3=1$ and $a\ne 1$.

In the section notation,Thus $b^3=(a^3,1,b^3)=(1,1,b^3)$. An automorphism $c$ satisfying $c=(1,1,c)$ fixes every finite word: repeatedly entering the third subtree eventually reaches the end of the word. Hence $b^3=1$. Since $b(00)=01$, it is nonidentity, and both $a$ and $b$ have order three.

The action on the first level defines a surjective homomorphismthat sends $a$ to the displayed cycle and $b$ to the identity. Its kernel is therefore the normal closure of $b$, proving that $\{b\}$ normally generates ${\mathrm{Stab}}_G(1)$.

Apply the Reidemeister–Schreier theorem with transversal $\{1,a,a^{-1}\}$. The generators arising from $a$ are trivial, while those arising from $b$ areConsequently these three elements generate ${\mathrm{Stab}}_G(1)$.

An element $g$ fixing the first level restricts to an automorphism $g_i$ on each rooted subtree, and composition is coordinatewise. Thereforeis a homomorphism. If all three sections are trivial, $g$ fixes every word, so $\varphi$ is injective.

Directly from the recursions,The generators found in the preceding part therefore have every section in $G$, so $\varphi ({\mathrm{Stab}}_G(1))\subseteq G^3$.

Moreover, every coordinate projection of this image contains both $a$ and $b$, and is therefore onto $G$. Since $a$ acts transitively on the first level, induction shows that $G$ acts transitively on every level of the rooted tree. The $n$th level has $3^n$ vertices, so the orders of these finite orbits are unbounded. Hence $G$ is infinite.

Use the convention $[g,h]=g^{-1}{h}^{-1}gh$. In $G/K$ the images of $a$ and $b$ commute and both have order three, so $G/K$ is a quotient of $C_3\times C_3$. Thus

Put $d=a^{-1}ba$. From the preceding section calculations,Since both elements fix the first level, their commutator is computed coordinatewise, andThe element $[d,x]$ belongs to $K$. The third-coordinate projection of $\varphi ({\mathrm{Stab}}_G(1))$ is onto $G$, so conjugating this element inside the stabilizer shows that $\varphi (K)$ contains $(1,1,x^g)$ for every $g\in G$. Because the conjugates $x^g$ generate $K$, it contains $1\times 1\times K$. Conjugation by $a$ cyclically permutes the coordinates; hence it also contains $K\times 1\times 1$ and $1\times K\times 1$. These coordinate subgroups commute, givingInjectivity of $\varphi$ identifies its inverse image with a subgroup of $K$ isomorphic to $K^3$.

A group $G$ is residually finite when, for every $1\ne g\in G$, there are a finite group $Q$ and a group homomorphism $q:G\to Q$ such that $q(g)\ne 1$. Equivalently, the intersection of all finite-index normal subgroups of $G$ is trivial.