Past exam of the mathematics course of the University of Cambridge / 2026 / iii

A minimal primary decomposition is an expression

I = Q_{1} \cap \dots \cap Q_{s}

in which every

Q_{i}

is a primary ideal, the prime ideals

Q_{i}

are pairwise distinct, and the decomposition is irredundant: deleting any

Q_{i}

changes the intersection.

Solved by gpt-5.6-sol high.

Solution

The Second uniqueness theorem for primary decomposition says that in a minimal primary decomposition of an ideal in a Noetherian ring, every primary component belonging to an isolated prime is unique. Here an isolated prime is a minimal member of the set

{Q_{i}}

Let

p = Q_{i}

be isolated and apply localization at a prime ideal. If

j \neq = i

, minimality of

p

gives

Q_{j} ⊈ p

, so some element of

Q_{j}

becomes a unit in

R_{p}

. Consequently

I R_{p} = Q_{i} R_{p} .

Because

Q_{i}

p

-primary, multiplication by any

s \in / p

cannot carry an element outside

Q_{i}

into

Q_{i}

. Therefore

Q_{i} = Q_{i} R_{p} \cap R = I R_{p} \cap R .

The right side depends only on

I

and

p

, proving uniqueness.

Solved by gpt-5.6-sol high.

Solution

No. Take the Noetherian ring

R = k [x]

, the finitely generated

R

-module

M = R \oplus R / (x),

and

N = 0

. Its annihilator of a module is

Ann_{R} (M) = 0,

which is a prime ideal and hence a primary ideal. But with

r = x

and

m = (0, \overline{1})

we have

m \neq = 0

and

r m = 0

, while

x^{k} M \neq = 0

for every

k

because the free summand survives. Thus

N

is not a primary submodule.

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Solution

The inverse limit is the submodule of the direct product consisting of compatible families:

lim_{i \in I} N_{i} = {(x_{i}) \in \prod_{i \in I} N_{i} : g_{ij} (x_{j}) = x_{i} whenever i \leq j} .

Its projection to

N_{i}

sends

(x_{j})

x_{i}

; these projections satisfy the universal property of an inverse limit.

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Solution

Two decreasing filtrations of a module

(M_{n})

and

(M_{n}^{'})

are equivalent when each contains a fixed shift of the other: there are

a, b \geq 0

such that

M_{n + a} \subseteq M_{n}^{'}, M_{n + b}^{'} \subseteq M_{n}

for every

n \geq 0

For an

I

-filtration,

I M_{n} \subseteq M_{n + 1}

, so

I^{n} M_{0} \subseteq M_{n}

. If it is stable from

s

onward, then

M_{s + n} = I^{n} M_{s} \subseteq I^{n} M_{0} .

It is therefore equivalent to the I-adic filtration. Any two stable

I

-filtrations are consequently equivalent to each other.

Solved by gpt-5.6-sol high.

Solution

No. Give

R = k [x]

the x-adic filtration

R_{n} = (x^{n})

, take

M = R

, and let

N = (x)

. At

n = 1

the intersection filtration gives

M_{1} \cap N = (x),

whereas the induced filtration of

N

gives

R_{1} N = (x^{2}) .

Thus intersection with a submodule need not equal the induced filtration.

Solved by gpt-5.6-sol high.

Solution

Yes. Scalar multiplication in the quotient module gives directly

R_{n} (M / N) = {r (m + N) : r \in R_{n}, m \in M} = (R_{n} M + N) / N = (M_{n} + N) / N .

Hence the displayed filtration is precisely the induced filtration.

Solved by gpt-5.6-sol high.

Solution

Work modulo

I

. Put

C = R / (I + J),

and let

A

be the image of

I^{'} / I

C

. The separating condition

I^{'} = I^{'} \cap (I + J)

modulo

I

says that

I^{'} / I \to C

is injective, so we may regard

A

as a submodule of

C

Since

r \in (I : I^{'})

, some power

r^{q}

annihilates

A

. Apply the Artin-Rees lemma to

A \subseteq C

and the principal ideal

(r)

. There is

s

such that for every

m \geq s

A \cap r^{m} C = r^{m - s} (A \cap r^{s} C) .

For

m \geq s + q

, the right side is zero. Pulling the equality

A \cap r^{m} C = 0

back to

R

gives

I^{'} \cap (I + J + (r^{m})) = I,

as required.

Solved by gpt-5.6-sol high.

Solution

For an

m

-primary ideal

I

in a Noetherian local ring, the function

n ⟼ length_{R} (R / I^{n})

agrees for all sufficiently large

n

with a polynomial in

n

. This is the Hilbert-Samuel polynomial, also called here the characteristic polynomial of

I

. Using

R / I^{n + 1}

instead merely shifts its variable.

Solved by gpt-5.6-sol high.

Solution

The denominator has degree one, independently of the number

n

of variables. This reflects the fact that all mixed monomials vanish and each component of the ring supports only one polynomial direction; equivalently, the Krull dimension of this graded ring is one.

Solved by gpt-5.6-sol high.

Paper 102 (Lie Algebras and Their Representations)

www.maths.cam.ac.uk/postgrad/part-iii/files/pastpapers/2026/III%20Paper%20102%20updated.pdf

1

i

Solution

For this Heisenberg Lie algebra,

[g, g] = C c, [g, C c] = 0,

because

c

belongs to the center. Thus its Lower central series of a Lie algebra is

g, C c, 0

, so

g

is a two-step Nilpotent Lie algebra.

Solved by gpt-5.6-sol high.

ii

Solution

Let

A, B, C

represent

a, b, c

. Since

c

is central,

C

commutes with

A

and

B

. Over the complex number field

C

C

has an eigenvalue

γ

, and its corresponding eigenspace is invariant under all three operators. The irreducibility of

V

therefore makes this eigenspace all of

V

, so

C = γ I

. Taking the trace of

C = [A, B] = A B - B A

gives

(dim V) γ = 0

by the cyclic property of the trace; hence

C = 0

The remaining operators

A

and

B

commute. Two commuting operators on a nonzero finite-dimensional complex vector space have a common eigenvector, whose span is invariant. Irreducibility therefore forces

dim V = 1

. Conversely, every pair

(α, β) \in C^{2}

defines a one-dimensional irreducible representation by

a ⟼ α, b ⟼ β, c ⟼ 0.

These are all the finite-dimensional irreducible representations.

Solved by gpt-5.6-sol high.

iii

Solution

For a finite-dimensional Lie algebra representation

ρ

V

, the Trace form of a Lie algebra representation is

(x, y)_{V} = tr (ρ (x) ρ (y)) .

Write again

A = ρ (a)

B = ρ (b)

, and

C = ρ (c) = [A, B]

. The operator

C

commutes with both

A

and

B

. Direct use of the cyclic property of the trace gives

tr (C A) tr (CB) tr (C^{2}) = tr (A B A - B A^{2}) = 0, = tr (A B^{2} - B A B) = 0, = tr (A BC - B A C) = 0.

In the last line, cyclicity and

A C = C A

turn

tr (A BC)

into

tr (B A C)

. Thus the nonzero vector

c \in g

is orthogonal to the basis

a, b, c

, and hence to all of

g

. The bilinear form is therefore degenerate.

Solved by gpt-5.6-sol high.

iv

Solution

Use the Polynomial representation of the Heisenberg Lie algebra on the infinite-dimensional polynomial ring

C [x]

a \cdot f = f^{'}, b \cdot f = x f, c \cdot f = f .

The product rule gives

[d / d x, x] = 1

, so this is a Lie algebra representation. It is a Faithful Lie algebra representation: if

α (d / d x) + β x + γ

is the zero operator, applying it first to

1

gives

β x + γ = 0

, and then applying the remaining operator to

x

gives

α = 0

To prove irreducibility, let

W

be a nonzero invariant subspace and choose a nonzero polynomial in

W

of least degree. If its degree were positive, repeated differentiation would produce a nonzero element of smaller degree, so

W

contains a nonzero constant. Invariance under multiplication by

x

then puts every monomial

x^{n}

W

, and hence

W = C [x]

Solved by gpt-5.6-sol high.

2

i

Solution

Choose the Borel subalgebra

b = t \oplus n^{+}

determined by the positive roots. Regard the one-dimensional space

C_{λ}

as a

b

-module on which

n^{+}

acts by zero and

h \in t

acts by

λ (h)

. The Verma module is

M_{λ} = U (g) \otimes_{U (b)} C_{λ} .

The Poincare-Birkhoff-Witt theorem identifies it as a vector space with

U (n^{-})

acting on a highest-weight vector

v_{λ}

For each positive root

α

, arbitrary powers of a negative-root vector contribute the geometric series

1 + e^{- α} + e^{- 2 α} + \dots

. Consequently the formal character of a weight module is

ch M_{λ} = e^{λ} \prod_{α \in R^{+}} (1 - e^{- α})^{- 1} .

This product is interpreted in the completion of the group algebra in the negative-root direction; the PBW basis proves that every coefficient is the correct finite weight space dimension.

Solved by gpt-5.6-sol high.

ii

Solution

The weights of

M_{λ}

lie below

λ

in the positive-root order, and every weight space is finite-dimensional. A nonzero submodule

V

is stable under the Cartan subalgebra, so it is a direct sum of its weight spaces. Choose a maximal weight

μ

occurring in

V

. Every positive-root operator would raise its weight; maximality therefore makes it kill any nonzero

v \in V_{μ}

. Thus

v

is a singular vector.

The Casimir element is central and acts throughout

M_{λ}

∣ λ + ρ ∣^{2} - ∣ ρ ∣^{2} .

The same element acts on the highest-weight vector

v

of weight

μ

∣ μ + ρ ∣^{2} - ∣ ρ ∣^{2} .

Both are the action of one operator on the same module, so the scalars agree and

∣ μ + ρ ∣ = ∣ λ + ρ ∣.

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iii

Solution

For the sl2 Lie algebra, let

v_{λ}

be the highest-weight vector. The Poincare-Birkhoff-Witt theorem gives the basis

(f^{r} v_{λ})_{r \geq 0}

, and the defining Lie brackets imply

e f^{r} v_{λ} = r (λ - r + 1) f^{r - 1} v_{λ} .

A positive-degree basis vector is singular exactly when

r = λ + 1

is a positive integer. Therefore

M_{λ}

is irreducible when

λ \in / Z_{\geq 0}

λ = m \in Z_{\geq 0}

, the vector

f^{m + 1} v_{λ}

has weight

- m - 2

and generates a submodule isomorphic to

M_{- m - 2}

. The latter is irreducible because

- m - 2 \in / Z_{\geq 0}

. Every nonzero submodule contains a singular vector by the preceding part, and the displayed coefficient shows that this is the only possible proper singular vector. Hence

U (sl_{2}) f^{m + 1} v_{λ} ≅ M_{- m - 2}

is the unique proper nonzero submodule, as summarized by the Reducibility of an sl2 Verma module.

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3

i

Solution

After ordering a symplectic basis in two blocks, write

h = diag (t_{1}, \dots, t_{n}, - t_{1}, \dots, - t_{n}), ε_{i} (h) = t_{i} .

Matrices in the Symplectic Lie algebra have block form

(A C B - A^{T}), B = B^{T}, q u a d C = C^{T} .

The root-space decomposition is

sp_{2 n} = t \oplus ⨁_{i \neq = j} g_{ε_{i} - ε_{j}} \oplus ⨁_{i < j} (g_{ε_{i} + ε_{j}} \oplus g_{- ε_{i} - ε_{j}}) \oplus ⨁_{i} (g_{2 ε_{i}} \oplus g_{- 2 ε_{i}}) .

For example, these one-dimensional spaces are spanned respectively by

E_{ij} - E_{n + j, n + i}, E_{i, n + j} + E_{j, n + i}, E_{n + i, j} + E_{n + j, i}, E_{i, n + i}, E_{n + i, i} .

Thus this is the Cn root system

R = {\pm ε_{i} \pm ε_{j} : i < j} \cup {\pm 2 ε_{i} : 1 \leq i \leq n} .

(5)

The upper-triangular choice gives

R^{+} = {ε_{i} - ε_{j} : i < j} \cup {ε_{i} + ε_{j} : i < j} \cup {2 ε_{i} : 1 \leq i \leq n} .

(6)

Its simple roots, highest root, fundamental weights, and half-sum of positive roots are

α_{i} θ ρ = ε_{i} - ε_{i + 1} = 2 ε_{1}, = n ε_{1} + (n - 1) ε_{2} + \dots + ε_{n} . ω_{k} (1 \leq i < n), = ε_{1} + \dots + ε_{k} α_{n} = 2 ε_{n}, (1 \leq k \leq n),

(7)

Using the notation requested in the paper, the root lattice

P

and weight lattice

Q

are

P = {(m_{1}, \dots, m_{n}) \in Z^{n} : \sum_{i} m_{i} \equiv 0 (mod 2)}, Q = Z^{n},

(8)

Q / P ≅ Z /2 Z

. This reverses the common notation in which the root lattice is called

Q

and the weight lattice is called

P

Since a multiple-edge arrow in a Dynkin diagram points toward the shorter root, the finite and extended diagrams are

α_{1} - α_{2} - \dots - α_{n - 2} - α_{n - 1} ⟸ α_{n}

(9)

and

α_{0} ⟹ α_{1} - α_{2} - \dots - α_{n - 2} - α_{n - 1} ⟸ α_{n}, α_{0} = - 2 ε_{1} .

(10)

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ii

Solution

The Weyl reflection formula

s_{α} (x) = x - ⟨ x, α^{\lor} ⟩ α

becomes especially concrete on

x = (x_{1}, \dots, x_{n})

s_{ε_{i} - ε_{j}} s_{ε_{i} + ε_{j}} s_{2 ε_{i}} : (x_{i}, x_{j}) ⟼ (x_{j}, x_{i}), : (x_{i}, x_{j}) ⟼ (- x_{j}, - x_{i}), : x_{i} ⟼ - x_{i},

with all unlisted coordinates fixed. The Weyl group is therefore the group of signed permutations

W ≅ (Z /2 Z)^{n} ⋊ S_{n} .

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iii

Solution

With

J = (0 - 1 10), A = (p r q s),

the equation

A J + J A^{T} = 0

reduces to

p + s = 0

. Hence

sp_{2} = {(p r q - p) : p, q, r \in C} = sl_{2}

as matrix Lie algebras, with the same commutator bracket.

Solved by gpt-5.6-sol high.

4

i

Solution

Let

R^{+}

be the positive roots,

W

the Weyl group,

ℓ (w)

its Coxeter length,

ρ

the half-sum of positive roots, and

α^{\lor}

a coroot. For a dominant integral highest weight

λ

, the Weyl character formula is

ch L_{λ} = \frac{\sum _{w \in W} ( - 1 ) ^{ℓ (w)} e ^{w (λ + ρ)}}{\sum _{w \in W} ( - 1 ) ^{ℓ (w)} e ^{wρ}} = \frac{\sum _{w \in W} ( - 1 ) ^{ℓ (w)} e ^{w (λ + ρ)}}{e ^{ρ} \prod _{α \in R^{+}} ( 1 - e ^{- α} )} .

Taking the value at the identity gives the Weyl dimension formula

dim L_{λ} = \prod_{α \in R^{+}} \frac{⟨ λ + ρ , α ^{\lor} ⟩}{⟨ ρ , α ^{\lor} ⟩} .

For the q-character convention relevant to the Principal sl2 subalgebra, set

h_{pr} = 2 ρ^{\lor}

, so

α_{i} (h_{pr}) = 2

for every simple root, and define

ch_{q} L_{λ} = \sum_{μ} (dim L_{λ} [μ]) q^{μ (h_{pr})} .

The q-character formula is the principal specialization of the Weyl character formula:

ch_{q} L_{λ} = \frac{\sum _{w \in W} ( - 1 ) ^{ℓ (w)} q ^{⟨ w (λ + ρ), 2 ρ^{\lor} ⟩}}{\sum _{w \in W} ( - 1 ) ^{ℓ (w)} q ^{⟨ wρ, 2 ρ^{\lor} ⟩}} .

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ii

Solution

Choose a short simple root

α_{1}

and a long simple root

α_{2}

, with

∣ α_{2} ∣^{2} = 3∣ α_{1} ∣^{2}

and angle

15 0^{\circ}

. The six positive roots of the G2 root system are

α_{1}, α_{2}, α_{1} + α_{2}, 2 α_{1} + α_{2}, 3 α_{1} + α_{2}, 3 α_{1} + 2 α_{2},

and their negatives complete the two concentric hexagons of short and long roots. The fundamental weights are

ω_{1} = 2 α_{1} + α_{2}, ω_{2} = 3 α_{1} + 2 α_{2},

ω_{1}

is itself a short root and

ω_{2}

is the highest root.

The seven-dimensional representation

L_{ω_{1}}

has weight set

0, \pm α_{1}, \pm (α_{1} + α_{2}), \pm (2 α_{1} + α_{2}),

each with weight multiplicity one. Their positive heights are

1, 2, 3

, so the q-character of a highest-weight representation is

ch_{q} L_{ω_{1}} = q^{6} + q^{4} + q^{2} + 1 + q^{- 2} + q^{- 4} + q^{- 6} .

This is one

sl_{2}

weight string, hence

L_{ω_{1}} ↓ sl_{2}^{pr} ≅ V_{6} .

(5)

The representation

L_{ω_{2}}

is the fourteen-dimensional Adjoint representation. Its nonzero weights are the twelve roots, and its zero-weight space is the two-dimensional Cartan subalgebra. The positive root heights are

1, 1, 2, 3, 4, 5

, so

ch_{q} L_{ω_{2}} = q^{10} + q^{8} + q^{6} + q^{4} + 2 q^{2} + 2 + 2 q^{- 2} + q^{- 4} + q^{- 6} + q^{- 8} + q^{- 10} .

(6)

Splitting this into ordinary

sl_{2}

strings gives

L_{ω_{2}} ↓ sl_{2}^{pr} ≅ V_{10} \oplus V_{2} .

(7)

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5

Solution

Use the B2 root system convention

α_{1} = ε_{2}, α_{2} = ε_{1} - ε_{2}, ω_{1} = \frac{ε _{1} + ε _{2}}{2}, ω_{2} = ε_{1} .

Thus

V = L_{ω_{2}}

is the five-dimensional vector representation of the Special orthogonal Lie algebra

so_{5}

. Label its weight vertices

A = ε_{1}, q u a d B = ε_{2}, q u a d C = 0, q u a d D = - ε_{2}, q u a d E = - ε_{1} .

The crystal basis is the colored chain

A 2 B 1 C 1 D 2 E,

because each Kashiwara operator

f_{i}

subtracts

α_{i}

For the tensor product of crystals, write

X Y

for

X \otimes Y

. The complete colored-arrow graph is compactly specified by

color 1 : color 2 : A B \to A C \to A D, B A \to C A \to D A, BB \to CB \to D B \to D C \to DD, BC \to CC \to C D, BE \to CE \to D E, EB \to EC \to E D; AA \to B A \to BB, A C \to BC, A D \to B D \to BE, C A \to CB, C D \to CE, D A \to E A \to EB, D C \to EC, DD \to E D \to EE .