Past exam of the mathematics course of the University of Cambridge / 2025 / iii

No. Take

R = Z

M = Q

, and

N_{i} = Z / i Z (i \geq 2) .

Every

N_{i}

is a torsion group, so localization at the nonzero integers gives

Q \otimes_{Z} N_{i} = 0

and hence

\prod_{i} (Q \otimes_{Z} N_{i}) = 0

By contrast, the element

x = (1 mod i)_{i \geq 2}

\prod_{i} N_{i}

has infinite order: no positive integer is divisible by every

i

. The canonical description

Q \otimes_{Z} \prod_{i} N_{i} ≅ (Z ∖ {0})^{- 1} (\prod_{i} N_{i})

shows that

1 \otimes x

is nonzero, because an element dies in this localization only if one nonzero integer annihilates it. Thus the left module is nonzero while the right module is zero, giving the tensor product and infinite direct product counterexample.

Solved by gpt-5.6-sol high.

iv

a

Solution

Assume tensor products of two nonzero modules never vanish. If

m \neq = n

were distinct maximal ideals, then

m + n = R

and

(R / m) \otimes_{R} (R / n) ≅ R / (m + n) = 0,

although both residue fields are nonzero. Hence

R

has one maximal ideal

m

and is a local ring.

For every module

M

(R / m) \otimes_{R} M ≅ M / m M .

m M = M

, this tensor product vanishes. Since

R / m \neq = 0

, the assumed property forces

M = 0

. Thus condition (a) implies condition (b).

Solved by gpt-5.6-sol high.

b

Solution

Conversely, assume condition (b), and let

M, N

be nonzero. The stated property gives

M / m M \neq = 0, N / m N \neq = 0.

These are nonzero vector spaces over the residue field

k = R / m

, so their tensor product over

k

is nonzero. Associativity and base change give

(M \otimes_{R} N) \otimes_{R} k ≅ (M \otimes_{R} k) \otimes_{k} (N \otimes_{R} k) ≅ (M / m M) \otimes_{k} (N / m N) \neq = 0.

Therefore

M \otimes_{R} N \neq = 0

. This proves the reverse implication and the local tensor nonvanishing criterion.

Solved by gpt-5.6-sol high.

v

a

Solution

There is an exact sequence of

R

-modules

0 ⟶ I \cap J x \mapsto (x, x) I \oplus J (x, y) \mapsto x - y I + J ⟶ 0.

A

is a flat module over

R

, tensoring this sequence with

A

preserves its left exactness. The image of each tensor product inside

A

is the corresponding extension of an ideal, so

I^{e} \cap J^{e} = (I \cap J)^{e} .

Both displayed inclusions therefore hold. This is the flat extension preserves finite ideal intersections property.

Solved by gpt-5.6-sol high.

b

Solution

The inclusion

(I \cap J)^{e} \subseteq I^{e} \cap J^{e}

always holds: each generator coming from

I \cap J

lies in both extended ideals.

The reverse inclusion can fail. Let

R = k [x, y], A = R / (x - y), I = (x), J = (y)

for a field

k

. Under

A ≅ k [x]

I^{e} = J^{e} = (x), I^{e} \cap J^{e} = (x),

whereas

I \cap J = (x y)

R

, so

(I \cap J)^{e} = (x^{2}) ⊊ (x) .

Thus statement (2) is true in general and statement (1) is false in general.

Solved by gpt-5.6-sol high.

2

i

Solution

A multiplicative subset

S \subseteq R

contains

1

and satisfies

s, t \in S \Rightarrow s t \in S

. The localization of a ring consists of fractions

r / s

modulo the relation

\frac{r}{s} = \frac{r ^{'}}{s ^{'}} ⟺ some u \in S satisfies u (r s^{'} - r^{'} s) = 0.

Its structure map

ι : R \to S^{- 1} R

makes every

s \in S

invertible. The universal property of localization says that if

f : R \to T

is any ring homomorphism for which every

f (s)

is a unit, there is one unique homomorphism

f : S^{- 1} R \to T

satisfying

f ι = f

, namely

f (r / s) = f (r) f (s)^{- 1} .

Solved by gpt-5.6-sol high.

ii

Solution

The local criterion states that an

R

-module

M

is flat if and only if

M_{p}

is flat over

R_{p}

for every prime ideal

p

. It is enough equivalently to test maximal ideals.

M

is flat, localization of an exact sequence and the natural isomorphism

(M \otimes_{R} N)_{p} ≅ M_{p} \otimes_{R_{p}} N_{p}

show immediately that every

M_{p}

is flat.

Conversely, let

N^{'} \to N

be injective and let

K

be the kernel of

M \otimes_{R} N^{'} ⟶ M \otimes_{R} N .

After localization at any prime

p

, flatness of

M_{p}

gives

K_{p} = 0

. A module whose localization at every maximal ideal is zero must itself be zero: if

0 \neq = x \in K

, its annihilator is contained in a maximal ideal

m

, and then

x /1 \neq = 0

K_{m}

. Hence

K = 0

, tensoring by

M

preserves every injection, and

M

is flat. This proves that flatness is local.

Solved by gpt-5.6-sol high.

iii

a

Solution

Put

S = A ∖ p

and define

B_{p} = S^{- 1} B,

where

A

acts on

B

through the given inclusion. By the prime ideal correspondence for localization, primes of

B_{p}

correspond to primes

q

B

satisfying

q \cap S = \emptyset

, equivalently

q \cap A \subseteq p

The localized extension

A_{p} \subseteq B_{p}

remains integral. If

q \cap A = p

, then

B_{p} / S^{- 1} q

is an integral domain integral over the field

A_{p} / p A_{p}

. An integral domain integral over a field is a field, so

S^{- 1} q

is maximal.

Conversely, if

n

is maximal in

B_{p}

, the contraction of a maximal ideal under an integral extension is maximal in the local ring

A_{p}

, hence equals

p A_{p}

. Contracting once more to

A

gives

p

. Extension and contraction are inverse under localization, proving the required fiber primes of an integral extension bijection.

Solved by gpt-5.6-sol high.

b

Solution

Take

A = Z \subseteq B = Q, p = (2) .

This extension is not integral. Since every nonzero integer is already invertible in

Q

B_{p} = Q

has the maximal ideal

(0)

. On the other hand, the only prime ideal of

B

(0)

, whose contraction to

Z

(0)

rather than

(2)

. The set of primes of

B

lying over

p

is therefore empty while

MaxSpec B_{p}

is not.

Solved by gpt-5.6-sol high.

iv

Solution

Let

m

be maximal in

R / I

and let

n \subseteq R

be its contraction. The ideal

n

is maximal and contains

I

. It is disjoint from

S = 1 + I

: if

1 + i \in n

with

i \in I \subseteq n

, then

1 \in n

. The prime ideal correspondence for localization therefore defines the proper ideal

S^{- 1} n

, and

S^{- 1} R / S^{- 1} n ≅ R / n

is a field. Thus

f (m) = S^{- 1} n

is maximal.

Conversely, let

q

be the contraction to

R

of a maximal ideal of

S^{- 1} R

. Then

q

is maximal among ideals disjoint from

S

. Since

q + I

is disjoint from

S

—otherwise an equation

q + i = 1 + j

would put

1 + (j - i) \in q \cap S

—maximality gives

I \subseteq q

. If a proper ideal strictly contained

q

in a maximal ideal

n

R

, then

n

would also contain

I

and hence remain disjoint from

S

, a contradiction. Thus

q

is maximal and contains

I

The two standard extension-contraction bijections, first for

R \to R / I

and then for

R \to S^{- 1} R

, now give inverse maps

MaxSpec (R / I) ≅ MaxSpec (S^{- 1} R) .

Solved by gpt-5.6-sol high.

3

i

Solution

For an ideal

I \subseteq k [T_{1}, \dots, T_{n}]

, define

V (I) = {a \in k^{n} : f (a) = 0 for every f \in I} .

For

X \subseteq k^{n}

, define

I (X) = {f : f (a) = 0 for every a \in X},

and recall that

I = {f : f^{r} \in I for some r \geq 1}

is the radical of an ideal.

For an algebraically closed field

k

, the Weak Hilbert Nullstellensatz says that every maximal ideal of

k [T_{1}, \dots, T_{n}]

(T_{1} - a_{1}, \dots, T_{n} - a_{n})

for a unique

a \in k^{n}

, equivalently every proper ideal has a common zero. The Strong Hilbert Nullstellensatz says

I (V (I)) = I .

To prove the weak form, let

m

be maximal. The residue field

K = k [T_{1}, \dots, T_{n}] / m

(5)

is a field finitely generated as a

k

-algebra. By the Zariski lemma,

K / k

is finite algebraic; algebraic closedness gives

K = k

. If

a_{i}

is the image of

T_{i}

, the quotient map is evaluation at

a = (a_{1}, \dots, a_{n})

and its kernel is

(T_{1} - a_{1}, \dots, T_{n} - a_{n})

. This proves the assertion.

Solved by gpt-5.6-sol high.

ii

Solution

Let

m

be maximal in

Q [T_{1}, \dots, T_{n}]

and put

K = Q [T_{1}, \dots, T_{n}] / m .

The Zariski lemma makes

K

a finite extension of

Q

. Base change gives

C [T_{1}, \dots, T_{n}] / m C [T_{1}, \dots, T_{n}] ≅ C \otimes_{Q} K .

This ring is nonzero because the field extension makes

C

a faithfully flat module over

Q

. Choose a maximal ideal

n

of the quotient, or equivalently a maximal ideal of

C [T_{1}, \dots, T_{n}]

containing

m C [T_{1}, \dots, T_{n}]

. Its contraction to the rational polynomial ring contains

m

and is proper, so maximality of

m

forces

n \cap Q [T_{1}, \dots, T_{n}] = m .

Solved by gpt-5.6-sol high.

iii

Solution

Let

a = (X Y^{3}, X^{2} (Y - 3)) .

Certainly

a \subseteq (X)

, so

a \subseteq (X)

. Conversely, cubing the second generator shows that

a

contains

X^{6} (Y - 3)^{3}

, while multiplying the first generator by

X^{5}

shows that it contains

X^{6} Y^{3}

. The ideals generated by

Y^{3}

and

(Y - 3)^{3}

are comaximal ideals, so a polynomial combination of the two polynomials is

1

. Multiplication by

X^{6}

gives

X^{6} \in a

, hence

X \in a

. Therefore

a = (X),

generated by the single polynomial

X

Solved by gpt-5.6-sol high.

iv

Solution

Write the finitely generated algebra as

A ≅ C [T_{1}, \dots, T_{n}] / I .

By the Weak Hilbert Nullstellensatz,

C

-algebra homomorphisms

A \to C

correspond exactly to the points of the affine algebraic set

V (I)

V (I)

is finite, its cardinality is finite. If it is infinite, the complex affine algebraic set cardinality dichotomy gives

∣ V (I) ∣ = ∣ C ∣,

which is uncountable. This also covers the zero algebra, whose homomorphism set is empty. Therefore

∣ Hom_{C} (A, C) ∣ \neq = ℵ_{0} .

Solved by gpt-5.6-sol high.

4

i

Solution

For a prime ideal

p

, its height is the supremum of lengths

r

of strict chains

p_{0} ⊊ p_{1} ⊊ \dots ⊊ p_{r} = p .

The height of an ideal

I

, without assuming

I

prime, is

ht (I) = p \supseteq I in f ht (p) .

The Krull dimension of a ring

R

dim R = sup {r : p_{0} ⊊ \dots ⊊ p_{r} are prime ideals of R} .

Solved by gpt-5.6-sol high.

ii

a

Solution

For the maximal ideal

m

(A, m)

, the associated graded ring is

G_{m} (A) = gr_{m} (A) = ⨁_{n \geq 0} m^{n} / m^{n + 1} .

x + m^{r + 1}

and

y + m^{s + 1}

are homogeneous classes, their product is

x y + m^{r + s + 1} .

It is a graded algebra over the residue field

k = A / m

The Hilbert series is

H_{G_{m} (A)} (t) = \sum_{n \geq 0} dim_{k} (m^{n} / m^{n + 1}) t^{n} .

The Hilbert-Serre theorem makes this rational. The number

d (G_{m} (A))

is the order of its pole at

t = 1

, as recorded by the pole dimension of an associated graded ring.

The Dimension theorem for Noetherian local rings states

dim A = dim G_{m} (A) = d (G_{m} (A)) .

Solved by gpt-5.6-sol high.

b

Solution

Put

\overline{A} = A / (x)

and

\overline{m} = m / (x)

. Applying the dimension theorem to

A

and

\overline{A}

gives

d (G_{\overline{m}} (\overline{A})) = dim \overline{A}, d (G_{m} (A)) = dim A .

Since

x

is a non-zero-divisor, it belongs to no minimal prime of the Noetherian ring

A

. Any chain of primes in

A / (x)

lifts to a chain

p_{0} ⊊ \dots ⊊ p_{r}

of primes of

A

containing

x

. A minimal prime

q \subseteq p_{0}

cannot contain

x

, so the inclusion is strict. Prepending

q

gives a chain of length

r + 1

A

. Thus the dimension drop by a non-zero-divisor gives

dim (A / (x)) \leq dim A - 1.

Combining these equalities proves

d (G_{m / (x)} (A / (x))) \leq d (G_{m} (A)) - 1 .

Solved by gpt-5.6-sol high.

iii

Solution

The polynomial ring

C [X, Y]

is a two-dimensional unique factorization domain. Let

p

be a prime ideal containing

(f, g)

. Since

f \neq = 0

, the prime

p

is nonzero. If it were not maximal, it would have height one, so the height-one prime in a unique factorization domain would give

p = (h)

for an irreducible polynomial

h

. Then

h

would divide both

f

and

g

, contrary to the hypothesis. Every prime of

R = C [X, Y] / (f, g)

is therefore maximal.

The Hilbert basis theorem makes

R

Noetherian, and it has Krull dimension zero by the preceding paragraph. The Noetherian dimension-zero criterion for an Artinian ring now shows that

R is Artinian .

Solved by gpt-5.6-sol high.

5

i

Solution

A discrete valuation on a field

K

is a surjective group homomorphism

v : K^{\times} ⟶ Z

satisfying

v (x + y) \geq min {v (x), v (y)}

whenever

x + y \neq = 0

. Its discrete valuation ring is

R_{v} = {0} \cup {x \in K^{\times} : v (x) \geq 0} .

One standard characterization defines a Dedekind domain as a Noetherian integrally closed domain of Krull dimension one. Equivalently, all its localizations at nonzero prime ideals are discrete valuation rings.

Solved by gpt-5.6-sol high.

ii

Solution

Choose

0 \neq = a \in m

. Since the one-dimensional local domain has no nonzero prime ideal other than

m

, one has

(a) = m

. Finite generation of

m

therefore gives some

n \geq 1

with

m^{n} \subseteq (a)

. Choose

n

minimal and

x \in m^{n - 1} ∖ (a) .

In the fraction field of

A

, put

y = x / a

. Then

y \in / A

but

y m \subseteq A .

y m \subseteq m

, multiplication by

y

would preserve the nonzero finitely generated faithful

A

-module

m

. The determinant trick would make

y

integral over

A

, contradicting that

A

is integrally closed and

y \in / A

. Hence some

t \in m

satisfies

y t \in / m

. Since

y t \in A

and

A

is local,

y t

is a unit.

For any

z \in m

, one has

yz \in A

, and therefore

\frac{z}{t} = \frac{yz}{y t} \in A .

Thus

m \subseteq (t)

, while the reverse inclusion follows from

t \in m

. Consequently

m = (t) .

This proves the principal maximal ideal in a one-dimensional normal local domain result.

Solved by gpt-5.6-sol high.

iii

a

Solution

Let

J

be an ideal of

S^{- 1} A

and contract it to an ideal

I

A

. Every element

a / s \in J

has

a /1 = (s /1) (a / s) \in J

, so

a \in I

and

J = S^{- 1} I .

Because

A

is a Noetherian ring, write

I = (a_{1}, \dots, a_{r})

. Then

J = (a_{1} /1, \dots, a_{r} /1),

so every ideal of

S^{- 1} A

is finitely generated. Hence every localization of a Noetherian ring is Noetherian, proving the Localization of a Noetherian ring theorem.

Solved by gpt-5.6-sol high.

b

Solution

Because

0 \in / S

, the localization

S^{- 1} A

remains an integral domain. Part (a) makes it Noetherian. Integral closedness is preserved by localization: if

x

in the common fraction field is integral over

S^{- 1} A

, clearing the finitely many denominators in a monic equation shows that

s x

is integral over

A

for some

s \in S

, whence

s x \in A

and

x \in S^{- 1} A

The prime ideal correspondence for localization shows that every chain of primes in

S^{- 1} A

comes from a chain in

A

, so

dim S^{- 1} A \leq 1.

If its dimension is one, it is a Noetherian integrally closed domain of dimension one and hence a Dedekind domain. If its dimension is zero, its zero ideal is maximal, so the domain is a field. This proves the Localization of a Dedekind domain alternative.

Solved by gpt-5.6-sol high.

Paper 102 (Lie Algebras and Their Representations)

www.maths.cam.ac.uk/postgrad/part-iii/files/pastpapers/2025/III%20Paper%20102.pdf

1

a

Solution

The Adjoint representation of a Lie algebra is

ad : g ⟶ gl (g), ad_{x} (y) = [x, y] .

The Killing form is the symmetric invariant bilinear form

κ (x, y) = tr (ad_{x} ad_{y}) .

Solved by gpt-5.6-sol high.

b

Solution

A vector subspace

I \subseteq g

is an ideal of a Lie algebra when

[g, I] \subseteq I

. A Nilpotent Lie algebra is one whose lower central series

g = γ_{1} (g), γ_{r + 1} (g) = [g, γ_{r} (g)]

eventually vanishes.

By the Engel theorem, the operators

ad_{x}

for a nilpotent complex Lie algebra can be represented simultaneously by strictly upper triangular matrices. Their products are strictly upper triangular and have zero trace. Hence

κ (x, y) = 0 for every x, y \in g .

Solved by gpt-5.6-sol high.

c

Solution

A Solvable Lie algebra is one whose derived series

g^{(0)} = g, g^{(r + 1)} = [g^{(r)}, g^{(r)}]

eventually vanishes. By the Lie theorem, the adjoint operators of a solvable complex Lie algebra are simultaneously upper triangular. If

x \in [g, g]

, then

ad_{x}

is a sum of commutators of upper triangular matrices and is therefore strictly upper triangular. For every

y \in g

, the product

ad_{x} ad_{y}

is strictly upper triangular, so

κ (x, y) = 0.

Thus

[g, g] \subseteq g^{⊥} .

For a nonzero example, let

g

have basis

h, e

with

[h, e] = e

. It is solvable because

[g, g] = C e

is abelian, but in the ordered basis

(h, e)

ad_{h} = (0001), κ (h, h) = 1.

Solved by gpt-5.6-sol high.

d

Solution

Invariance of the Killing form gives

κ ([a, x], y) = κ (x, [y, a]) .

x \in g^{⊥}

, the right-hand side vanishes for every

y

, so

[a, x] \in g^{⊥}

. Thus

g^{⊥}

is an ideal.

Let

h = ad (g^{⊥}) \subseteq gl (g)

. For

x \in [g^{⊥}, g^{⊥}]

and

y \in g^{⊥}

tr (ad_{x} ad_{y}) = κ (x, y) = 0.

The Cartan solvability criterion makes

h

solvable. The kernel of the adjoint map on

g^{⊥}

lies in its center and is abelian, so

g^{⊥}

is itself solvable. This proves the Solvability of the radical of the Killing form.

Solved by gpt-5.6-sol high.

e

Solution

The Killing form of a complex Simple Lie algebra is nondegenerate. Since

B

is also nondegenerate, there is a unique endomorphism

T

g

satisfying

κ (x, y) = B (T x, y) .

Invariance of both forms gives

T ([z, x]) = [z, T x],

T

intertwines the adjoint representation. That representation is irreducible because its invariant subspaces are ideals. The Schur lemma therefore gives

T = c I

. Since

κ

is nondegenerate,

c \neq = 0

, and

κ = c B .

This is the uniqueness of an invariant bilinear form on a simple Lie algebra.

Solved by gpt-5.6-sol high.

f

Solution

For

sl_{3} (C)

, the Killing form of the special linear Lie algebra is

κ (X, Y) = 6 tr (X Y) .

For

H_{1} = E_{11} - E_{22}

and

H_{2} = E_{22} - E_{33}

, its matrix on the Cartan part is

(12 - 6 - 6 12),

whose determinant is

108

. Each

E_{ij}

pairs only with

E_{ji}

, with value

6

. In the stated ordering, the remaining block is

(0 6 I_{3} 6 I_{3} 0),

whose determinant is

- 6^{6}

. Therefore

det κ = - 108 \cdot 6^{6} = - 5038848 .

Solved by gpt-5.6-sol high.

2

a

Solution

A Weyl chamber is a connected component of

E ∖ ⋃_{α \in Φ} α^{⊥} .

A root basis

Δ

is a basis of

E

made of roots such that every root is an integer combination of

Δ

whose nonzero coefficients all have one sign.

Choose a regular vector

γ

, meaning

(γ, α) \neq = 0

for every root. Declare

Φ_{γ}^{+} = {α \in Φ : (γ, α) > 0} .

The indecomposable roots in

Φ_{γ}^{+}

form a root basis

Δ_{γ}

, and every root basis arises in this way. Its chamber is the component containing

γ

Solved by gpt-5.6-sol high.

b

Solution