 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-1/18g/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

K \subseteq E \subseteq L,

one has

[L : K] = [L : E] [E : K] .

To prove it, let

(u_{i})_{i = 1}^{m}

be a

K

E

and

(v_{j})_{j = 1}^{n}

E

L

. Every

x \in L

can be written as

\sum_{j} a_{j} v_{j}

with

a_{j} \in E

, and expanding each

a_{j}

in the basis

(u_{i})

shows that the

mn

products

u_{i} v_{j}

span

L

over

K

\sum_{i, j} c_{ij} u_{i} v_{j} = 0, c_{ij} \in K,

then independence of the

v_{j}

over

E

gives

\sum_{i} c_{ij} u_{i} = 0

for every

j

, and independence of the

u_{i}

over

K

gives every

c_{ij} = 0

. Thus

(u_{i} v_{j})

is a

K

L

, proving the formula.

 New to topics? Read the docs here!