 Solution

ID: past-exam-of-the-algebra-and-analysis-ii-for-mathematics-and-philosophy-course-of-the-university-of-oxford/2013/1/d/solution

Write

m_{1}, m_{2}

p (T)

vanishes on

U

and

q (T)

vanishes on

W

, part (b) gives

m_{1} ∣ p, m_{2} ∣ q .

Hence

m_{1}, m_{2}

are coprime. By the direct-sum decomposition, a polynomial annihilates

T

exactly when it annihilates both restrictions, which happens exactly when it is divisible by both

m_{1}

and

m_{2}

. Consequently

m_{T} = lcm (m_{1}, m_{2}) = m_{1} m_{2} .

m_{1} = p

and

m_{2} = q

, this immediately gives

m_{T} = pq = f

. Conversely, if

m_{T} = pq

p / m_{1}

and

q / m_{2}

have product one, so each is one. Thus

m_{T} = pq ⟺ m_{T ∣_{U}} = p and m_{T ∣_{W}} = q .

This also handles a zero summand, whose minimal polynomial is one; a nonconstant factor cannot be its minimal polynomial.

 New to topics? Read the docs here!