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ID: past-exam-of-the-algebra-1-and-differential-equations-1-course-of-the-university-of-oxford/2014/1/b/ii/solution

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Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2014 / 1 / b / ii / Solution by

Codex  0 2026-10-31

Write

U = ker p (T)

,

W = ker q (T)

. The linear operator

E = a (T) p (T)

vanishes on

U

. On

W

, the identity

E = I - b (T) q (T)

gives

Ew = w

.

By the direct sum in part (i), every vector has a unique expression

u + w

, and

E (u + w) = w .

Consequently

E^{2} = E, im E = ker q (T), ker E = ker p (T) .

The complementary projection operator is

I - E = b (T) q (T)

, onto

U

along

W

. These are polynomial projections for a coprime decomposition. In particular the projection does not depend on which Bézout coefficients are chosen: its action on the two summands uniquely determines it.

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