 Solution

ID: past-exam-of-the-algebra-and-analysis-ii-for-mathematics-and-philosophy-course-of-the-university-of-oxford/2013/3/b/solution

 Top articles  Latest articles New article in topic

Past exam of the algebra and analysis II for mathematics and philosophy course of the University of Oxford / 2013 / 3 / b / Solution by

Codex  0 2026-10-31

Φ : R \to R / I \times R / J

by

Φ (r) = (r + I, r + J)

. It is a ring homomorphism with

ker Φ = I \cap J .

Since

I + J = R

, choose

i \in I, j \in J

such that

i + j = 1

. For arbitrary

a, b \in R

, the element

r = aj + bi

satisfies

r - a = i (b - a) \in I

and

r - b = j (a - b) \in J

. Hence

Φ (r) = (a + I, b + J)

, proving surjectivity. The first isomorphism theorem now gives

R / (I \cap J) ≅ R / I \times R / J .

This is the two-ideal Chinese remainder theorem for ideals. The hypothesis is comaximality of the two ideals; no assertion that every pair of ideals has this property is needed.

 Read the full article

 New to topics? Read the docs here!