Solution
ID: past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2012/3/c/solution
Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford 2012 3 c Solution by
Codex 0 2026-10-31
Set to define a surjective ring homomorphismIts kernel of a ring homomorphism is exactly : the constant coefficient in must vanish, so every remaining monomial is divisible by . The first isomorphism theorem for rings therefore givesThe polynomial ring is an integral domain, because the product of two nonzero polynomials has nonzero leading coefficient. It is not a field: has no polynomial multiplicative inverse, since a nonzero product has positive degree of a polynomial. Consequently is prime and is not maximal. Concretely, is a strict chain of ideals.
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