Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2012/3/c/solution

Set to define a surjective ring homomorphism
Its 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 gives
The 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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