(a) Prove that every maximal ideal of has the form , where is an integer prime and is a primitive integer polynomial that is irreducible modulo
(b) Let be an ideal of generated by two polynomials and that have no common factors other than . Prove that is finite.
Part A
Let be a maximal ideal of . Suppose does not contain an integer. Lifting this ideal to , it will become principal with generator, say . Let , where is primitive, and be a prime that does not divide any coefficients of . Then , and is not maximal. Thus by contradiction, must contain an integer.
Let . If is not prime, then we can write for integers . Then , suggesting the existence of a zero divisor and contradiction the property of integral domain. Thus must be prime.
Let be a prime, and consider the ring homomorphism obtained by modding out . is a PID, so all maximal ideals are of the form , where is an irreducible polynomial. There is a bijective correspondence between maximal ideals of containing and the maximal ideals of . Therefore the maximal ideals of containing a prime are precisely the of the form where is irreducible modulo .
Part B
The unique factorization of and must not have any shared terms. Lifting these polynomials to , we have . With Bezout’s Identity, there exist such that . Then there exists primitives such that for some constant
the rest of the proof is left as an exercise
Student Question
Question
Show that .
We have previously shown that , so has degree 4.
Note that , so . Suppose , implying . This gives a contradiction, as these two field extensions have different degrees.
