We know that if is a field and if is a variable over then is a PID and a non-zero ideal of is maximal if and only if is prime if and only if is generated by an irreducible element of If is a PID which is not a field, then could have prime ideals which are not maximal. For example, in the ideal is prime but not maximal. In this two-part post, we will find prime and maximal ideals of when is a PID.
Notation. Throughout this post, is a PID and is the polynomial ring in the variable over Given a prime element we will denote by the natural ring homomorphism
Definition Let be a prime element of An element is called irreducible modulo if is irreducible in Let be the natural ring homomorphism. Then, since an element is irreducible modulo if and only if is irreducible in Note that is a field because is a PID.
Problem 1. Prove that if is prime and if is irreducible modulo then is a maximal ideal of If then is a prime but not a maximal ideal of
Solution. Clearly So is a maximal ideal of because is irreducible in and is a field. So is a maximal ideal of If then and so is a domain which implies that is prime. Finally, is not maximal because, for example,
Problem 2. Prove that a non-zero ideal of is prime if and only if either for some irreducible element or for some prime and some which is either zero or irreducible modulo
Solution. If is irreducible, then is a prime ideal of because is a UFD. If or is irreducible modulo a prime then is a prime ideal of by Problem 1.
Conversely, suppose that is a non-zero prime ideal of We consider two cases.
Case 1. : Let Then is clearly not a unit because then wouldn’t be a proper ideal of So, since and is a prime ideal of there exists a prime divisor of such that So and hence is a prime ideal of Thus we have two possibilities. The first possibility is that which gives us and therefore The second possibility is that for some irreducible element which gives us because
Case 2. : Let be the field of fractions of and put Then is a non-zero prime ideal of because is a prime ideal of Note that So, since is a PID, for some irreducible element Obviously, we can write where and is irreducible and the gcd of the coefficients of is one. Thus and, since we have for some and But then and so because is prime and Hence We will be done if we prove that To prove this, let So for some Therefore, since the gcd of the coefficients of is one, we must have by Gauss’s lemma. Hence and the solution is complete.
See the next part here!