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Prime Ideals and Irreducible Polynomials


1. Let F be a field and p(x) and irreducible polynomial over F. Prove that (p(x)) is a prime ideal of F[x].

2. If R has no divisors of zero, then neither does R[x].

Solution Preview

1. Proof:
Suppose f(x), g(x) is in F[x] and f(x)g(x) belongs to (p(x)). This means that p(x) divides f(x)g(x). But we know p(x) is irreducible, then p(x) divides f(x) or p(x) divides g(x). If p(x) divides f(x), then f(x) belongs to (p(x)); if p(x) divides g(x), then g(x) belongs to (p(x)). Therefore, (p(x)) is a prime ideal over ...

Solution Summary

Prime Ideals and Irreducible Polynomials are investigated. The solution is detailed and well presented. The response received a rating of "5" from the student who originally posted the question.