Explore BrainMass

Explore BrainMass

    Prime Ideals and Irreducible Polynomials

    Not what you're looking for? Search our solutions OR ask your own Custom question.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!


    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].

    © BrainMass Inc. brainmass.com March 4, 2021, 6:18 pm ad1c9bdddf

    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 and discussed The solution is detailed and well presented.