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Prime Ideals and Irreducible Polynomials

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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].

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Solution Summary

Prime ideals and irreducible polynomials are investigated and discussed The solution is detailed and well presented.

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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 ...

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