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Answer to Question #20014 in Abstract Algebra for Matthew Lind

Question #20014
Let F be a field and let J be an ideal in F[X]. Prove that J is prime if and only if its generator is irreducible over F.
Expert's answer
Let F be a field and let J be an ideal in F[X]. Prove that J is prime if and only if its generator is irreducible over F.

Answer.
Since F is a field, each ideal J in F[X] is principal, so J = < p(X) >, where p(X) is a certain polynomial called generator of J, and so every element of J has the form c(X)*p(X), where c(X) runs over all F[X].

Suppose the generator p(X) of J is not reducible, so p(X)=a(X)*b(X) for some polynomials a(X) and b(X) such that deg a, deg b > 0 Then a(X) and b(X) does not belong to J, while their product p(X) does so.
Hence J is not prime.

Conversely, if J is not prime, that is there are polynomials a(X) and b(X) with deg a, deg b > 0 and such that
a(X)*b(X)
belongs to J, so
& a(X)*b(X) = c(X)*p(X)

Every element of F[X] can be uniquely (up to order) written as a product of irreducible polynomials.
Therefore if p(X) is prime, the either a or b is divided by p, and so either a or b belongs to J, which contradicts to the assumption.
Hence p is not prime.

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