Let $f(x)$ be irreducible in $F[x]$, where $F$ is a field. If $f(x) | p(x)q(x)$, prove that either $f(x) | p(x)$ or $f(x) | q(x)$.

Question

Let $f(x)$ be irreducible in $F[x]$, where $F$ is a field. If $f(x) | p(x)q(x)$, prove that either $f(x) | p(x)$ or $f(x) | q(x)$.

Is this true because every irreducible element in a field is prime?

Answer 1

Prove it by contradiction(using factorisation of $p(x)\ and\ q(x)$),i.e.,

Case 1: Both of p(x) and q(x) are irreducible

This is not possible because then $p(x)$ and $q(x)$ are the only factors of $p(x)q(x)$.

Case 2: Exactly one of them is reducible

In this case, $f(x)$ must be a factor of the reducible one($\because$ only factors of $p(x)q(x)$ are the irreducible polynomial and the factors of the reducible one)

Case 3: Both of them are reducible

Say $$p(x) = \prod_{i=1}^mg_i^{k_i}(x)$$ $$q(x) = \prod_{i=1}^nh_i^{l_i}(x)$$ Then, $f(x)$ being a factor of $p(x)q(x)$, must be factored in terms of $g_i(x)$ and $h_i(x)$,which is a contradiction to its irreducbility.

Hope it is helpful

Answer 2

Yes, every irreducible element in $F[x]$ geberates a prime ideal, since $F[x]$ (polynomial ring in one indeterminate over a field) is a P.I.D.

The same would be true for several indeterminates since $F[x_1,\dots ,x_n]$ is a U.F.D.

Answer 3

$F[X]$ with $F$ a field is a unique factorisation domain, so the irreducible elements are prime and the conclusion is straightforward.