Question about ideals

Question

I believe the following is a true statement, but I am unsure, so I wanted to check with people.

If $p$ is an irreducible polynomial in $n$ indeterminates then $(p)$, the ideal generated by it, is prime.

My idea is: Let $gh \in (p)$ this implies that $f$ divides $g$ or $h$. Thus one of them is in $(p)$, the problem is that I never used the irreducible part.

Answer 1

The fact that $f$ divides either $g$ or $h$ uses that $f$ is irreducible. Otherwise its irreducible components might be distributed amongst $g$ and $h$.

Answer 2

This depends on the fact that the ring of polynomials in $n$ indeterminates (over a field) is a unique factorization domain (you find the proof in every starter's textbook on ring theory).

If $p$ divides $gh$, then $gh=pq$, so, by unique factorization, $p$ must be among the irreducible factors of $g$ or of $h$ (up to multiplications by nonzero constants).