$p(x)$ irreducible implies $(p(x))$ radical

Question

Reviewing a bit of algebra, the following concern arose me:

Let $p(x)$ be a irreducible polynomial in $K[x_1,x_2,..., x_n]$. Is $(p(x))$ radical?

I think the answer is yes, however I have not been able to find a proof.

I know that to prove this I just must to verify that, if $p|qh$, then $p|q$ or $p|h$ by irreducibility. But this is not clear for me...

Answer 1

If $p(x)$ is irreducible, then $(p(x))$ is prime. Prime ideals are radical. Let $P$ be a prime ideal of a ring $R$. Then $x^n\in P$ implies $x\in P$ as $P$ is prime, so $P$ is radical.