# Irreducible ideals are prime in polynomial rings

Let $k$ be an algebraically closed field and $R$ the polynomial ring in $n$ variables over $k$. If $J$ is an irreducible ideal of $R$ then it is a prime ideal as well.

To establish this statement I have succeeded to show that J would be primary and it will follow if we can show that it is radical as well... but I am not seeing anything to establish it even using Nullstellensatz... any help is appreciated.

For example $(X^2)\subset k[X]$ is irreducible but not prime.
Remark The correct implications (valid in any noetherian ring) are:$$\text {prime} \implies\text {irreducible} \implies \text {primary}$$
• @tandra There are only two ideals above $(X^2)$: just $(X)$ and the entire ring. You can't find a pair that intersects to $(X^2)$. – rschwieb Aug 11 '14 at 19:45