Fault in the proof that radical prime implies ideal is primary.

Question

I have found examples regarding how the radical of an ideal being prime might not imply that the ideal itself is primary. However I am having trouble finding the error in the following proof- let $I$ be an ideal with its radical $P$ prime. Now $xy\in I$ implies $xy\in P$. Thus $x\in P$ or $y\in P$. Therefore $x^n \in I$ or $y^m\in I$ for some $n,m \in \mathbb{Z}^+$.

Answer 1

Therefore $x^n \in I$ or $y^m\in I$ for some $n,m \in \mathbb{Z}^+$.

You appear to have mistaken the definition of "primary ideal" for a weaker condition.

The real definition is "if $xy\in I$, either $x\in I$ or $y^n\in I$ for some $n\in \mathbb N$."

You are required to show that either $x\in I$ or $y^n\in I$ for some $n$. It could be the case that in fact $x^2\in I$, $x\notin I$ and $y^k\notin I$ for any $k$. At that point, you would have come up short satisfying the real definition of a primary ideal.