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 $n,m \in \mathbb{Z}^+$.
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If you do have an example as you say in the first sentence, then it will break the logic of your proof somewhere. If you're still not convinced, you should mention what example you have in mind. – rschwieb Feb 09 '24 at 12:45
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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.
rschwieb
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