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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.

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You cannot "establish this statement" because it is false!
For example $(X^2)\subset k[X]$ is irreducible but not prime.

Remark The correct implications (valid in any noetherian ring) are:$$ \text {prime} \implies\text {irreducible} \implies \text {primary} $$

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  • $\begingroup$ +1 very simple. Maybe the OP was hoping that the "irreducible implies prime" statement for elements in such a polynomial ring also held for ideals... but they're quite different. $\endgroup$ – rschwieb Aug 11 '14 at 15:37
  • $\begingroup$ Dear @rschwieb, you are right: it is indeed confusing that non irreducible elements may generate irreducible principal ideals... $\endgroup$ – Georges Elencwajg Aug 11 '14 at 15:54
  • $\begingroup$ Thank you..could you just tell me how can we argue here that (X^2) is irreducible ideal..it is primary that I have understood!!... @ Georges Elencwajg $\endgroup$ – tandra Aug 11 '14 at 18:16
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    $\begingroup$ @tandra There are only two ideals above $(X^2)$: just $(X)$ and the entire ring. You can't find a pair that intersects to $(X^2)$. $\endgroup$ – rschwieb Aug 11 '14 at 19:45

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