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I am reading currently Michael Atiyah. I encountered the following proposition. It seems to me that the following condition characterizes prime ideals geometrically that is it tells us how ideal fits between bunch of prime ideals. Does the condition characterizes prime ideals ? That is if ideals satisfy the condition below then it is necessarily prime ?

Conditions:

1) Let $p_1,...,p_n$ be prime ideals and let $a$ be an ideal contained in $\cup p_i$. Then $a \subset p_i$ for some i.

2) Let $a_1,...,a_n$ be ideals and let $p$ be a prime ideal containing $\cap a_i$, then $a_i \subset p$ for some i.

Follow up question: Is there a different description of prime ideals by maybe how it sits among other ideals if the above condition doesn't characterize it ?

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    $\begingroup$ The first part is the prime avoidance theorem, and it is true if at most two ideals aren't prime, which means it is true for any two ideals $\;p_1,p_2\;$ ...The second one is more elementary as the intersection contains the product... $\endgroup$ – DonAntonio Feb 8 '17 at 17:16
  • $\begingroup$ If the ring is PID, the first is true for any ideals, not only primes. $\endgroup$ – Wolfram Feb 8 '17 at 17:17
  • $\begingroup$ It is a famous (unsolved) problem of Kaplansky to characterize the posets that are order isomorphic to ${\rm Spec}(R)$ for Noetherian rings $R$. Some links to the literature: R. Wiegand, Prime ideal in Noetherian rings: a survey, 2009 and S. Sarussi, Totally ordered sets and the prime spectra of rings, 2014 $\endgroup$ – Bill Dubuque Feb 8 '17 at 18:14
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  1. does not characterize prime ideals: there are some rings in which the prime avoidance lemma works for nonprime ideals.

  2. does not characterize prime ideals: Let $R$ be a commutative ring with linearly ordered ideals, but is not a field. Denote its maximal ideal by $M$. Suppose additionally that $M\neq M^2$. (Take $F[x]/(x^3)$ where $F$ is a field, for example.)
    For any two ideals, $A\cap B=\min(A,B)$ according to the linear order of the ideals. Now, $\cap_{i=1}^nA_i\subseteq M^2$ clearly implies that one of $A_i$ is contained in $M^2$ (since the intersection is equal to one of the $A_i$.) But $M$ is obviously not prime since $M\nsubseteq M^2$.
    Note by the same arguments, every ideal in $R$ satisfies the proposed condition, prime or not!

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  • $\begingroup$ If you exchange "intersection" for "ideal product," then you do have a characterization for primes. As this example shows, ideal products have the chance to "get smaller" than intersections do. $\endgroup$ – rschwieb Feb 8 '17 at 21:21
  • $\begingroup$ Do you have a concrete example for 2)? $\endgroup$ – user26857 Feb 8 '17 at 22:54
  • $\begingroup$ @user26857 The ring I describe in 2)? Like $F[x]/(x^3)$? $\endgroup$ – rschwieb Feb 8 '17 at 23:18

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