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Consider the following question where the ring is assumed to be commutative.

For ideals $I, J$ of a ring $R$, their product $I J$ is defined as the ideal of $R$ generated by the elements of the form $x y$ where $x \in I$ and $y \in J$.

  1. Prove that, if a prime ideal $P$ of $R$ contains $I J$, then $P$ contains either $I$ or $J$.
  2. Give an example of $R, I$ and $J$ such that the two ideals $I,J$ and $I \cap J$ are different from each other.

The first part is easy enough. For the second part I came up with the example $I=J=3\mathbb{Z}$ as then $IJ = 9\mathbb{Z}$ and $I \cap J = 3\mathbb{Z}$. However, I am worried that the question may be asking for distinct $I$ and $J$. The only prime ideals on top of my head are ones of the form $n\mathbb{Z}$ in the context of the integers and I am convinced that finding an example in the integers with $I \neq J$ is not possible. I am unsure were to look for one with $I \neq J$ . Could someone guide me in the right direction?

rschwieb
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Maths Wizzard
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  • First, please write the integers as $\mathbb{Z}$. Also, why would you need the ideals to be prime? Take $2\mathbb{Z}$ and $4\mathbb{Z}$. – Mark Apr 01 '23 at 18:37
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    Easy counterexamples are in PIDs where $(i)\cap (j) = ({\rm lcm}(i,j)),,$ which $=(i)(j)\iff i,j$ are coprime. – Bill Dubuque Apr 01 '23 at 18:39
  • This yields: if $IJ = I\cap J$ for all ideals in a domain $R$ then $R$ is a field. – Bill Dubuque Apr 01 '23 at 18:46
  • @Mark The question asked for prime ideals. I think if it didn't then your example would be fine. – Maths Wizzard Apr 01 '23 at 19:53
  • @BillDubuque thank you for the comment. I am afraid I don't see how this easily gives us the counter example as I need varify that the ideals are also prime. Is there any PID that has obvious such prime ideals? – Maths Wizzard Apr 01 '23 at 19:55
  • But $(2)$ does not include any primality hypothesis. – Bill Dubuque Apr 01 '23 at 20:02
  • Oh, I see now. Just to undestand your point, what PID would have been a candidate for you? I am still struggling with being able to have standard easy to work with examples of the different domains. – Maths Wizzard Apr 01 '23 at 20:03
  • @MathsWizzard Well, you didn't write it in the statement of the question for some reason. – Mark Apr 01 '23 at 20:07
  • @Mark Apologies, Bill has clarified for me that this wasn't in the second part. – Maths Wizzard Apr 01 '23 at 20:08
  • The most common PIDs are $\Bbb Z$ and $,F[x] = $ polynomials in $x$ with coefficients in a field $F,,$ which are PIDs because they enjoy Euclidean Division (with smaller remainder). – Bill Dubuque Apr 01 '23 at 20:53

1 Answers1

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An easy thing to do is to find a ring with a nonzero ideal $I$ such that $I^2=\{0\}$, containing a nonzero ideal $J$.

You could use, for example, $I=4\mathbb Z/16\mathbb Z$ and $J=8\mathbb Z/16\mathbb Z$ in $R=\mathbb Z/16\mathbb Z$.

There you have it: $\{0\}=IJ\neq J\cap I=J$

rschwieb
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