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I'm trying to think of an example of a ring $A$ and ideals $I$,$J$ s.t. $I \cup J$ is not an ideal.

And what is the smallest ideal containing $I$ & $J$?

Will $A \mathbb{Z}$, $I = 2\mathbb{Z}$, and $J = 4\mathbb{Z}$ work? and the smallest ideal containing them be $\mathbb{Z}$, the integers?

Can someone add some explanation as to why this works? Thanks.

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$2\mathbb{Z} \cup 4\mathbb{Z}=2\mathbb{Z}$ – Chris Eagle May 29 '12 at 21:21
$2\mathbf Z$ and $3\mathbf Z$ would work, though. [Check whether the union is closed under addition.] – Dylan Moreland May 29 '12 at 21:24
Ironically, most choices of $I$ and $J$ would work with $A = \mathbb{Z}$; you had the misfortune of making one of the special choices that doesn't. – Hurkyl May 29 '12 at 21:26

Hint 1: Try $2\mathbb{Z}\cup 3\mathbb{Z}$ in $\mathbb{Z}$.

Hint 2: Ideals are additively closed, so at the very least the smallest ideal containing $I$ and $J$ also has to contain $I+J$...

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You could try the ideals $I=3\mathbb{Z}$ and $J=5\mathbb{Z}$. Then their union is not an ideal since it is not closed under addition. You have $3,5\in I\cup J$, but $8\notin I\cup J$, for example.

Secondly, the smallest ideal containing $I$ and $J$ is $I+J$. A brief proof can be found here as problem 2a.

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