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Let $\mathbb Q$ be the set of all rational numbers. I would like to know what the ideal for $\mathbb Q$ as ring is. I think the ideal of $\mathbb Q$ is $\mathbb Q$, Am I right?

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We can't say the ideal since $\{0\}$ is also an ideal. – Davide Giraudo Feb 1 '12 at 18:49
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Hint: $\mathbb{Q}$ is a field. – user5137 Feb 1 '12 at 18:50
A much more interesting question is: What are the subrings of $\mathbb Q$? – lhf Feb 1 '12 at 19:06
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See also this question: A ring is a field iff the only ideals are (0) and (1) – Martin Sleziak Feb 1 '12 at 19:29

1 Answer

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An ideal must enjoy the property that if you multiply any of its members by any rational number, what you get is still in the ideal. But if a member $x$ of the supposed ideal is not $0$, then it's easy to show that if you take all products of $x$ with rational numbers, the set that you get, which is $\{xy:y\in\mathbb{Q}\}$, is all of $\mathbb{Q}$. So nothing smaller than $\mathbb{Q}$ can be an ideal in this ring, except $\{0\}$.

Here's the proof: suppose $w$ is any member of $\mathbb{Q}$. Then $w/x$ is rational. So $w/x$ is the value of $y$ that will serve.

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