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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3We can't say the ideal since ${0}$ is also an ideal. – Davide Giraudo Feb 01 '12 at 18:49
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13Hint: $\mathbb{Q}$ is a field. – Feb 01 '12 at 18:50
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1A much more interesting question is: What are the subrings of $\mathbb Q$? – lhf Feb 01 '12 at 19:06
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3See also this question: A ring is a field iff the only ideals are (0) and (1) – Martin Sleziak Feb 01 '12 at 19:29
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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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Let $I$ be an ideal of $\mathbb{Q}$ with unity s.t. $1 \in I$.
We have $q \in \mathbb{Q} \Rightarrow q = 1•q ∈ I \Rightarrow \mathbb{Q} ⊆ I \Rightarrow \mathbb{Q} = I$
Hence $\mathbb{Q}$ has only two ideals i.e, $\mathbb{Q}$ itself and $\{0\}$.
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