What does $\mathbb{Q}(\sqrt{2})$ mean? Is it $\{ a+b\sqrt{2} \mid a, b \in \mathbb{Q} \}$ ?
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1My humble opinion: yes. – manooooh May 08 '19 at 23:17
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See here for ring and field adjunctions. – Bill Dubuque May 08 '19 at 23:18
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Yes, but be careful, $\Bbb Q(\sqrt[3]{2})\neq{a+b\sqrt[3]{2} \mid a,b\in \Bbb Q }$. – Dietrich Burde May 09 '19 at 14:52
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hmm, so what is that – Jonny May 09 '19 at 16:33
1 Answers
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Yes. Formally, $\mathbb{Q}(a)$ means “the smallest field containing $a$ and $\mathbb{Q}$". It turns out that the smallest field containing both $\mathbb{Q}$ and $\sqrt{2}$ is precisely the field you wrote.
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Why they are not "The smallest field"? Or am I misunderstanding something? – manooooh May 08 '19 at 23:32
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4Good answer by the way. It is important to note that the definition is what you wrote in the answer. For example $\mathbb{Q}(\pi)$ is not ${a+b\pi: a,b\in\mathbb{Q}}$. It's the specific element $\sqrt{2}$ for which $\mathbb{Q}(\sqrt{2})$ can be described in such a simple way. – Mark May 08 '19 at 23:36
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