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What does $\mathbb{Q}(\sqrt{2})$ mean? Is it $\{ a+b\sqrt{2} \mid a, b \in \mathbb{Q} \}$ ?

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    $\begingroup$ My humble opinion: yes. $\endgroup$ – manooooh May 8 '19 at 23:17
  • $\begingroup$ See here for ring and field adjunctions. $\endgroup$ – Bill Dubuque May 8 '19 at 23:18
  • $\begingroup$ Yes, but be careful, $\Bbb Q(\sqrt[3]{2})\neq\{a+b\sqrt[3]{2} \mid a,b\in \Bbb Q \}$. $\endgroup$ – Dietrich Burde May 9 '19 at 14:52
  • $\begingroup$ hmm, so what is that $\endgroup$ – Jonny May 9 '19 at 16:33
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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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  • $\begingroup$ Good answer!! Why not $\sqrt1$ or $\sqrt0$? $\endgroup$ – manooooh May 8 '19 at 23:28
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    $\begingroup$ @manooooh what about them? $\endgroup$ – Yuval Gat May 8 '19 at 23:30
  • $\begingroup$ Why they are not "The smallest field"? Or am I misunderstanding something? $\endgroup$ – manooooh May 8 '19 at 23:32
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    $\begingroup$ @manooooh Do you know what a field is? $\endgroup$ – Mark May 8 '19 at 23:32
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    $\begingroup$ Good 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. $\endgroup$ – Mark May 8 '19 at 23:36

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