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If we have the field extension $\Bbb{R}:\Bbb{Q}$, we define the set $$\Bbb{R} ^{\text{alg}}:=\{a\in \Bbb{R}: a \text{ is algebraic over }\Bbb{Q} \}$$ We want to show that $\Bbb{R}^{\text{alg}}$ is a subfield of $\Bbb{R}$ and find the $[\Bbb{R} ^{\text{alg}}:\Bbb{Q}]$.

Answer:

If we consider that $a,b\in \Bbb{R} ^{\text{alg}} \iff a,b$ are algebraic over $\Bbb{Q} $, we have that the extension $\Bbb{Q}(a,b):\Bbb{Q}$ is algebraic over $\Bbb{Q}$. We know that $\Bbb{Q}(a,b)$ is a field, so $a-b\in \Bbb{Q}(a,b)$ and if $b\neq 0,\ ab^{-1} \in \Bbb{Q}(a,b)$, and from the previous observation are algebraic. So, $a-b,\ ab^{-1} \in \Bbb{R}^{\text{alg}}$. And this means $\Bbb{R}^{\text{alg}} \leq\mathbb{R}$.

Right? And how can we find the $[\Bbb{R} ^{\text{alg}}:\Bbb{Q}]$?

Thank you in advance.

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  • $\begingroup$ Do you have any guesses about what $[\Bbb R^{\rm alg}:\Bbb Q]$ might be? $\endgroup$ – arctic tern Apr 6 '17 at 0:08
  • $\begingroup$ @arctictern Thank you for your comment. Maybe $\infty$? But I can't find a way to determine a basis! $\endgroup$ – Chris Apr 6 '17 at 0:13
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    $\begingroup$ Who said anything about determining a basis? $\endgroup$ – arctic tern Apr 6 '17 at 0:18
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    $\begingroup$ See this question: math.stackexchange.com/questions/552220/… $\endgroup$ – Jonas Apr 6 '17 at 0:23
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    $\begingroup$ Find a lower bound on the degree of the extension by finding appropriate subfields. Show that you can take this lower bound to be as large as you want. $\endgroup$ – Asvin Apr 6 '17 at 5:32

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