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We just started field extensions in my algebra course, so this is all relatively new to me, and I'm not sure where to even begin with this proof. For example, I'm trying to show that given a subfield $F$ of $\mathbb{C}$, if we define $E$ as $E=F(\sqrt{a}+\sqrt{b})$, $\sqrt{a}$ and $\sqrt{b}$ also belong to $E$. It seems obvious, but I'm not sure where to start. Thanks for any help

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  • $\begingroup$ Well what do the elements of $F(\sqrt{a}+\sqrt{b})$ look like? $\endgroup$ – Brandon Thomas Van Over Nov 17 '15 at 2:06
  • $\begingroup$ Hint: Hard to know what square root means. We will assume that we cannot have $\sqrt{b}=-\sqrt{a}$. Fields are closed under reciprocal, so $\frac{a-b}{\sqrt{a}+\sqrt{b}}$ is in $E$. $\endgroup$ – André Nicolas Nov 17 '15 at 2:23
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We will assume $a \neq b$.

It suffices to show that $\sqrt{a} \in E$, since then $\sqrt{b} = (\sqrt{a} + \sqrt{b}) - \sqrt{a}$ surely is, as well.

Note that: $(\sqrt{a} + \sqrt{b})^3 = a\sqrt{a} + 3a\sqrt{b} + 3b\sqrt{a} + b\sqrt{b} = (a+3b)\sqrt{a} + (3a + b)\sqrt{b}$.

Thus $(\sqrt{a} + \sqrt{b})^3 - (3a + b)(\sqrt{a} + \sqrt{b}) = 2(b-a)\sqrt{a}$.

Since $a \neq b$, $\dfrac{1}{2(b-a)} \in F$, thus:

$\sqrt{a} = \dfrac{1}{2(b -a)}[(\sqrt{a} + \sqrt{b})^3 - (3a+b)(\sqrt{a} + \sqrt{b})] \in F(\sqrt{a} + \sqrt{b})$.

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