Suppose I have a field $\Bbb{Q}(\sqrt{2})$. And as I know that $\Bbb{Q}(\sqrt{2})=\{a+b\sqrt{2}: a,b\in \Bbb{Q}\}$ as a vector space basis would be $\{a,b\}$. I wanna know that how would elements be in $\Bbb{Q}(\sqrt{2},\sqrt{3})$? How to generalize the elements in $\Bbb{Q}(\sqrt{p_1},\sqrt{p_2},\dots,\sqrt{p_n})$? where each $p_i$ is prime number.
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"as a vector space basis would be {a,b}" -- This is unclear, as certainly $a,b$ isn't a basis (it's not even defined in the present scope). Did you mean the coordinates would be $(a,b)$ with respect to the ordered basis $(1,\sqrt{2})$? – Brian Moehring Aug 08 '20 at 19:04
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2A typical element of $\Bbb Q(\sqrt2,\sqrt3)$ is $a+b\sqrt2+c\sqrt3+d\sqrt6$ for $a,\ldots,d\in\Bbb Q$ etc. – Angina Seng Aug 08 '20 at 19:13
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@BrianMoehring Ohh sorry! basis would be ${1,\sqrt{2}}$ of $\Bbb{Q}(\sqrt{2})$ as a vector space. – Unknown Aug 09 '20 at 01:59
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See this old thread for many approaches. – Jyrki Lahtonen Aug 10 '20 at 09:56
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$\Bbb{Q}(\sqrt{2},\sqrt{3})=\Bbb{Q}[\sqrt{2},\sqrt{3}] = $ image of the map $\Bbb{Q}[x,y] \to \Bbb{R}$ induced by $x \mapsto \sqrt{2}$, $y \mapsto \sqrt{3}$.
Therefore $\Bbb{Q}(\sqrt{2},\sqrt{3})$ is the set of all polynomials expressions in $\sqrt{2}$ and $\sqrt{3}$. Consider a typical monomial $\sqrt{2}^i \sqrt{3}^j$. Conclude that $\Bbb{Q}(\sqrt{2},\sqrt{3})$ is the set of all linear combinations of $\sqrt{2}$, $\sqrt{3}$, $\sqrt{6}$.
lhf
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