$\mathbb{Q}\left ( \sqrt{2},\sqrt{3},\sqrt{5} \right )=\mathbb{Q}\left ( \sqrt{2}+\sqrt{3}+\sqrt{5} \right )$

Question

Prove that $$\mathbb{Q}\left ( \sqrt{2},\sqrt{3},\sqrt{5} \right )=\mathbb{Q}\left ( \sqrt{2}+\sqrt{3}+\sqrt{5} \right )$$ I proved for two elements, ex, $\mathbb{Q}\left ( \sqrt{2},\sqrt{3}\right )=\mathbb{Q}\left ( \sqrt{2}+\sqrt{3} \right )$, but I can't do it by the similar method.

Answer 1

This is a boring but purely algebraic derivation.

For any three distinct positive integers $a,b,c$, let $x = \sqrt{a}+\sqrt{b}+\sqrt{c}$, we have

$$\begin{align} & (x-\sqrt{a})^2 = (\sqrt{b}+\sqrt{c})^2 = b + c + 2\sqrt{bc}\\ \implies & (x^2 + a - b - c) - 2\sqrt{a}x = 2\sqrt{bc}\\ \implies & (x^2 + a - b - c)^2 - 4\sqrt{a}x(x^2 + a - b - c) + 4ax^2 = 4bc\\ \end{align} $$ It is easy to check $x^2 + a - b - c \ne 0$. This leads to $$\sqrt{a} = \frac{(x^2+a-b-c)^2 + 4(ax^2-bc)}{4x(x^2+a-b-c)} \in \mathbb{Q}(x)$$ By a similar argument, we have $\sqrt{b}, \sqrt{c} \in \mathbb{Q}(x)$ and hence

$$\mathbb{Q}(\sqrt{a}, \sqrt{b}, \sqrt{c}) \subset \mathbb{Q}(x) = \mathbb{Q}(\sqrt{a}+\sqrt{b}+\sqrt{c})$$

Since $\sqrt{a} + \sqrt{b} + \sqrt{c} \in \mathbb{Q}(\sqrt{a},\sqrt{b},\sqrt{c})$, we have $$\mathbb{Q}(\sqrt{a}+\sqrt{b}+\sqrt{c}) \subset \mathbb{Q}(\sqrt{a}, \sqrt{b}, \sqrt{c})$$ Combine these two result, we get $$\mathbb{Q}(\sqrt{a}+\sqrt{b}+\sqrt{c}) = \mathbb{Q}(\sqrt{a}, \sqrt{b}, \sqrt{c})$$

Answer 2

Notice that $\mathbf{Q}(\sqrt{5})$ is not a subfield of $K=\mathbf{Q}(\sqrt{2},\sqrt{3})$ so $x^{2}-5$ is irreducible in $\mathbf{Q}(\sqrt{2},\sqrt{3})[x]$ and hence $[K(\sqrt{5}):\mathbf{Q}]=8$. This extension is Galois, as the splitting field over $\mathbf{Q}$ of $(x^{2}-5)(x^{2}-3)(x^{2}-2)$. Hence there are 8 distinct automorphisms, sending $\sqrt{5} \to \pm \sqrt{5}$, $\sqrt{3}\to \pm \sqrt{3}$ and $\sqrt{2} \to \pm \sqrt{2}$, to chose an automorphism you pick what it does to $\sqrt{2}$, $\sqrt{5}$ and $\sqrt{3}$. Now $\sqrt{2}+\sqrt{3}+\sqrt{5}$ generates this field if and only if it does not belong to any proper subfield if and only if it's only fixed by the identity automorphism (Galois correspondence).And in fact it is (and it s not hard to check).

Answer 3

mich95's answer is best and most general. But as a consequence, it does mean that $\sqrt{2}$ is some rational expression in $\sqrt2+\sqrt3+\sqrt5$. If you really want to extend your method, consider powers of this element, and how they are all a linear combination of elements from $\{1,\sqrt2,\sqrt3,\sqrt5,\sqrt6,\sqrt{10},\sqrt{15},\sqrt{30}\}$. If the first $8$ powers are linearly independent, then linear algebra will provide you with how to represent $\sqrt2$ (and the other square roots).

$$\begin{align} (\sqrt2+\sqrt3+\sqrt5)^0&=1&&=[1,0,0,0,0,0,0,0]\\ (\sqrt2+\sqrt3+\sqrt5)^1&=\sqrt2+\sqrt3+\sqrt5&&=[0,1,1,1,0,0,0,0]\\ \frac12(\sqrt2+\sqrt3+\sqrt5)^2&=5+\sqrt6+\sqrt{10}+\sqrt{15}&&=[5,0,0,0,1,1,1,0]\\ \frac12(\sqrt2+\sqrt3+\sqrt5)^3&=13\sqrt2+12\sqrt3+10\sqrt5+3\sqrt{30}&&=[0,13,12,10,0,0,0,3]\\ \frac18(\sqrt2+\sqrt3+\sqrt5)^4&=28+10\sqrt6+8\sqrt{10}+7\sqrt{15}&&=[28,0,0,0,10,8,7,0]\\ \frac18(\sqrt2+\sqrt3+\sqrt5)^5&=98\sqrt2+83\sqrt3+65\sqrt5+25\sqrt{30}&&=[0,98,83,65,0,0,0,25]\\ \frac1{16}(\sqrt2+\sqrt3+\sqrt5)^6&=385+153\sqrt6+119\sqrt{10}+99\sqrt{15}&&=[385,0,0,0,153,119,99,0]\\ \frac1{16}(\sqrt2+\sqrt3+\sqrt5)^7&=1439\sqrt2+1186\sqrt3+920\sqrt5+371\sqrt{30}&&=[0,1439,1186,920,0,0,0,371] \end{align}$$

You may have noticed the bipartite nature of these vectors. $\sqrt{2}$ is represented by $[0,1,0,0,0,0,0,0]$. So if you do the linear algebra on the 1st, 3rd, 5th, and 7th powers of $\sqrt2+\sqrt3+\sqrt5$, you have a system of equations that (luckily?) has a unique solution, showing you what coefficients give $$\sqrt2=a(\sqrt2+\sqrt3+\sqrt5)+\frac{b}{2}(\sqrt2+\sqrt3+\sqrt5)^3+\frac{c}{8}(\sqrt2+\sqrt3+\sqrt5)^5+\frac{d}{16}(\sqrt2+\sqrt3+\sqrt5)^7$$

Solving $$ \begin{bmatrix} 0&0&0&0\\1&13&98&1439\\1&12&83&1186\\1&10&65&920\\0&0&0&0\\0&0&0&0\\0&0&0&0\\1&3&25&371\\ \end{bmatrix} \begin{bmatrix} a\\b\\c\\d \end{bmatrix} = \begin{bmatrix} 0\\1\\0\\0\\0\\0\\0\\0 \end{bmatrix} $$ yields $a=-\frac{10}{39}$, $b=\frac{379}{468}$, $c=-\frac{191}{234}$, and $d=\frac{23}{468}$. So $$\sqrt2=-\frac{10}{39}(\sqrt2+\sqrt3+\sqrt5)+\frac{\frac{379}{468}}{2}(\sqrt2+\sqrt3+\sqrt5)^3-\frac{\frac{191}{234}}{8}(\sqrt2+\sqrt3+\sqrt5)^5+\frac{\frac{23}{468}}{16}(\sqrt2+\sqrt3+\sqrt5)^7$$

Answer 4

It is worth noting from alex.jordan and achille hui's responses that if $\theta = \sqrt{2} + \sqrt{3} + \sqrt{5}$, we explicitly have $$\begin{align*} \sqrt{2} &= \tfrac{5}{3} \theta - \tfrac{7}{72} \theta^3 - \tfrac{7}{144} \theta^5 + \tfrac{1}{576} \theta^7, \\ \sqrt{3} &= \tfrac{15}{4} \theta - \tfrac{61}{24} \theta^3 + \tfrac{37}{96} \theta^5 - \tfrac{1}{96} \theta^7, \\ \sqrt{5} &= -\tfrac{53}{12} \theta + \tfrac{95}{36} \theta^3 - \tfrac{97}{288} \theta^5 + \tfrac{5}{576} \theta^7. \end{align*}$$