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.
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1How did you prove it for two elements? We might be able to extend your particular method if you show it to us. – Milo Brandt May 31 '15 at 03:59
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It is sufficient to show that the left hand side is contained in the right hand side. We have $\sqrt{2} = \frac{(\sqrt{2}+\sqrt{3})^2+2-3}{2(\sqrt{2}+\sqrt{3})}$. So it belongs to the right. – TheKiet May 31 '15 at 04:06
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2Notice that you're in a Galois extension of $\mathbf{Q}$. Proving the statement is equivalent to showing that no non identity automorphism of $\mathbf{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ fixes $\sqrt{2}+\sqrt{3}+\sqrt{5}$. – mich95 May 31 '15 at 04:10
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You can do it the same way, but the work involved will be substantial, as you will wind up looking at higher powers of $\sqrt{2} + \sqrt{3} + \sqrt{5}$. mich95's method is short and sweet. – David Wheeler May 31 '15 at 04:33
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@DavidWheeler Thanks! :$ – mich95 May 31 '15 at 04:39
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See this question for a more general formula for the dimension of fields of this type, and the comments to this answer for a Galois theoretic proof for the fact that (the analogue of) this element generates the field extension. If you are not familiar with Galois theory yet (a typical time for this question is just before you get into Galois theory), then you can try the tricks below. – Jyrki Lahtonen May 31 '15 at 12:48
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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).
mich95
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I don’t find this convincing, since I think one needs a proof that $\Bbb Q(\sqrt5,)$ is not contained in $\Bbb Q(\sqrt2,\sqrt3,)$, even if your proof is only a few words in length. If you know about ramification, it is indeed obvious; but from an elementary standpoint, I think it needs a proof. – Lubin May 31 '15 at 04:37
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I do not know about ramifications. But we know that the degree 2 subfields of $\mathbf{Q}(\sqrt{2},\sqrt{3})$ are $\mathbf{Q}(\sqrt{2})$, $\mathbf{Q}(\sqrt{6})$ and $\mathbf{Q}(\sqrt{3})$, and it's clear / not hard to show that $\mathbf{Q}(\sqrt{5})$ is not any of those. – mich95 May 31 '15 at 04:38
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I guess that you’d have to argue that $X^2-5$ is irreducible over each of those fields. Their rings of integers all are PIDs, so I guess there’s not much work involved. – Lubin May 31 '15 at 04:40
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1Assume that $\sqrt{5}$ belongs to $\mathbf{Q}(\sqrt{2})$, then we can find rationals a and b such that $a+b\sqrt{2}=\sqrt{5}$. Squaring both sides, $a^{2}+2b^{2}+2ab\sqrt{2}=5$. Since 5 is rational, but $\sqrt{2}$ is not, $a=0$ or $b=0$. Now b cant be zero as this implies $a=\sqrt{5}$ so $a$ must be zero, and this implies that $\frac{\sqrt{5}}{\sqrt{2}}$ is rational.Contradiction. – mich95 May 31 '15 at 04:45
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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 $$\sqrt2=-\frac{10}{39}(\sqrt2+\sqrt3+\sqrt5)+\frac{\frac{379}{468}}{2}(\sqrt2+\sqrt3+\sqrt5)^3-\frac{\frac{191}{234}}{8}(\sqrt2+\sqrt3+\sqrt5)^5+\frac{\frac{23}{468}}{16}(\sqrt2+\sqrt3+\sqrt5)^7$$
2'5 9'2
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One can also note that if we simply show that those powers are linearly independent (over the rationals) - which we could do via calculating a determinant - then we have that $\mathbb Q(\sqrt{2}+\sqrt{3}+\sqrt{5})$ is a $8$ dimensional vector space over $\mathbb Q$ contained in $\mathbb Q(\sqrt{2},\sqrt{3},\sqrt{5})$ which has dimension at most $8$. By the standard theorems of linear algebra, the two are therefore equal. – Milo Brandt May 31 '15 at 13:56
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@Meelo Definitely, but the point of this answer is to explicitly find a linear combination that results in $\sqrt{2}$. – 2'5 9'2 Jun 01 '15 at 22:24
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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 $$\mathbb{Q}(\sqrt{a}+\sqrt{b}+\sqrt{c}) = \mathbb{Q}(\sqrt{a}, \sqrt{b}, \sqrt{c})$$
achille hui
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It is worth noting from alex.jordan and achille hui's responses that if $\theta = \sqrt{2} + \sqrt{3} + \sqrt{5}$, we explicitly have $$\begin{align*} \sqrt{2} &= \tfrac{5}{3} \theta - \tfrac{7}{72} \theta^3 - \tfrac{7}{144} \theta^5 + \tfrac{1}{576} \theta^7, \\ \sqrt{3} &= \tfrac{15}{4} \theta - \tfrac{61}{24} \theta^3 + \tfrac{37}{96} \theta^5 - \tfrac{1}{96} \theta^7, \\ \sqrt{5} &= -\tfrac{53}{12} \theta + \tfrac{95}{36} \theta^3 - \tfrac{97}{288} \theta^5 + \tfrac{5}{576} \theta^7. \end{align*}$$
heropup
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