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Is $\sqrt{2+\sqrt{5}}$ included in the extension field $\mathbb{Q}(\sqrt5)$?

The motivation for this question is the understanding that the square root of a Complex Number $(\sqrt{a_0+b_0i})$ is itself a Complex Number, $(a_1+b_1i)$.

(I know. I'm comparing $\mathbb{C}$, an extension of $\mathbb{R} $, to $\mathbb{Q}(\sqrt5)$, an extension of $\mathbb{Q}$.)

I've tried the following:

$$m+n\sqrt5 = \sqrt{2+\sqrt{5}}$$

and solving for $m$ and $n$.

In my journey, I've found the minimal polynomial $x^4-4x^2-1$. I don't think there exist Rational numbers, $m$ and $n$.

Am I correct in saying so?

MathAdam
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    Do you know about the norm map, $N(a + b \sqrt 5) = (a + b \sqrt 5)(a - b \sqrt 5) = a^2 - 5b^2$? – Ravi Fernando Mar 12 '22 at 03:58
  • If you have found that the minimal polynomial is of degree $4$ then it can't lie in a quadratic extension of rationals. – Paramanand Singh Mar 13 '22 at 08:05
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    Also square root of a complex number is a complex number, but square root of $a+bi$ with $a, b$ rational is not necessarily of same form $c+di$ with $c, d$ rational. – Paramanand Singh Mar 13 '22 at 08:07

1 Answers1

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Taking from where you left off: $(m+n\sqrt{5})^2 = 2+\sqrt{5}\implies m^2+2mn\sqrt{5}+5n^2=2+\sqrt{5}\implies m^2+5n^2=2, 2mn=1$. Thus $m^2n^2=\dfrac{1}{4}$, and $m^2+\dfrac{5}{4m^2} = 2\implies 4m^4-8m^2+5=0\implies 4(m^2-1)^2=-1$. This equation has no solutions in $\mathbb{R}$, and thus has no solution in $\mathbb{Q}$. So back to the main question,$\sqrt{2+\sqrt{5}}\notin \mathbb{Q}(\sqrt{5})$.

Wang YeFei
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