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Which of these are constructible numbers?

$$-\frac35\quad,\quad27^\frac16+2i\quad,\quad2^\frac13\quad,\quad e^{\frac{\pi i}{10}}$$

Please tell me how you came to the answer too. Thanks!

Aditya Hase
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Eric Kim
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1 Answers1

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Ok, so you already know what numbers are constructible and what not. Now, why? Be sure to complete the following arguments, and remember: a number is constructible iff it belongs to a tower of field extension each of which is of degree two, i.e. $\;z\in\Bbb C\;$ is constructible iff $\;z\in K\;,\;\;K\;$ a subfield of $\;\Bbb C\;$ , and such that

$$\Bbb Q=K_0\subset K_1\subset K_2\subset\ldots\subset K_n =K\;,\;\text{and}\;\;[K_i:K_{i-1}]=2\;\;\;\forall\; i=1,2,...,n\;,\;\;or\;\; K=\Bbb Q$$

So:

$$\begin{align*}&-\frac35\in\Bbb Q\implies\color{red}{-\frac35\;\;\text{constructible}}\\{}\\ &\begin{cases}27^{1/6}=\sqrt3\in\Bbb Q{\sqrt3}\;\;\text{and}\;\;[Q(\sqrt3:\Bbb Q]=2\\{}\\2i\in\Bbb Q(i)\;\;\text{and}\;\;[\Bbb Q(i):\Bbb Q]=2\end{cases}\implies\color{red}{\sqrt[6]{27}+2i\in\Bbb Q(\sqrt3,i)\;\;\text{constructible}}\\{}\\ &e^{\pi i/10}=e^{2\pi i/20}\;\;\text{and}\;\;\varphi(20)=8=2^3\implies\;\color{red}{e^{\pi i/10}\;\;\text{constructible, or follows also from:}}\\{}\\&\text{the fact that the minimal polynomial over the rationals of}\;\;e^{\pi i/10}\;\;\text{is}\;\;x^8-x^6+x^4-x^2+1\end{align*}$$

This last result follows from factoring $\;x^{10}+1=\left(x^2\right)^5+1\;$ .

Finally

$$\sqrt[3]2\in\Bbb Q(\sqrt[3]2)\;,\;\;\text{and}\;\;[\Bbb Q(\sqrt[3]2:\Bbb Q]=3\implies \color{green}{\sqrt[3]2\;\;\text{isn't constructible}}$$

Timbuc
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  • Thanks Timbuc, I am having hard time understanding. What is [Ki : Ki-1] ? and why is it = 2 for sqrt(3) and 3 for 2^1/3 ? Also, how is p(20)=8 ? – Eric Kim Dec 02 '14 at 08:53
  • @EricKim, you know some fields, fields extensions and Galois Theory? that $;\varphi;$ is Euler's totient Function. – Timbuc Dec 02 '14 at 10:51