Exercise 10.1.5

In this exercise you will give two proofs that ζ 3 = e 2 πi 3 is constructible.

(a)
Give a direct geometric construction of ζ 3 with each step justified by citing C 1 , C 2 , P 1 , P 2 or P 3 .
(b)
Use Theorem 10.1.6 to show that ζ 3 is constructible.

Answers

Proof.

(a)
As Re ( ζ 3 ) = 1 2 , ζ 3 is on the perpendicular bisector of ( 1 , 0 ) .

We gives the details of the construction: C ( 0 , 1 ) and C ( 1 , 1 ) are constructible by C 2 , so the two intersection points are constructible by P 3 , and these two points are ζ 3 , ζ ¯ 3 , so ζ 3 is constructible.

(b)
As the minimal polynomial of ζ 3 over is x 2 + x + 1 , the extension ( ζ 3 ) is quadratic, so ζ 3 is constructible by Theorem 10.1.6.
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2022-07-19 00:00
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