Constructability of algebraic elements over $\mathbb{Q}$

Asked May 05 '21 at 11:17

Active May 05 '21 at 11:17

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I'm currently reading upon the classical straightedge and compass constructions and I came across the following theorem:

lf a real number $c$ is constructible, then $c$ is algebraic of degree a power of $2$ over the field $\mathbb{Q}$.

My question is, is the converse to the above theorem true? That is, if $c$ is algebraic of degree a power of $2$ over $\mathbb{Q}$, is $c$ constructible?

asked May 05 '21 at 11:17

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You need a tower of quadratic extensions for $c$ to be constructible. – Parcly Taxel May 05 '21 at 11:22
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All you can do with constructions is field operations and square roots. – Arthur May 05 '21 at 11:27
Got it, thanks for all the helpful responses! – May 05 '21 at 11:30

Constructability of algebraic elements over $\mathbb{Q}$

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