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Let $F\subseteq\mathbb{C}$ be a Galois extension of $\mathbb{Q}$ such that $[F:\mathbb{Q}]=2^n$; then all elements in $F$ are constructible.


Added.

Here is what I have so far.

Since there exist the finite normal extension $F\subseteq\mathbb{C}$ over $\mathbb{Q}$ which contains an element say, s, then s has a degree of power of 2. From group theory we have 2 propositons we could use. 'A group $G$ is solvable iff it has a normal series whose factors are abelian', and also 'If $G$ is solvable and $N\triangleleft G$, then $G/N$ is solvable'. Then we can say that the galois group F has a normal series $\{e\}=G_0\triangleleft G_1\triangleleft\cdots\triangleleft G_n=\mathrm{Gal}(F/\mathbb{Q})$, whose factors have order 2. Thus there must exist intermediate fields $\mathbb{Q}=F_0 \subseteq F_1 \subseteq F_2 \subseteq\cdots\subseteq F_n=F$, such that $[F_{j+1}:F_j]=2$ for all $j$. Now this is where I got stuck. Next I need to somehow show that $F_{j+1}=F_j(s_j)$, where $s^2_j$ is an element in $F_j$. Then I think I can conclude that if $s_j$ is constructible than every element in $F_{j+1}$ is constructible... then with $F_n$ is constructible...

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    First use geometry to show that if $K$ is an extension of $\mathbb{Q}$ with all elements constructible, and $L$ is a 2-dimensional extension of $K$, then $L$ also has all elements constructible. Then use Galois theory to show that a $2^n$-dimensional extension can be decomposed as a series of $n$ $2$-dimensional extensions.2011-02-15
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    @ Chris thanks.I need to show that all the elements in F are constructible. I can't assume that they already are, can I?2011-02-15
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    No. The point of the first part is that with induction (using the base case $K=\mathbb{Q}$, which certainly has all elements constructible) and the second part, it proves the result.2011-02-15
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    @ Chris, thanks..I'll try to do that.2011-02-15

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