Is it possible to find a extension of degree $2^n$ of $\mathbb Q$ which is not constructible?

Question

It is well known that if a number $\lambda$ is constructible, then $\mathbb [Q[\lambda]:\mathbb Q]=2^n$ for some $n.$

Therefore, I am wondering whether there exist a counterexample for the converse?

In other words, equivalently, is it possible to find a field $F$ such that $\mathbb Q \subseteq F \subseteq \mathbb R,$ and $[F:\mathbb Q ] = 2^n$ for some $n,$ but there does NOT exist a sequence $\mathbb Q\subseteq F_1 \subseteq F_2\subseteq \ldots \subseteq F_{n-1} \subseteq F_n= F$ such that $[F_{i+1} : F_i]=2$ for each $i$?

Many thanks.

Yes, it is. I think we have covered this a few times. Gimme a minute. — Jyrki Lahtonen, Sep 05 '22 at 14:28
Suppose $L$ is the splitting field of the minimal polynomial of $\lambda$ over $\mathbb{Q}$. Then an equivalent condition is that $\lambda$ is constructible if and only if $[L:\mathbb{Q}]$ is a power of $2$. You can see a proof here: http://www2.math.uu.se/~svante/papers/sjN8.pdf — Mark, Sep 05 '22 at 14:29
So for example, if $\lambda$ is a root of $x^4-4x+2$ then $[\mathbb{Q}(\lambda):\mathbb{Q}]=4$, but $\lambda$ is not constructible. (this polynomial has Galois group $S_4$, so the splitting field is of degree $24$ over $\mathbb{Q}$) — Mark, Sep 05 '22 at 14:30
Without the condition that $F\subset\mathbb R$, there is an easy counter-example. Let $K/\mathbb Q$ have Galois group $S_8$ and let $L$ be the fixed field of $S_7\subset S_8$. Then $L/\mathbb Q$ is a degree $8$ extension which does not factor into degree $2$ extensions, since the only index $2$ subgroup in $S_8$ is $A_8$, which does not contain $S_7$. — Kenta S, Sep 05 '22 at 14:31
Here I presented roots of $x^4-ax-1$, $a\in\Bbb{Z}, a\equiv1\pmod{14}$, as examples of such numbers. They generate an extension of degree four, but are not constructible. The argument is more or less the same Mark used. — Jyrki Lahtonen, Sep 05 '22 at 14:32
A more general variant. Still not an ideal duplicate target. — Jyrki Lahtonen, Sep 05 '22 at 14:34

Is it possible to find a extension of degree $2^n$ of $\mathbb Q$ which is not constructible?

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