A property of constructible elements over $\mathbb{Q}$

Question

We say $a$ is constructible over $\mathbb{Q}$,when there exists a chain of quadratic field extension from $\mathbb{Q}$ ending up with a field contains $a$.

I feel it difficult to prove this property:

If $a$ constructible, then there exists a chain of quadratic extension

$$\mathbb{Q} \rightarrow \mathbb{F}_1\rightarrow…\rightarrow\mathbb{F}_n=\mathbb{Q}(a).$$

The most difficult part is that the last field is required to be $\mathbb{Q}(a)$.

And note that this is not really true:

$$[K:F]≥[K∩L:F∩L] \text{ for the field } K,L,F$$

So, how to prove this?

Answer 1

Hint: suppose that $\mathbb{Q} \rightarrow \mathbb{K}_1\rightarrow\cdots\rightarrow\mathbb{K}_m$ is a chain of quadratic extensions with $a\in\mathbb K_m$; then consider the fields $\mathbb{F}_1=\mathbb{K}_1\cap\Bbb Q(a)$, $\mathbb{F}_2=\mathbb{K}_2\cap\Bbb Q(a)$, ..., $\mathbb{F}_m=\mathbb{K}_m\cap\Bbb Q(a)$.

Answer 2

Let $\mathbb{K}$ be the splitting field of minimal polynomial of $a$, Consider the group $Aut(\mathbb{Q}(a)/\mathbb{Q})$ which is of cardinality $2^{\ell}$ for some $\ell$. If the group $Aut(\mathbb{Q}(a)/\mathbb{Q})$ is non-trivial then there exists an element $\sigma \in Aut(\mathbb{Q}(a)/\mathbb{Q})$ of order $2$ and if you consider the fixed field of $\sigma$, call it $L_1$ then we have $[\mathbb{Q}(a):L_1] = 2$. We can keep proceeding by trying to find an element of order $2$ in $Aut(L_1/\mathbb{Q})$ and finding $L_2$ which is the fixed field of this order $2$ automorphism and so on until we find $L_i \subset \mathbb{Q}(a)$ such that $Aut(L_i/\mathbb{Q}) = \{e\}$. This is because for $j>i$ if $[L_j:L_{j-1}] > 2$ then wlog $L_j = L_{j-1}(a_1,...,a_{\ell})$ and all $a_i$ has minimal polynomial of degree $2$ over $L_{j-1}$ and hence $L_j/L_{j-1}$ extension can be written as series of quadratic extensions. Take $r \in L_i \setminus \mathbb{Q}$ then $r$ is constructible and hence if $i \geq 1$, by induction we write it as a series of quadratic extension all of which has trivial automorphism group as otherwise you can extend it to a non-trivial automorphism of $L_i$. But then in any non-trivial quadratic extension Automorphism group is non-trivial since both roots are present in the extension. Hence a contradiction. Hence we have $Aut(\mathbb{Q}(a)/\mathbb{Q}) = \{e\}$. But then by the other answer by wsh there exists an element $b$ such that $\mathbb{Q} \subsetneq \mathbb{Q}(b) \subsetneq \mathbb{Q}(a)$. But then $b$ is constructible and by induction has a chain of quadratic extensions and hence non-trivial automorphism and hence a contradiction.

Answer 3

I have proved my question by Galois theory.

We say $a$ is constructible over field $\mathbb{F}$ if there exist a chain of quadratic extensions begin from $\mathbb{F}$ and contains $a$ somewhere.

Firstly,if $a$ is constructible over arbitrary field $\mathbb{F}$ with character 0 and $f_a$ is minimal polynomial of $a$ over $\mathbb{F}$, then all roots of $f_a$ is constructible.So Galois group of $f_a$,denoted by $G$, is a 2-group,thus nilpotent.Denote the splitting field of $f_a$ by $\mathbb{K}$

Secondly, do induction of the $log_2$ of the degree of $a$.

$n=1$ case is done automorphically for $a$ is degree 2 over $\mathbb{F}$.

Assume if $n\leq k$,for any character zero field $\mathbb{F}$ and any $a$ constructible and of degree $2^n$ over $\mathbb{F}$,there exists a chain of quadratic extension beginning from $\mathbb{F}$ ending with $\mathbb{F(a)}$.

When $n=k+1$ for any character 0 field $\mathbb{F}$,only need to show that there exists $b$ such that $b$ is not in $\mathbb{F}$ and $\mathbb{F(b)} \subsetneq \mathbb{F(a)}$.By Galois theory and primitive element theorem,this is equal to say that Galois group of $\mathbb{K}/\mathbb{F(a)}$,denoted by $H$, is a proper but not maximal subgroup of $G$ unless $a$ itself a square root over $\mathbb{F}$.

If so,take $\mathbb{F}=\mathbb{Q}$ and note that if $a$ is real,then any subextension of $\mathbb{Q(a)}$ over $\mathbb{Q}$ is real automorphically, done.

Finally we claim the property we said about $H$ is true and it follows from the fact that every maximal subgroup of nilpotent group is of index $p$, which is a prime number and in particular $p$-group is nilpotent.

For the proof of the claim,see Every proper maximal subgroup of a $p$-group $P$ is normal and has index $p$.