Suppose there is given a tower of fields in $\mathbb{C}$:
$\mathbb{Q} =L_0 \subseteq L_1 \subseteq ... \subseteq L_n \supseteq \mathbb{Q}(\alpha)$ with an $\alpha \in \mathbb{C}$
such that all extensions are quadratic:
$[L_{j+1} : L_j] = 2$.
Then I define the intersections with $\mathbb{Q}(\alpha)$ as follows:
$M_j := L_j \cap \mathbb{Q}(\alpha)$
and get a new tower of fields:
$\mathbb{Q} =M_0 \subseteq M_1 \subseteq ... \subseteq M_n = \mathbb{Q}(\alpha)$.
My question: Is the following claim correct:
It is either $M_{j+1} = M_j$ or $[M_{j+1} : M_j] = 2$.
Ian Stewart claims this in his book "Galois Theory" (4th edition, page 97) in a proof of one theorem. I don't know why this claim should be obvious. Probably this has something to do with the "Tower law": $[M:K] = [M:L][L:K]$ for $K \subseteq L \subseteq M$.
That claim is not true in general. (And I checked that Stewart doesn't have any additional hypotheses; this is simply an incorrect proof in that book.) For example, let $\alpha\in\mathbb R$ satisfy $\alpha^4=2$ and let $L_3=\mathbb Q(i,\alpha)$, $L_2=\mathbb Q(\alpha+i\alpha)$, and $L_1=\mathbb Q(i\alpha^2)$. Then $\mathbb Q=L_0\subset L_1\subset L_2\subset L_3\supset\mathbb Q(\alpha)$ and $[L_{j+1}:L_j]=2$, but the fields $M_j:=L_j\cap\mathbb Q(\alpha)$ satisfy $[M_3:M_2]=4$.
To justify the above assertions, first note that $(i\alpha^2)^2=-2$, so that $[L_1:L_0]=2$. Next, $(\alpha+i\alpha)^2=2i\alpha^2$ so that $L_1\subseteq L_2$, and also $2/(\alpha+i\alpha)$ is a root of $x^4+2$, which is irreducible over $\mathbb Z$ by Eisenstein's criterion and hence irreducible over $\mathbb Q$, so $[L_2:L_1]=2$. Likewise $x^4-2$ is irreducible over $\mathbb Q$, so $[\mathbb Q(\alpha):\mathbb Q]=4$. Since $\mathbb Q(\alpha)\subset\mathbb R$, the minimal polynomial of $i$ over $\mathbb Q(\alpha)$ is $x^2+1$, so $[L_3:\mathbb Q(\alpha)]=2$, and thus $[L_3:\mathbb Q]=8$. Since clearly $L_2\subseteq L_3$, and $[L_2:\mathbb Q]=[L_2:L_1]\cdot [L_1:L_0]=4$, it follows that $[L_3:L_2]=[L_3:L_0]/[L_2:L_0]=2$. Plainly $M_3=\mathbb Q(\alpha)$ satisfies $[M_3:\mathbb Q]=4$, so it remains to show that $M_2=\mathbb Q$, i.e., $L_2\cap \mathbb Q(\alpha)=\mathbb Q$. Suppose otherwise. Then $\sum_{k=0}^3 c_k \alpha^k = \sum_{\ell=0}^3 d_\ell (\alpha+i\alpha)^\ell$ for some $c_k,d_\ell\in\mathbb Q$ where $d_1,d_2,d_3$ are not all zero. Expand the powers of $\alpha+i\alpha$ to rewrite this as $$\sum_{k=0}^3 c_k \alpha^k = d_0 + d_1 (\alpha+i\alpha) + d_2 (2i\alpha^2) + d_3 (-2\alpha^3+2i\alpha^3).$$ But $1,\alpha,\alpha^2,\alpha^3,i,i\alpha,i\alpha^2,i\alpha^3$ are linearly independent over $\mathbb Q$ (since $1,i$ is a basis for $\mathbb Q(\alpha,i)$ as a $\mathbb Q(\alpha)$-vector space, and $1,\alpha,\alpha^2,\alpha^3$ is a basis for $\mathbb Q(\alpha)$ as a $\mathbb Q$-vector space), so by considering coefficients of $i\alpha,i\alpha^2,i\alpha^3$ in the displayed equation we see that $d_1=d_2=d_3=0$, contradiction.
More generally, one could let $L_3/\mathbb Q$ be any Galois extension whose Galois group $G$ is dihedral of order $8$, and write $G=\langle \sigma,\tau\rangle$ where $\sigma^2=1$ and $\sigma\tau\sigma=\tau^{-1}$. Then put $\mathbb Q(\alpha)=L_3^{\langle\sigma\rangle}$ and $L_2=L_3^{\langle\sigma\tau\rangle}$ and $L_1=L_2^{\langle \sigma\tau,\sigma^2\rangle}$. Again with $M_j:=L_j\cap\mathbb Q(\alpha)$, we have $[M_3:M_2]=4$ and $\mathbb Q=L_0\subset L_1\subset L_2\subset L_3\supset\mathbb Q(\alpha)$ where $[L_{j+1}:L_j]=2$.