2
$\begingroup$

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 $K \subseteq L \subseteq M$.

$\endgroup$
  • $\begingroup$ Is $\mathbb{Q}(\alpha)$ assumed to be Galois? $\endgroup$ – Dean Young Apr 20 at 2:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.