Suppose $L'$ is a fixed cyclic galois extension over $\mathbb {Q} $ of degree $4$.Now we know that there exists also a degree $k$ extension $L$ over $L'$ but the extension $(L/\mathbb{Q})$ may not be the galois extension. So my question is, can we always find such cyclic extension $(L/L')$ of degree $k$ such that $(L/\mathbb Q) $ is also a cyclic galois extension of degree $4k$?
Number Theory – Common Galois Extension Over ?
fieldsgalois-groupsgalois-theorynt.number-theorynumber-fields
Best Answer
If $k$ is odd, then yes. If $L'/\mathbb{Q}$ is a cyclic extension of degree $4$, choose an extension $M/\mathbb{Q}$ that is cyclic of degree $k$. Then the compositum $L'M/\mathbb{Q}$ will have ${\rm Gal}(L'M/\mathbb{Q}) \cong {\rm Gal}(L'/\mathbb{Q}) \times {\rm Gal}(M/\mathbb{Q})$ which will be cyclic of order $4k$.
If $k$ is even, the answer is no. In particular, if $L'/\mathbb{Q}$ is a cyclic extension that is not totally real (like $L' = \mathbb{Q}(\zeta_{5})$), then it is not possible to find a degree $8$ extension $L/\mathbb{Q}$ with ${\rm Gal}(L/\mathbb{Q}) \cong \mathbb{Z}/8\mathbb{Z}$. If such an extension were to exist and $c \in {\rm Gal}(L/\mathbb{Q})$ is complex conjugation, then $c^{2} = 1$ and this implies that $\langle c \rangle = {\rm Gal}(L/L')$, which forces $c|_{L'} = 1$ and hence $L'$ must be totally real.
It should be possible to give necessary and sufficient conditions on when it is possible to find such an extension $L$. It is well-known that a quadratic extension $K/\mathbb{Q}$ can be embedded in a $\mathbb{Z}/4\mathbb{Z}$ extension if and only if $K$ is totally real and the odd primes that ramify in $K$ are all $\equiv 1 \pmod{4}$. There is an extensive literature on Galois embedding problems. For a short intro, see Section 3.4 of Klüners and Malle's paper "A database for field extensions of the rationals". For much more detail, read Chapter 4 of Malle and Matzat's "Inverse Galois Theory".