Suppose I have fields $F \leq L_1 < L_2 \leq E$, where $E/F$ is separable, $[E : F] = p^n$ for a prime number $p$, and $L_2 / L_1$ is Galois and of degree $p$. In addition, I have a subfield $F \leq K \leq E$. Is it true then that $L_2 \cap K / L_1 \cap K$ is Galois (and of degree $1$ or $p$)?

I have proven this in the analogous case for groups, where you replace Galois-ness with normality. However, that proof relies on the Second Isomorphism Theorem, and I cannot find a way of adapting it to the case of fields. As a partial result, $E/F$ is separable, so $L_2 \cap K / L_1 \cap K$ is separable as well, so we only need to show it is also normal to get Galois.

Edit: This is my attempt at solving a more general question I asked here, so it may turn out that this statement is false if this isn't the correct approach.

  • $\begingroup$ Is the question really to classify the Galois $p^m$ extensions ? $\endgroup$ – reuns Nov 22 '17 at 4:55
  • $\begingroup$ @reuns It's my attempt to solve another question (I just edited the question to add a link to it). $\endgroup$ – Mr Bingley Nov 22 '17 at 5:00

$L_2 = \mathbb{Q}(\sqrt[3]{2},\zeta_3),L_1 = \mathbb{Q}(\zeta_3)$, $K= \mathbb{Q}(\sqrt[3]{2})=L_2\cap K, L_1 \cap K = \mathbb{Q}$ isn't a counter-example because you asked $[L_2:K \cap L_1] = p^m$

Assume $L_2 \cap K/ L_1\cap K$ is of degree $p$ . With $L_3/K \cap L_1$ its Galois closure, then $Gal(L_3/K \cap L_1)$ embeds in $S_p$ (permutation group) thus $[L_3:K \cap L_1] \ | \ p!$. But since $[E:L_3 :L_1 \cap K: F]$ is a power of $p$, then so is $[L_3:K \cap L_1]$ and $L_3 = L_2 \cap K$.

Thus it leaves us with the problem of classifying Galois extensions $L_3/L_1 \cap K$ of degree $p^{3+m}$ with a non-Galois subextension $L_2 \cap K/L_1 \cap K$ of degree $p^{2+l}$

  • $\begingroup$ I'm sorry, I don't know that much about the Galois closure, and don't really understand this argument. $\endgroup$ – Mr Bingley Nov 22 '17 at 5:37
  • $\begingroup$ @MrBingley If $[L_2\cap K: L_1 \cap k] = p^2$ it doesn't work. If $M/k$ is separable of degree $p$ then $M = k(\alpha)$, where $f \in k[x]$ the minimal polynomial of $\alpha$ is of degree $p$, and $N/k$ is Galois with $N = k(\alpha_1,\ldots,\alpha_p)$ and $\alpha_j$ are all the roots of $f$. Thus $Gal(N/k)$ acts by permuting the $\alpha_j$ so it embeds in $S_p$ and $[N:k] \ | \ p!$ $\endgroup$ – reuns Nov 22 '17 at 5:59
  • $\begingroup$ Ahhh, I see. Now if it is true that $[L_2 \cap K : L_1 \cap K] \leq [L_2 : L_1]$, then we could ignore the $> p$ cases and I think that would do it. $\endgroup$ – Mr Bingley Nov 22 '17 at 6:35

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