I have a sequence of field extensions $F\subseteq L\subseteq K$ and I need to compute the Galois group of K over L. If $F=\mathbb{Q}$ then FixedGroup(K, L) does exactly this, but I was wondering if there was an easy way to do it when $F\neq \mathbb{Q}$ or do I need to write a code to do this?
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$\begingroup$ This question has nothing to do with F, right? So you can just use AbsoluteField() to reduce to the case $F = \mathbf{Q}$. $\endgroup$– David LoefflerSep 14, 2016 at 19:35
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$\begingroup$ In theory, David's suggestion should work. However, the FixedGroup command only works when $K/\mathbb{Q}$ is normal. Depending on how you have the fields $L$ and $K$, it might be easier to build $L$, write down a polynomial $g(y) \in L[y]$ whose splitting field is $K$, and then use GaloisGroup to compute the Galois group of $g(y)$. $\endgroup$– Jeremy RouseSep 14, 2016 at 20:18
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$\begingroup$ Unfortunately both my fields K and L are naturally defined over F and while L is a subfield of K, I'm having trouble getting magma to define K as an extension of L. For example "RelativeField" requires both fields to simple but my L is not simple over F. So far I haven't thought of a way to obtain g(y) unless I can get K/L to be relative extension or unless I can find a single generator for a given ideal which magma expresses with multiple generators (but I believe it should be principal). $\endgroup$– Christine McMeekinSep 15, 2016 at 19:12
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$\begingroup$ One other thing you could try is to have Magma compute ${\rm Gal}(K/F)$, and then run through the subgroup lattice of this group, call ${\rm GaloisSubgroup}$ on each subgroup to get a polynomial defining the corresponding subfield between $F$ and $K$, and check to see which is isomorphic to $L$. $\endgroup$– Jeremy RouseSep 15, 2016 at 19:34
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$\begingroup$ sadly GaloisGroup(K/F) also does not work because K/F is not simple, but thank you for the suggestion. I do expect K/L to be simple though. $\endgroup$– Christine McMeekinSep 20, 2016 at 14:43
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