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Say we have the polynomial $x^4-2$, the splitting field of this over $\Bbb Q$ is $\Bbb Q(\alpha, i)$, $\alpha=\sqrt[4]{2}$, and its Galois group is isomorphic to $D_8$.

Now I know a way to find the lattice of subfields of $\Bbb Q(\alpha, i)$ in the Galois correspondence.

If we know the subgroup structure of $D_8$ then we can find the corresponding subfield for each subgroup.

Say for example $h$ is an automorphism which maps $\alpha \rightarrow i\alpha$ and $i \rightarrow i$ and another $g$ which maps $\alpha \rightarrow \alpha$ and $i \rightarrow -i$. Then these generate the galois group and a subgroup is $<g,h^2>$ which has as its fixed field the intersection of the fixed fields of $g$ and $h^2$. Thos in turn are found by noting that g has order 2, so its fixed field must be degree 8/2=4 over Q and it must contain $\Bbb Q(\alpha) $ therefore it must be $\Bbb Q(\alpha)$, similarly $h^2$ has fixed field $\Bbb Q(i,\sqrt{2})$, the intersection (fixed field of $g, h^2$) is then $\Bbb Q(\sqrt{2})$

This method works just fine but it requires that you fully know the subgroup structure of the Galois group in order to find the subfields of the galois extension. My questions are

1)Is there a way to find the subfields without knowing anything about the structure of the group

2) Can we then work in the other direction and construct the subgroups from the lattice of subfields.

If I haven't quite asked the right question here but you know of some way of doing these problems without having to know subgroup structures , please feel free to answer anyway.

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  • $\begingroup$ Well, you must know one way or the other, mustn't you? If you cannot point subgroups and also not subfields, how will you find the subgroups/subfields lattices? $\endgroup$ – DonAntonio Jul 13 at 10:18
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    $\begingroup$ So we have a nice elegant way of solving a problem, and you want to know whether there is some other (probably more difficult) way of solving it? The point is that computing with finite groups is a priori much easier than computing subfields of fields. $\endgroup$ – Derek Holt Jul 13 at 12:13
  • $\begingroup$ @DerekHolt I had seen a method which goes like if, say we had $x^4-2$ then we that its splitting field is $\Bbb Q(\sqrt[4]{2},i)$ so we know that we need to $\sqrt[4]{2}$. Its min poly is $x^4-2$, roots are $+_-\sqrt[4]{2}, +_i\sqrt[4]{2}$, but we haven't adjoined I yet so any automorphism can only map to $+_\sqrt[4]{2}$ giving two automorphisms of the galois group , we know though that there should be an isomorphism between fields that adjoin roots of the min poly, so $\BbbQ(i\sqrt[4]{2})$ is another subfield, we also know that $\Bbb Q(i)$ should be a subfield for the same reason, but I $\endgroup$ – excalibirr Jul 13 at 19:54
  • $\begingroup$ don't see how to find using this method say the subfield $\Bbb Q(\sqrt[4]{2}+i\sqrt[4]{2}$, so I suppose my question boils down to how can we extend the method I mentioned in the above comment to find such subfields ? I know it may be more difficult but I'd like to know how to work both ways $\endgroup$ – excalibirr Jul 13 at 19:56
  • $\begingroup$ @DonAntonio I added a method (incomplete ) on how to find subfields given the splitting field, could you tell me how to extend it ? $\endgroup$ – excalibirr Jul 13 at 19:57

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