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I am trying to find the Galois group of $x^{4}+7$ over $\mathbb{Q}$ and all of the intermediate extensions between $\mathbb{Q}$ and the splitting field of this polynomial. I have found that the splitting field is $\mathbb{Q}(i, \sqrt[4]{7}\sqrt{2})=\mathbb{Q}(i,\sqrt[4]{28})$.

Edit: (Justifying the splitting field stated above)

The roots of the polynomial are $\sqrt[4]{7}\big(\pm \frac{\sqrt{2}}{2}\pm i\frac{\sqrt{2}}{2}\big)$ so the spliting field is $E=\mathbb{Q}\bigg(\sqrt[4]{7}\big(\pm \frac{\sqrt{2}}{2}\pm i\frac{\sqrt{2}}{2}\big) \bigg)=\mathbb{Q}\bigg(\sqrt[4]{7}(\pm\sqrt{2}\pm i\sqrt{2} ) \bigg)$.

We have $\sqrt[4]{7}(\sqrt{2}+i\sqrt{2}) \in E$ and $(\sqrt[4]{7}(\sqrt{2}-i\sqrt{2}) \in E$, hence their sum $=2\sqrt[4]{7}\sqrt{2} \in E \implies \sqrt[4]{7}\sqrt{2} \in E$. Then we also have $\sqrt[4]{7}\sqrt{2}+i\sqrt[4]{7}\sqrt{2}-\sqrt[4]{7}\sqrt{2} = i\sqrt[4]{7}\sqrt{2} \in E \implies \frac{i\sqrt[4]{7}\sqrt{2}}{\sqrt[4]{7}\sqrt{2}}=i \in E$. So $\mathbb{Q}(i,\sqrt{4}\sqrt{7}) \subseteq E$. But all of the roots of $x^{4}+7$ are contained in $\mathbb{Q}(i,\sqrt[4]{7}\sqrt{2})$ so we in fact have $E=\mathbb{Q}(i,\sqrt[4]{7}\sqrt{2})=\mathbb{Q}(i,\sqrt[4]{28})$.

I'm pretty confident this is correct and I have determined the Galois group is the automorphisms of the form $ \sigma: \bigg\{ \begin{array}{ll} i \mapsto \pm i\\ \sqrt[4]{28} \mapsto \pm i^{k}\sqrt[4]{28}& k \in \{0,1\} \end{array} $

which will be isomorphic to $D_{8}$, so there should be 10 intermediate subfields.

Edit 2: I think I found a systematic way to find the ten intermediate subfields.

Still trying to figure out what the last three intermediate fields are in terms of my generators.

$\mathbb{Q}(i,\sqrt[4]{28})\\ \mathbb{Q}(i,\sqrt{28})\\ \mathbb{Q}(i)\\ \mathbb{Q}(\sqrt[4]{28})\\ \mathbb{Q}(i\sqrt[4]{28})\\ \mathbb{Q}(i\sqrt{28})\\ \mathbb{Q}\\ $

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    $\begingroup$ Why did you delete your earlier post about this? $\endgroup$ – Zev Chonoles Jul 9 '15 at 1:10
  • $\begingroup$ I deleted my earlier post about this because I initially just asked about the splitting field but decided to ask about the Galois group because I'm also struggling with that. So the initial question was a subquestion of this one and didn't see a good reason to have both of them being asked. Why such concern? $\endgroup$ – TuoTuo Jul 9 '15 at 1:13
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    $\begingroup$ In you pre-edit post: Did you mean that the splitting field was $\mathbb{Q}(i,\sqrt[4]{7},\sqrt{2})$ -- i.e., did you forget a comma? Multiplying yields $\sqrt[4]{28}$, which you have in your edits, but I think you need both or those elements to be separate... which makes it look like $[K:\mathbb{Q}]=16$. I'm learning this too, so hopefully someone will correct me if I'm wrong. $\endgroup$ – dmk Jul 9 '15 at 2:03
  • $\begingroup$ @dmk no, I there is purposely not a comma there. You can conclude that $\sqrt[4]{7}\sqrt{2}=\sqrt[4]{28}$ is in the splitting field because $\sqrt[4]{7}(\sqrt{2}+i\sqrt{2})-\sqrt[4]{7}(-\sqrt{2}+i\sqrt{2})$ is in the splitting field (multiply this out). Once you know that $\sqrt[4]{7}\sqrt{2}$ is in the field you can easily conclude that $i$ is in the splitting field. Since all of the roots of the original polynomial are contaned in the field generated by $\sqrt[4]{7}\sqrt{2}$ and $i$ over $\mathbb{Q}$ these work as generators for the splitting field. $\endgroup$ – TuoTuo Jul 9 '15 at 5:03
  • $\begingroup$ Okay, I've convinced myself that the splitting field is what you found. However, when you can say that $\alpha = \sqrt[4]{28}$ is in $K$ because, etc. ... I mean, that shows $\mathbb{Q}(\alpha) \subseteq K$, but not necessarily equality. An easier way to think of it, I think, is that, having found the roots at the outset -- namely, $\frac{1}{2}\left(\pm\alpha\pm i\alpha\right)$ -- we know we need $K=\mathbb{Q}(\alpha,i)$. $\alpha$ has a fourth-degree minimal polynomial; $\mathbb{Q}(\alpha) \subseteq \mathbb{R}$, so $[K:\mathbb{Q}(\alpha)]=2$. $\endgroup$ – dmk Jul 9 '15 at 13:18
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As you argue, the splitting field for $x^4+7$ is $\mathbb{Q}\bigl(i,\sqrt[4]{28}\bigr)$. Since $\bigl|\mathbb{Q}\bigl(i,\sqrt[4]{28}\bigr) :\mathbb{Q}(i)\bigr| = 4$ and $|\mathbb{Q}(i) :\mathbb{Q}| = 2$, the extension has degree $8$, so the Galois group has order $8$. It's not hard to see that the Galois group is dihedral, and is generated by the automorphisms $r$ and $s$ defined by $$ r(i) = i,\qquad r\bigl(\sqrt[4]{28}\bigr) = i\sqrt[4]{28},\qquad s(i)=-i,\qquad s\bigl(\sqrt[4]{28}\bigr)= \sqrt[4]{28}. $$ Note that $s$ is just complex conjugation.

The dihedral group of order $8$ has ten different subgroups. Here is a list of the corresponding intermediate fields:

  • The whole group corresponds to $\mathbb{Q}$.

  • The cyclic subgroup $\{1,r,r^2,r^3\}$ corresponds to $\mathbb{Q}(i)$.

  • The subgroup $\{1,r^2,s,r^2s\}$ contains $s$, so the corresponding field must be real. It is $\mathbb{Q}\bigl(\sqrt{7}\bigr)$.

  • The third subgroup $\{1,r^2,rs,r^3s\}$ of order 4 must then correspond to $\mathbb{Q}\bigl(\sqrt{-7}\bigr)$.

  • The subgroup $\{1,s\}$ corresponds to the real part of the splitting field, i.e. $\mathbb{Q}\bigl(\sqrt[4]{28}\bigr)$.

  • The subgroup $\{1,r^2\}$ corresponds to $\mathbb{Q}\bigl(i,\sqrt{7}\bigr)$.

  • The subgroup $\{1,rs\}$ corresponds to $\mathbb{Q}\bigl(\sqrt[4]{-7}\bigr)$, where $\sqrt[4]{-7}$ denotes the principle fourth root, i.e. $e^{i\pi/4}\sqrt[4]{7}$.

  • The subgroup $\{1,r^2s\}$ corresponds to $\mathbb{Q}\bigl(i\sqrt[4]{28}\bigr)$.

  • The subgroup $\{1,r^3s\}$ corresponds to $\mathbb{Q}\bigl(i\sqrt[4]{-7}\bigr)$.

  • The trivial subgroup corresponds to the whole splitting field $\mathbb{Q}\bigl(i,\sqrt[4]{28}\bigr)$.

In each case, you can check that the listed field is correct by checking that it has the right degree and that its elements are fixed by the action of the given elements of the Galois group.


Edit: I'd like to answer the question of how one finds all these subfields. It shouldn't take more than a few minutes to find all of them.

Given a subgroup $H$ of the Galois group $G$, one of the best ways of finding generators for the fixed field of $H$ is symmetrization. The way that this works is that you take any element $\alpha$ of the splitting field and compute $$ \sum_{h\in H} h(\alpha)\qquad\text{or}\qquad \prod_{h\in H} h(\alpha). $$ Either of these quantities is invariant under the action of $H$, and therefore lies in the fixed field.

For example, to compute the fixed field for the subgroup $\{1,rs\}$ above, we start with the element $\alpha = \sqrt[4]{28}$ and symmetrize: $$ \alpha + rs(\alpha) \;=\; \sqrt[4]{28} + i\sqrt[4]{28} \;=\; 2\sqrt[4]{-7}. $$ Thus $\sqrt[4]{-7}$ lies in the fixed field. Since $\mathbb{Q}\bigl(\sqrt[4]{-7}\bigr)$ has the right degree, it must be the fixed field for $\{1,rs\}$.

Similarly, to compute the fixed field for $\{1,r^2\}$, we start with $\alpha \sqrt[4]{28}$ and symmetrize multiplicatively: $$ \alpha\,r^2(\alpha) \;=\; \sqrt[4]{28} \bigl(-\sqrt[4]{28}\bigr) \;=\; -2\sqrt{7}. $$ Thus $\sqrt{7}$ lies in the fixed field for $\{1,r^2\}$. Clearly $i$ also lies in this fixed field, and $\mathbb{Q}\bigl(i,\sqrt{7}\bigr)$ has the right degree, so the fixed field for $\{1,r^2\}$ is $\mathbb{Q}\bigl(i,\sqrt{7}\bigr)$.

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  • $\begingroup$ Thanks for your very complete answer Jim. This question is off of one of our quals and it seems like it would take an excessive amount of time in a test situation to write down all of the subfields of $D_{8}$ and determine their fixed fields, unless there is a trick I don't know for doing it quickly. I'm not too keen on using $\sqrt[4]{-7}$ notation. I fixed my duplicate subfields in work above, but now I'm missing 3 of the subfields and am not sure how to get them. $\endgroup$ – TuoTuo Jul 10 '15 at 6:07
  • $\begingroup$ @Andrew I've added a bit of an explanation above. $\endgroup$ – Jim Belk Jul 10 '15 at 14:14
  • $\begingroup$ @ Jim Thanks for the additional explanation. $\endgroup$ – TuoTuo Jul 11 '15 at 16:18
  • $\begingroup$ Jim, I believe you left out a negative sign in the field where you said "Thus $\sqrt[4]{-7}$ lies in the fixed field. Since $\mathbb{Q}(\sqrt[4]{7}$ has the right degree..." $\endgroup$ – TuoTuo Jul 11 '15 at 17:05
  • $\begingroup$ @Andrew Yes, thanks. I have fixed it. $\endgroup$ – Jim Belk Jul 11 '15 at 17:29

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