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I've come to a fork in the road, and it is sending me on wild goose chases. This question comes from a final exam for an Intermediate Abstract Algebra course I just took this past Spring. I'm re-working the questions to which there were two problems that require using the Galois correspondence. The second I have yet to start, because I'm trying to finish the final aspect of the first, and I'm almost finished with this except I'm stuck between one of two ways to proceed (if, indeed, I'm correct regarding my work below).

Details & Preliminary Work (Somewhat Rough): We are working in $\mathbb{Q}\subset\mathbb{Q}\big(\sqrt{3+\sqrt{5}}\big)\subset\mathbb{R}$. Let $\alpha=\sqrt{3+\sqrt{5}}\in\mathbb{R}$, then the minimal/irreducible polynomial over $\mathbb{Q}$ is $f(x)=x^{4}-6x^{2}+4=\text{irr}(\alpha,\mathbb{Q})\in\mathbb{Q}[x]$. So, $[\mathbb{Q}(\alpha):\mathbb{Q}]=\text{deg}(\alpha,\mathbb{Q})=\text{deg}\big(f(x)\big)=4$. I've previously proved that $\sqrt{3-\sqrt{5}}\in\mathbb{Q}(\alpha)$, so $\mathbb{Q}(\alpha)$ is a splitting field for the separable polynomial $f(x)\in\mathbb{Q}[x]$ (as the roots of $f(x)\in\mathbb{Q}[x]$, mentioned below, are all simple [i.e. the roots are distinct]). Hence, the extension $\mathbb{Q}\subset\mathbb{Q}(\alpha)$ must be a Galois extension. Furthermore, $\sigma\in\text{Gal}\big(\mathbb{Q}(\alpha)/\mathbb{Q}\big)$ is uniquely determined by its values of $\sigma\big(\sqrt{3+\sqrt{5}}\big)\in\big\{\sqrt{3+\sqrt{5}},-\sqrt{3+\sqrt{5}},\sqrt{3-\sqrt{5}},-\sqrt{3-\sqrt{5}}\big\}$, so the $\big|\text{Gal}\big(\mathbb{Q}(\alpha)/\mathbb{Q}\big)\big|=4$ since all four possibilities must occur. Label the roots as such: $\alpha=\alpha_{1}=\sqrt{3+\sqrt{5}}=-\alpha_{2}$, and $\alpha_{3}=\sqrt{3-\sqrt{5}}=-\alpha_{4}$. We now define $\sigma,\tau\in\text{Gal}\big(\mathbb{Q}(\alpha)/\mathbb{Q}\big)$ as followed:

$\sigma(\alpha_{1})=\alpha_{2}$

$\sigma(\alpha_{2})=\sigma(-\alpha_{1})=-\alpha_{2}=\alpha_{1}$

$\sigma(\alpha_{3})=\sigma\bigg(\dfrac{\pm 2}{\alpha_{1}}\bigg)=\dfrac{\pm 2}{\alpha_{2}}=\alpha_{4}$; where $\alpha_{1}\alpha_{3}=\pm 2=\sqrt{4}=\alpha_{2}\alpha_{4}$

$\sigma(\alpha_{4})=\alpha_{3}$.

And:

$\tau(\alpha_{1})=\alpha_{3}$

$\tau(\alpha_{2})=\tau(-\alpha_{1})=-\alpha_{3}=\alpha_{4}$

$\tau(\alpha_{3})=\tau\bigg(\dfrac{\pm 2}{\alpha_{1}}\bigg)=\dfrac{\pm 2}{\alpha_{3}}=\alpha_{1}$

$\tau(\alpha_{4})=\alpha_{2}$.

Finally, above gives that:

$\sigma\tau(\alpha_{1})=\alpha_{4}$

$\sigma\tau(\alpha_{2})=\alpha_{3}$

$\sigma\tau(\alpha_{3})=\alpha_{2}$

$\sigma\tau(\alpha_{4})=\alpha_{1}$;

whereas the $\text{Gal}\big(\mathbb{Q}(\alpha)/\mathbb{Q}\big)=\big\{e,\sigma,\tau,\sigma\tau\big\}$. My issue resides with the mappings for $\sigma$ and $\tau$ above. The Galois group is isomorphic to a subgroup of $S_{4}$. Also, the Galois group is of order $4$, so it is isomorphic to either the Klein $4$-group, $V_{4}$, or the cyclic group $\mathbb{Z}/4\mathbb{Z}$; however we have the maps above give that the Galois group is isomorphic to $V_{4}$, and $\sigma\mapsto(1,2)(3,4)$, $\tau\mapsto(1,3)(2,4)$, $\sigma\tau\mapsto(1,4)(2,3)$ which are all (even) permutations in $S_{4}$. As $f(x)\in\mathbb{Q}[x]$ is irreducible, the action by the Galois group on the set $\big\{\alpha_{1},\alpha_{2}, \alpha_{3}, \alpha_{4}\big\}$ is a transitive action, and the maps above justify this transitive action. The Klein $4$-group has three, proper subgroups, so the Galois group must also have three, proper subgroups being $\langle\sigma\rangle$, $\langle\tau\rangle$, $\langle\sigma\tau\rangle\subset\text{Gal}\big(\mathbb{Q}(\alpha)/\mathbb{Q}\big)$.

If I am correct above, then the Galois correspondence gives that $\langle\sigma\rangle$ will map to $\mathbb{Q}(\sqrt{5})$, but here is my MAIN problem (again if everything above is correct - especially the automorphisms in the Galois group defined above) - what are the fixed fields that correspond to $\langle\tau\rangle$ and $\langle\sigma\tau\rangle$!? I keep re-working things, and I still can't arrive to even an educated guess, requiring verification, in regards to the fixed fields that $\langle\tau\rangle$ and $\langle\sigma\tau\rangle$ map to by the Galois correspondence. Any help, suggestions, recommendations, hints, tips, etc. will be greatly appreciated. In one attempt, I tried using a basis for $\mathbb{Q}(\alpha)=\mathbb{Q}\big(\sqrt{3+\sqrt{5}}\big)$ as a $\mathbb{Q}$-vector space, and I still can't seem to see what remains fixed - take any $\beta\in\mathbb{Q}(\alpha)$, for $\tau(\beta)$ to remain fixed we must have that $\tau(\beta)=\beta$ whereas $\beta=a_{0}+a_{1}\alpha_{1}+a_{2}\alpha_{1}^{2}+a_{3}\alpha_{1}^{3}$ for some $a_{i}\in\mathbb{Q}$, $i=0,1,2,3$, for example. Moreover, as the $\big|\text{Gal}\big(\mathbb{Q}(\alpha)/\mathbb{Q}\big)\big|=4$, the only other possibility is that $\text{Gal}\big(\mathbb{Q}(\alpha)/\mathbb{Q}\big)\cong\mathbb{Z}/4\mathbb{Z}$, and we are done in this case (?).

Lastly, I'm very happy to provide additional details, side-work, explanations, etc., on anything that is above if need be. The test gives this question with several sub-questions, and I'm stumped on the very last part. That being said, I tried to provide somewhat of a sketch on how I arrived to the last part of the question (which is exactly my Question stated in the title). Overall, thank you for your time, and anything provided to set me straight on this is GREATLY APPRECIATED!

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First of all all your work seems correct to me (there are some typos where you defined the automorphisms)

The intermediate fields:

We note first that $ord(\sigma)=ord(\tau)=ord(\sigma\tau)=2$, hence their index is also $2$.

1) For $\sigma$:

$\sigma(\alpha_1)=-\alpha_1$, hence the square of $\alpha_1$ lies in the fixed field of $\sigma$. We have $\alpha_1^2=3+\sqrt{5}$, hence $\mathbb Q(3+\sqrt{5})=\mathbb Q(\sqrt{5})$ is the fixed field of $\sigma$ because, $\sqrt{5}$ has degree $2$ over $\mathbb Q$.

2) For $\tau$:

$\tau(\alpha_1)=\alpha_3$, hence the sum of both lies in the fixed field of $\tau$: $\alpha_1+\alpha_3=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}=\sqrt{a}$. Squaring both sided yields that $a=3+\sqrt{5}+2\sqrt{4}+3-\sqrt{5}=6+2\sqrt{4}=10$. So the fixed field of $\tau$ is $\mathbb Q(\sqrt{10}$), because $\sqrt{10}$ has degree $2$ over $\mathbb Q$.

3) For $\sigma\tau$:

$\sigma\tau(\alpha_1)=\alpha_4$, hence the sum of both lies in the fixed field of $\sigma\tau:$ $\alpha_1+\alpha_4=\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}=\sqrt{b}$. Squaring both sided again yields: $3+\sqrt{5}-2\sqrt{4}+3-\sqrt{5}=6-4=2$. So the fixed field of $\sigma\tau$ is $\mathbb Q(\sqrt{2})$, because $\sqrt{2}$ has degree 2 over $\mathbb Q$.

I hope there are no mistakes, typos etc. (I am on the way) but thats a way you can solve some of this problems.

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  • $\begingroup$ M.O. - First of all, THANK YOU!! I can't stress enough how much time I spent pondering over this. I knew that the orders of each subgroup were 2, hence the degree of the intermediate field extensions must be 2 as well. That being said, I started trying (mostly) random square roots under the maps $\tau$ and $\sigma\tau$ to see where they map to, but I never got anywhere. I have one question: What gave you the idea to check the sums of both $\alpha_{1}+\alpha_{3}$ and $\alpha_{1}+\alpha_{4}$ under the maps $\tau$ and $\sigma\tau$, respectively??? Again, thank you so much for the help!!! $\endgroup$ – Procore Apr 30 '15 at 22:07
  • $\begingroup$ I also corrected the typos within the defined automorphisms above. M.O. I can thank you enough! My blessing goes out to you, and ignore my question in the comment above, I understand the reasoning now. What incredible theory! $\endgroup$ – Procore Apr 30 '15 at 22:45
  • $\begingroup$ Glad that I was able to help you :) $\endgroup$ – Marm May 3 '15 at 14:32

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