So I'm working with $\alpha = \sqrt{5+\sqrt{5}}$ and $E=\mathbb{Q}(\alpha)$. The minimal polynomial of $\alpha$ over $\mathbb{Q}$ is $f(x) = x^4 -10x^2 +20$ and I've determined that $E$ is its splitting field. Let $G={\rm Gal}(E/\mathbb{Q})$. Now, $E$ is a finite separable extension, and it's also normal since it's the splitting field of $f$, so we know then that $[E:\mathbb{Q}] = 4 = \left|G\right|$. So either $G\cong \mathbb{Z}_4$ or $G\cong\mathbb{Z}_2\times\mathbb{Z}_2$. Finding the actual automorphisms in $G$ is my problem. I've looked at this question and the last answer given by oliverjones, but I don't understand how he determines the automorphisms like that. I would appreciate any help, either explaining the answer of oliverjones or a standalone answer/hint to my problem.
2 Answers
As in the linked question, it is not hard to determine that the roots of $f$ are $$\alpha_1:=\sqrt{5+\sqrt{5}},\qquad\alpha_2:=\sqrt{5-\sqrt{5}},\qquad\alpha_3:=-\sqrt{5+\sqrt{5}},\qquad\alpha_4:=-\sqrt{5-\sqrt{5}}.$$ You noted that $|\operatorname{Gal}(E/\Bbb{Q})|=4$ so there are $4$ field automorphisms of $E$, and they are determined by permutations of these roots. We have the identities $$\alpha_1+\alpha_3=\alpha_2+\alpha_4=0\qquad\alpha_1\alpha_2=\alpha_3\alpha_4=2\sqrt{5},\qquad\alpha_1\alpha_4=\alpha_2\alpha_3=-2\sqrt{5},$$ and it is easily checked that $\alpha_1\alpha_3\neq\pm2\sqrt{5}$ and $\alpha_2\alpha_4\neq\pm2\sqrt{5}$, and similarly $$\alpha_1+\alpha_2\neq0,\qquad\alpha_1+\alpha_4\neq0,\qquad\alpha_2+\alpha_3\neq0,\qquad\alpha_3+\alpha_4\neq0.$$ So then the first chain of equalities tells us that $\sigma(\alpha_1)+\sigma(\alpha_3)=0$, and hence $$\sigma(\alpha_1)=\alpha_1\qquad\Rightarrow\qquad\sigma(\alpha_3)=\alpha_3,$$ $$\sigma(\alpha_1)=\alpha_2\qquad\Rightarrow\qquad\sigma(\alpha_3)=\alpha_4,$$ $$\sigma(\alpha_1)=\alpha_3\qquad\Rightarrow\qquad\sigma(\alpha_3)=\alpha_1,$$ $$\sigma(\alpha_1)=\alpha_4\qquad\Rightarrow\qquad\sigma(\alpha_3)=\alpha_2.$$ In the first and third cases we see that $\alpha_1\alpha_3$ is fixed by $\sigma$, meaning that $$\sigma(-5-\sqrt{5})=\sigma(\alpha_1\alpha_3)=\alpha_1\alpha_3=-5-\sqrt{5},$$ from which it follows that $\sigma(\sqrt{5})=\sqrt{5}$. So by the second chain of equalities above we get $$\sigma(\alpha_2)=\sigma\left(\frac{2\sqrt{5}}{\alpha_1}\right)=\frac{2\sqrt{5}}{\sigma(\alpha_1)}\qquad\text{ and }\qquad\sigma(\alpha_4)=\sigma\left(-\frac{2\sqrt{5}}{\alpha_1}\right)=-\frac{2\sqrt{5}}{\sigma(\alpha_1)},$$ which shows that $\sigma$ is determined entirely by where it sends $\alpha_1$. Similarly, in the second and fourth cases we see that $\sigma(\alpha_1\alpha_3)=\alpha_2\alpha_4$, hence that $$\sigma(-5-\sqrt{5})=\sigma(\alpha_1\alpha_3)=\alpha_2\alpha_4=-5+\sqrt{5},$$ from which it follows that $\sigma(\sqrt{5})=-\sqrt{5}$. So just as before we get $$\sigma(\alpha_2)=\sigma\left(\frac{2\sqrt{5}}{\alpha_1}\right)=-\frac{2\sqrt{5}}{\sigma(\alpha_1)}\qquad\text{ and }\qquad\sigma(\alpha_4)=\sigma\left(-\frac{2\sqrt{5}}{\alpha_1}\right)=\frac{2\sqrt{5}}{\sigma(\alpha_1)},$$ which again shows that $\sigma$ is determined entirely by where it sends $\alpha_1$.
Hence these are the four automorphisms of $E$ over $\Bbb{Q}$. Now it remains to figure out what $\sigma(\alpha_2)$ and $\sigma(\alpha_4)$ are in each of these cases, but that is a matter of some handy computation.
Most likely you will soon learn some more advanced methods for determining the structure of the Galois group of a field extension. But looking at symmetric expressions in the roots of a generating polynomial that take values in the base field (or intermediate extensions) is a very concrete approach that also makes very clear what the automorphisms are, in stead of only determining the Galois group up to isomorphism. So if I may I'd recommend trying this approach on another extension for yourself.
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Interesting, I hadn't thought about this approach. I'll be sure to give it a try. Btw, in the second and third chain of equalities, did you mean squareroots of $20$ and $-20$? – Auclair May 30 '16 at 21:24
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Ah you're right, that should be $\sqrt{20}$ and $-\sqrt{20}$. Unfortunately that makes the argument a bit less smooth, though a similar reasoning will work. I'll fix it up in a bit. – Servaes May 30 '16 at 21:25
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Wow, this is fantastic. I'll have a go at it tomorrow, thank you! – Auclair May 30 '16 at 23:51
The roots of this polynomial are $\sqrt{5+\sqrt{5}}$, $-\sqrt{5+\sqrt{5}}$, $\sqrt{5-\sqrt{5}}$ and $-\sqrt{5-\sqrt{5}}$ the discriminant of the polynomial is not squart so th galois group is cyclic.
other method is to look at how many are there intermediate extension:
the intermediate extensions are given by the coefficients of the following polynomial
$(x-\sqrt{5+\sqrt{5}})(x+\sqrt{5+\sqrt{5}})=x^2-5-\sqrt{5}$
$(x-\sqrt{5+\sqrt{5}})(x-\sqrt{5-\sqrt{5}})= x^2-x\sqrt{\left( 5-\sqrt{5}\right) }-\sqrt{\left( 5+\sqrt{5}\right) }x+\sqrt{\left( 5+\sqrt{5}\right) }\sqrt{\left( 5-\sqrt{5}\right) }$
$(x-\sqrt{5+\sqrt{5}})(x-\sqrt{5-\sqrt{5}})=x^2-x\sqrt{\left( 5-\sqrt{5}% \right) }-\sqrt{\left( 5+\sqrt{5}\right) }x+\sqrt{\left( 5+\sqrt{5}\right) }% \sqrt{\left( 5-\sqrt{5}\right) }$.
The first polynomial give $\Bbb{Q}(\sqrt{5})$
the second $\Bbb{Q}(\sqrt{( 5-\sqrt{5}) }-\sqrt{( 5+\sqrt{5}) },\sqrt{(5+\sqrt{5}) }\sqrt{( 5-\sqrt{5}) })$ then contain $\sqrt{( 5-\sqrt{5}) }+\sqrt{( 5+\sqrt{5}) }$ and so contian $\sqrt{( 5+\sqrt{5}) }$ the field is the same as the whole field $\Bbb{Q}(\alpha)$
for the third polynomial, the field is also the whole field $\Bbb{Q}(\alpha)$, and so there are only an field intemidiate and the galois group for the whole field is cyclique $Z_4$
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in the other answer not the perfect automorphism, because if $\alpha$ send to $\alpha$ this is the identity – m.idaya May 30 '16 at 21:24
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the four automorphisms are : $\alpha\mapsto\alpha, \alpha\mapsto -\alpha, \alpha\mapsto\beta $ and $\alpha\mapsto -\beta$ : – m.idaya May 30 '16 at 21:35