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I'm trying find the Galois Group of $f(x)=x^4+5x^2+5$.

I've finded the roots:

$\alpha_1=i \sqrt{\frac{5-\sqrt5}{2}}$; $\alpha_2=-i \sqrt{\frac{5-\sqrt5}{2}}$; $\alpha_3=i \sqrt{\frac{5+\sqrt5}{2}}$; $\alpha_4=-i \sqrt{\frac{5+\sqrt5}{2}}$;

And i've finded that:

$\alpha_1=i \sqrt{\frac{5-\sqrt5}{2}}$; $\alpha_2=- \alpha_1$; $\alpha_3=\frac{\sqrt5}{\alpha_1}$; $\alpha_4=-\frac{\sqrt5}{\alpha_1}$;

But, the problem is defining the automorphisms. If i define the automorphisms like this:

$\sigma_1(\alpha_1)=\alpha_1$;

$\sigma_2(\alpha_1)=-\alpha_1$;

$\sigma_3(\alpha_1)=\frac{\sqrt5}{\alpha_1}$;

$\sigma_4(\alpha_1)=-\frac{\sqrt5}{\alpha_1}$;

The solution of the exercise says that I should get $\mathbb{Z_4}$, but none of the automorphisms give me a generator of all group. Am I defining automorphisms well?

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1 Answer 1

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The automorphisms is well defined, but i ignored that$\sqrt5 \in \mathbb{Q}(\alpha_1)$ and $\sqrt5=2 \alpha_1^2+5$, then the automorphisms $\sigma_3, \sigma_4$ have order $4$

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