2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $4$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.