First of all, I proved that $\mathbb{Q}(\sqrt{5}+\sqrt{7}+i)=\mathbb{Q}(\sqrt{5},\sqrt{7},i)$.Found that $|G|=|Gal(\mathbb{Q}(\sqrt{5},\sqrt{7},i)|=[\mathbb{Q}(\sqrt{5},\sqrt{7},i):\mathbb{Q}]=8$. That's because $\mathbb{Q}(\sqrt{5},\sqrt{7},i)$ is a splitting field of $f(x)=(x^2-5)(x^2-7)(x^2+1)/\mathbb{Q}$.Finally I showed that every $\sigma \in G $ has an order of 2 and that G is abelian.

Do the above really prove that $G \simeq Z_2 \times Z_2 \times Z_2 ? $

Edit: here is a link on how i showed that $\mathbb{Q}(\sqrt{5}+\sqrt{7}+i)=\mathbb{Q}(\sqrt{5},\sqrt{7},i)$ https://www.overleaf.com/read/bsrbxwnxnkcg

  • $\begingroup$ oops... i meant 1.Thank you for the point out $\endgroup$
    – GGG
    May 23, 2022 at 16:48
  • $\begingroup$ What have you computed for the minimal polynomial of $\sqrt{5}+\sqrt{7}+i$? $\endgroup$ May 23, 2022 at 16:56
  • $\begingroup$ @DietrichBurde I haven't yet , i really thought it wasn't necessary to be honest.I would probably use the formula $Irr(\sqrt{5}+\sqrt{7}+i,\mathbb{Q})=\prod_{\sigma \in G}(x-\sigma(\sqrt{5}+\sqrt{7}+i))$ $\endgroup$
    – GGG
    May 23, 2022 at 17:08
  • 1
    $\begingroup$ Just taking the last sentence alone, the structure of finite abelian groups would say there is only one (the one you identify) isomorphism class of abelian groups of size 8 in which every element is its own inverse. $\endgroup$
    – irh
    May 23, 2022 at 17:14
  • 1
    $\begingroup$ The minimal polynomial is $x^8-44x^6+542x^4+964x^2+841$, and it has Galois group $C_2\times C_2\times C_2$. $\endgroup$ May 23, 2022 at 18:20


You must log in to answer this question.