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
$\begingroup$
$\endgroup$
14
-
$\begingroup$ oops... i meant 1.Thank you for the point out $\endgroup$– GGGMay 23, 2022 at 16:48
-
$\begingroup$ What have you computed for the minimal polynomial of $\sqrt{5}+\sqrt{7}+i$? $\endgroup$– Dietrich BurdeMay 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 $Irr(\sqrt{5}+\sqrt{7}+i,\mathbb{Q})=\prod_{\sigma \in G}(x-\sigma(\sqrt{5}+\sqrt{7}+i))$ $\endgroup$– GGGMay 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. $\endgroup$– irhMay 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 $C_2\times C_2\times C_2$. $\endgroup$– Dietrich BurdeMay 23, 2022 at 18:20
|
Show 9 more comments