Question

Show that ${\mathbf{Gal}}_{\mathbf{Q}}\mathbf{Q}\left(\sqrt{2},\sqrt{3},\sqrt{5}\right)\mathbf{\cong }{\mathbf{Z}}_{\mathbf{2}}\mathbf{\times }{\mathbf{Z}}_{\mathbf{2}}\mathbf{\times }{\mathbf{Z}}_{\mathbf{2}}$ .

Short answer

${\mathrm{Gal}}_{\mathrm{Q}}\mathrm{Q}\left(\sqrt{2},\sqrt{3},\sqrt{5}\right)\cong {\mathrm{Z}}_2\times {\mathrm{Z}}_2\times {\mathrm{Z}}_2$

Step-by-step solution

Automorphism

The Galols group is denoted by Gal_FK when the set of all F automorphism of K is an extension of F.

It is the Galols group of K over F.

To prove that GalQQ(2,3,5)≅Z2×Z2×Z2

The Galols group of $\mathrm{Q}\left(\sqrt{2},\sqrt{3},\sqrt{5}\right)$ is considered over Q.

The three minimal polynomial equations are ${\mathrm{x}}^2-2$, ${\mathrm{x}}^2-3$and ${\mathrm{x}}^2-5$.

Which takes a root $\sqrt{2}$ to $\sqrt{2}$or $-\sqrt{2}$and similarly for other polynomials.

There are atmost 4 automorphism in given ${\mathrm{Gal}}_{\mathrm{Q}}\mathrm{Q}\left(\sqrt{2},\sqrt{3},\sqrt{5}\right)$over Q such that there are three possibilities:

${\sqrt{2}}^{\mathrm{i}}\to \sqrt{2},{\sqrt{2}}^{\mathrm{r}}\to \sqrt{2},{\sqrt{2}}^{\mathrm{\alpha }}\to \sqrt{2},{\sqrt{2}}^{\mathrm{\beta }}\to -\sqrt{2}$

And,

${\sqrt{3}}^{\mathrm{i}}\to \sqrt{3},{\sqrt{3}}^{\mathrm{r}}\to \sqrt{3},{\sqrt{3}}^{\mathrm{\alpha }}\to \sqrt{3},{\sqrt{3}}^{\mathrm{\beta }}\to -\sqrt{3}$

Same as,

${\sqrt{5}}^{\mathrm{i}}\to \sqrt{5},{\sqrt{5}}^{\mathrm{r}}\to \sqrt{5},{\sqrt{5}}^{\mathrm{\alpha }}\to \sqrt{5},{\sqrt{5}}^{\mathrm{\beta }}\to -\sqrt{5}$

Now, the minimal polynomial is isomorphism.

So, $\mathrm{\sigma }:\mathrm{Q}\sqrt{2}\cong \mathrm{Q}\sqrt{-2}$

$\mathrm{\sigma }:\sqrt{2}\cong \sqrt{-2}$

For, ${\mathrm{x}}^3-3$ over $\mathrm{Q}\sqrt{2}$ and sigma extends to a Q automorphism.

$\mathrm{Q}\left(\sqrt{2},\sqrt{3}\right)=\mathrm{Q}\left(\sqrt{2},\sqrt{3}\right)$

Since, $\zeta \sqrt{3}=\sqrt{3}$

Therefore, $\mathrm{\zeta }\in {\mathrm{Gal}}_{\mathrm{Q}}\mathrm{Q}\left(\sqrt{2},\sqrt{3}\right)$

Further, each $\left(\mathrm{\zeta },\mathrm{\alpha },\mathrm{\beta }\right)$has order 2 in ${\mathrm{Gal}}_{\mathrm{Q}}\mathrm{Q}\left(\sqrt{2},\sqrt{3}\right)$

Then,

$\begin{array}{rcl}\left(\mathrm{\zeta }\circ \mathrm{\zeta }\right)\sqrt{2}& =& \sqrt{2}\\ & =& \mathrm{i}\sqrt{2}\end{array}$

Similarly,

$\begin{array}{rcl}\left(\mathrm{\zeta }\circ \mathrm{\zeta }\right)\sqrt{3}& =& \sqrt{3}\\ & =& \mathrm{i}\sqrt{3}\end{array}$

Now, similarly for $x^2-5$, sigma extends to a Q automorphism of

$Q\left(\sqrt{2},\sqrt{3},\sqrt{5}\right)=Q\left(\sqrt{2},\sqrt{3},\sqrt{5}\right)$

Since, $\zeta \sqrt{5}=\sqrt{5}$

Therefore, $\mathrm{\zeta }\in {\mathrm{Gal}}_{\mathrm{Q}}\mathrm{Q}\left(\sqrt{2},\sqrt{3},\sqrt{5}\right)$

Further, each $\left(\zeta ,\alpha ,\beta \right)$ has order 2 in$Ga{l}_{Q}Q\left(\sqrt{2},\sqrt{3},\sqrt{5}\right)$

Thus, it is proved that $Ga{l}_{Q}Q\left(\sqrt{2},\sqrt{3},\sqrt{5}\right)\cong Z_2\times Z_2\times Z_2$.