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12E

Expert-verified
Found in: Page 414

### Abstract Algebra: An Introduction

Book edition 3rd
Author(s) Thomas W Hungerford, David Leep
Pages 608 pages
ISBN 9781111569624

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

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

See the step by step solution

## Step 1: Automorphism

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

It is the Galols group of K over F.

## Step 2: 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 $Ga{l}_{Q}Q\left(\sqrt{2},\sqrt{3},\sqrt{5}\right)\cong {Z}_{2}×{Z}_{2}×{Z}_{2}$ .