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12E

Expert-verified

Abstract Algebra: An Introduction

Found in: Page 414

Book edition 3rd

Author(s) Thomas W Hungerford, David Leep

Pages 608 pages

ISBN 9781111569624

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Short Answer

Show that ${Gal}_{Q} Q (\sqrt{2}, \sqrt{3}, \sqrt{5}) ≅ Z_{2} \times Z_{2} \times Z_{2}$ .

${Gal}_{Q} Q (\sqrt{2}, \sqrt{3}, \sqrt{5}) ≅ Z_{2} \times Z_{2} \times Z_{2}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: 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.

Step 2: To prove that GalQQ(2,3,5)≅Z2×Z2×Z2

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

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

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

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

${\sqrt{2}}^{i} \to \sqrt{2}, {\sqrt{2}}^{r} \to \sqrt{2}, {\sqrt{2}}^{α} \to \sqrt{2}, {\sqrt{2}}^{β} \to - \sqrt{2}$

And,

${\sqrt{3}}^{i} \to \sqrt{3}, {\sqrt{3}}^{r} \to \sqrt{3}, {\sqrt{3}}^{α} \to \sqrt{3}, {\sqrt{3}}^{β} \to - \sqrt{3}$

Same as,

${\sqrt{5}}^{i} \to \sqrt{5}, {\sqrt{5}}^{r} \to \sqrt{5}, {\sqrt{5}}^{α} \to \sqrt{5}, {\sqrt{5}}^{β} \to - \sqrt{5}$

Now, the minimal polynomial is isomorphism.

So, $σ : Q \sqrt{2} ≅ Q \sqrt{- 2}$

$σ : \sqrt{2} ≅ \sqrt{- 2}$

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

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

Since, $ζ \sqrt{3} = \sqrt{3}$

Therefore, $ζ \in {Gal}_{Q} Q (\sqrt{2}, \sqrt{3})$

Further, each $(ζ, α, β)$ has order 2 in ${Gal}_{Q} Q (\sqrt{2}, \sqrt{3})$

Then,

$\begin{array}{rcl} (ζ \circ ζ) \sqrt{2} & = & \sqrt{2} \\ = & i \sqrt{2} \end{array}$

Similarly,

$\begin{array}{rcl} (ζ \circ ζ) \sqrt{3} & = & \sqrt{3} \\ = & i \sqrt{3} \end{array}$

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

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

Since, $ζ \sqrt{5} = \sqrt{5}$

Therefore, $ζ \in {Gal}_{Q} Q (\sqrt{2}, \sqrt{3}, \sqrt{5})$

Further, each $(ζ, α, β)$ has order 2 in $G a l_{Q} Q (\sqrt{2}, \sqrt{3}, \sqrt{5})$

Thus, it is proved that $G a l_{Q} Q (\sqrt{2}, \sqrt{3}, \sqrt{5}) ≅ Z_{2} \times Z_{2} \times Z_{2}$ .

Most popular questions for Math Textbooks

Let $K$ is Galois over $F$ show that there are only finitely many intermediate fields.

Question:

(a) Show that every automorphism of R maps positive elements to positive elements. [Hint: Every positive element of R is a square].

(b) If $a, b \in R, a < b, σ \in {Gal}_{Q} R$ , prove that $σ (a) < σ (b)$ . [Hint: a<b if and only if b-a>0].

(c) Prove that ${Gal}_{Q} R = < i >$ . [Hint: If $c < r < d$ , with $c, d \in Q$ , then data-custom-editor="chemistry" $c < σ (r) < d$ ; show that this implies $σ (r) = r$ .

Prove that the Galois group of an irreducible cubic polynomial is isomorphic to Z₃ or W S₃ .

Let K is a normal extension of Q and $[K : Q] = P$ with p prime show that $G a l_{Q} K ≅ Z_{P}$ .

If K is radical extension of F prove that $[K : F]$ is finite.

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