Exercise 6.2.3

Consider L = ( ζ 5 , 2 5 ) , where ζ 5 = e 2 πi 5 . By proposition 4.2.5, the minimal polynomial of ζ 5 over is x 4 + x 3 + x 2 + x + 1 .

(a)
Show that [ L : ] = 20 .
(b)
Show that L is the splitting field of x 5 2 over , and conclude that Gal ( L ) is a group of order 20.

We will describe the structure of this Galois group in section 6.4.

Answers

Proof. Write ζ = ζ 5 .

(a)
as L = ( ζ , 2 5 ) , Proposition 6.1.4(b) shows that σ Gal ( L ) is uniquely determined by σ ( ζ ) , σ ( 2 3 ) .

Moreover by Proposition 6.4.1(a), σ ( ζ ) is a root of f = x 4 + x 3 + x 2 + x + 1 , whose roots are ζ i , 1 i 4 , and σ ( 2 5 ) is a root of g = x 5 2 , whose roots are ζ j 2 5 , 0 j 4 .

Then Exercice 6.1.1 shows that

| Gal ( L ) | deg ( f ) deg ( g ) = 20 .

Moreover, by Exercise 5.1.8, [ L : ] = 4 × 5 = 20 .

(b)
L is the splitting field of the separable irreducible polynomial g = x 5 2 [ x ] over . Indeed, g is irreducible over by Schönemann-Eisenstein Criterion with p = 2 , and separable since its roots in are 2 5 , ζ 2 5 , ζ 2 2 5 , ζ 3 2 5 , ζ 4 2 5 which are distinct.

By theorem 6.2.1, | Gal ( L ) | = [ L : ] , therefore

| Gal ( L ) | = [ L : ] = 20 .

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2022-07-19 00:00
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