# Frobenius automorphism galois group of a cubic

• Galois group of \$x^3 2 \$ over \$\mathbb Q\$ Mathematics Stack Exchange
• galois theory finite fields, a cubic extension on finite fields. Mathematics Stack Exchange

• Frobenius classes in the Galois group of a polynomial.

. A field homomorphism is a ring homomorphism f: F → F between two fields F, F. Since It turns out that the splitting field of a cubic polynomial f = x3 +ax2 +bx+c ∈ F[​x] has. field of a cubic polynomial) may only have a Galois group isomorphic to S3 or .

Notice that the Frobenius automorphism fixes elements in the prime subfield. a Frobenius element Frobp, an element of the Galois group that reduces to For p = 2,3 the number of cube roots of 2 mod p determines whether Frobp.

its other roots ordered by the action of the Frobenius automorphism.
Marty Green Marty Green 1, 1 1 gold badge 15 15 silver badges 21 21 bronze badges.

Sign up or log in Sign up using Google. I recently found an answer to my own question that no one else has brought up, so I thought I'd post it. Billy Billy 4, 17 17 silver badges 19 19 bronze badges.

So we have a few options from what we've deduced so far.

Frobenius automorphism galois group of a cubic
Qiaochu Yuan Qiaochu Yuan k 34 34 gold badges silver badges bronze badges.

I recently found an answer to my own question that no one else has brought up, so I thought I'd post it. Billy Billy 4, 17 17 silver badges 19 19 bronze badges. Sign up using Facebook. Featured on Meta.

## Galois group of \$x^3 2 \$ over \$\mathbb Q\$ Mathematics Stack Exchange

Pages tells how to find the Galois group of a cubic and quartic.

cyclic Galois group generated by the Frobenius automorphism: [Math Processing Error]. of the Galois group is actually the degree of the field extension. Definition . extension, with cyclic Galois group, generated by the Frobenius automorphism . For example, we know (Fontana-Tartaglia, ) that for a cubic equation. Use Problem 3 to determine the Galois group of an irreducible quadratic is cyclic and is generated by the Frobenius automorphism σ(x) = xp, x ∈ E. Proof.
Contrast this with the seemingly analogous situation of, say, the 6-th roots of unity.

Sign up using Email and Password. Related 3. How are we going to extend this to an element of the Galois group?

Video: Frobenius automorphism galois group of a cubic 302.S7b: Field Automorphisms and Galois Groups

Well, let's try it. I have a general idea that based on the order of the group there's supposed to be at least a 3-cycle.

## galois theory finite fields, a cubic extension on finite fields. Mathematics Stack Exchange

Marty Green Marty Green 1, 1 1 gold badge 15 15 silver badges 21 21 bronze badges.

 Frobenius automorphism galois group of a cubic I just can't figure out a convincing, "constructive" argument to show that I can swap the "real" cube root with one of the imaginary cube roots. Sign up or log in Sign up using Google. Contrast this with the seemingly analogous situation of, say, the 6-th roots of unity.Video: Frobenius automorphism galois group of a cubic Frobenius EndomorphismFeedback post: Moderator review and reinstatement processes. Why did my reputation suddenly increase by points? But this doesn't feel very "constructive" to me.A brute-force way to see it is easy enough.

## 3 thoughts on “Frobenius automorphism galois group of a cubic”

1. Vudomuro:

If the adjoined roots are taken from a larger "pre-existing" field such as the complex numbers, an abstract choice of splitting field is equivalent to selecting one nontrivial cube root of 1 two choices and one cube root of 2 three choices. Well done and thanks!

2. Kagak:

Sign up using Facebook. Why did my reputation suddenly increase by points?

3. Zolozil: