But this doesn't feel very "constructive" to me. Sign up or log in Sign up using Google. Sign up using Email and Password. So, perhaps my argument is no argument after all The Galois group of a polynomial only exists with reference to a specified field containing the coefficients. I recently found an answer to my own question that no one else has brought up, so I thought I'd post it. Viewed 17k times. Feedback post: Moderator review and reinstatement processes. Why did my reputation suddenly increase by points?

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.

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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.

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

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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 Endomorphism Feedback 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. |

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!

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