Uploaded January 2025 | Updated September 2026, 2 weeks ago
An elegant proof that a separable polynomial is irreducible if and only if its Galois group acts transitively on its roots. This applies for polynomials with coefficients in any field. The proof utilizes the fundamental theorem of Galois theory and the Orbit-Stabilizer theorem to directly draw the desired equivalence.
At 1:59, there is one more implicit condition for the map to be a group action, which is that if ɸ ∈ G and r ∈ Z, then ɸ(r) ∈ Z. In other words, elements of the Galois group must map roots of f to other roots of f. I prove that condition in this video: youtu.be/SKlrzUcYvr0
Orbit-Stabilizer Theorem explanation: youtu.be/De09CSdzaeo
Group Theory playlist: youtube.com/playlist?list=PLug5ZIRrShJHDvvls4OtoBHi6cNnTZ6a6
0:00 Setup
3:30 Proof
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Music: C418 - Smooth Fall
An elegant proof that a separable polynomial is irreducible if and only if its Galois group acts transitively on its roots. This applies for polynomials with coefficients in any field. The proof utilizes the fundamental theorem of Galois theory and the Orbit-Stabilizer theorem to directly draw the desired equivalence.
At 1:59, there is one more implicit condition for the map to be a group action, which is that if ɸ ∈ G and r ∈ Z, then ɸ(r) ∈ Z. In other words, elements of the Galois group must map roots of f to other roots of f. I prove that condition in this video: youtu.be/SKlrzUcYvr0
Orbit-Stabilizer Theorem explanation: youtu.be/De09CSdzaeo
Group Theory playlist: youtube.com/playlist?list=PLug5ZIRrShJHDvvls4OtoBHi6cNnTZ6a6
0:00 Setup
3:30 Proof
Subscribe to see more new math videos!
Music: C418 - Smooth Fall
