So in Galois Theory, given a polynomial, say of the second degree, the permutation of the roots compose the Galois Group if their algebraic relation holds true when we transpose the roots. If I am understanding this correctly then given a polynomial with roots A,B:
A+B=1 & B+A=1 then (A B) & (B A) e Galois Group
I can't understand why in a rational root polynomial, the group generated is trivial. How does that work? Isn't the inverse still an inverse and different than the identity?
(sorry I don't get the math syntax here yet)