In the proof of (3) => (1) (around 8:30), the statement of (3) seems not quite correct: the automorphism does not really "fix L", but merely maps L to L. Correct? "Fixing L" would mean that the automorphism maps each element of L to itself, which is not what is stated at the beginning or proved in (2) => (3).
The definition of an algebraic extension K < L says, that for each a in L there has to be a p_a(x) ≠ 0 in K[x], such that p_a(a) = 0. (i.e. every element in L is algebraic over K). Of course, you can choose those p_a to be irreducible. I hope this helped.
Happy new year, my good sir.
Thanks for sharing the knowledge!
Great breakdown. thanks!
Galois: let's call these extenstions normal
Gauss: :(
In the proof of (3) => (1) (around 8:30), the statement of (3) seems not quite correct: the automorphism does not really "fix L", but merely maps L to L. Correct? "Fixing L" would mean that the automorphism maps each element of L to itself, which is not what is stated at the beginning or proved in (2) => (3).
he addresses this at 16:00. Fixing L just means that sigma(L)=L.
Happy New YEAR
Happy new year!
Happy new year Sir
For (1)->(2), what guarantees that there is a p_α in K[x] for any α in L?
The definition of an algebraic extension K < L says, that for each a in L there has to be a p_a(x) ≠ 0 in K[x], such that p_a(a) = 0. (i.e. every element in L is algebraic over K). Of course, you can choose those p_a to be irreducible. I hope this helped.
Early for the first time
ye
Exercices pls
Dummit and Foote
Definition of splitting fields for a “set of” polynomials is in lecture of algebraic closure. Now, I do not watch it.