Galois: let's call these extenstions normal Gauss: :(
@ching-tsunchou2655
3 жыл бұрын
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).
@Brien831
3 жыл бұрын
he addresses this at 16:00. Fixing L just means that sigma(L)=L.
@staj6236
3 жыл бұрын
Happy New YEAR
@clareliao3080
3 жыл бұрын
Happy new year!
@137mathe
3 жыл бұрын
Happy new year Sir
@big-lion
Жыл бұрын
For (1)->(2), what guarantees that there is a p_α in K[x] for any α in L?
@David-mu1eh
Жыл бұрын
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.
@f5673-t1h
3 жыл бұрын
Early for the first time
@migarsormrapophis2755
3 жыл бұрын
ye
@الأستاذنورالدينمداجلية
3 жыл бұрын
Exercices pls
@JuanPerez-qe5dy
Жыл бұрын
Dummit and Foote
@hausdorffm
3 жыл бұрын
Definition of splitting fields for a “set of” polynomials is in lecture of algebraic closure. Now, I do not watch it.
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