Galois group of R over Q

[@juliefinkjulesheartmagic1111]

Great vid!

[@darkshoxx]

Okay so Galois theory has been a while for me, but my first thought was that can't be true because you have a sequence of field extensions like R>Q(i)>Q, and [Q(i):Q]=2 so [R:Q]=[R:Q(i)]*[Q(i):Q] = [R:Q(i)]*2 > 1, so the group cannot be trivial. I'm assuming my thought is wrong because that property only holds for actual Galois extensions, i.e. normal and separable, and R/Q isn't even algebraic to begin with, so all of those rules are out of the door immediately, is that it? You can still always define a/the Galois group as autos in the big that fix the small, but they're not Groups of Galois field extensions in the original sense. Is that right?

[@coconutmath4928]

I'm assuming by Q(i) you mean Q adjoin i, in which case the inclusion R>Q(i)>Q doesn't hold because i is a complex number? It is definitely true that R is an infinite dimensional extension of Q, but that doesn't matter for the group of automorphisms because it's not a Galois extension (which I think is what you were getting at).

[@darkshoxx]

@@coconutmath4928 yeah excuse my monkey brain, I was looking for any extension of degree 2 from Q in R like sqrt(2), and took the one that doesn't even land in R, my bad. If anything it confirms my initial proposition: "Galois theory has been a while for me" 😄 So yeah imagine that entire argument with sqrt(2). And the answer is, the entire thing isn't galois, so we can't use arguments of degrees of towers..?

[@miki_lip]

but could there exist a polynomial whose splitting field is R?

[@penguinlord3918]

not a polynomial with coefficients in Q, since that polynomial would only have finite roots.

[@miki_lip]

@@penguinlord3918 yeah that makes sense

[@coconutmath4928]

I will agree with the existing comments :)

[@gqip]

The field extension R/Q is not Galois. It’s not even algebraic.

[@moyangwang]

Right... There is no Galois group R/Q. I guess the title means "the group of automorphism of R that fixes Q is trivial".

[@coconutmath4928]

Yeah... I am debating whether I should reupload the video to fix that haha. "Aut(R/Q)" is probably better