Hi prof. I have a question at 52:29. Why Is C algebreic over Q? For example π Is in C and It Is not algebreic.
@billkinneymath2 жыл бұрын
You're right. That was a mistake on my part. I think I was thinking about the fact that C is algebraic over R, but that example wouldn't work either because C is a finite (degree 2) extension of R (C = R(i)). A better example would be Q(2^(1/2), 3^(1/2), 5^(1/2),7^(1/2), 11^(1/2),...) as an infinite degree algebraic extension of Q.
@user-ci9oe7ws4l2 жыл бұрын
@@billkinneymath all right, thank you very much. Also this video was really helpfull.
@eddiejennings52626 ай бұрын
Thank you again; this is INSPIRING. Great connections between concepts and examples. I too when first learning did not work through enough concrete examples. Very respectfully!
@billkinneymath6 ай бұрын
Awesome to hear it's inspiring to you!
@aidanhennessey5586 Жыл бұрын
I’m confused. The FTGT only makes statements about galois extensions. Q(cuberoot(2)), for example, is a degree 3 extension of Q but has no non-trivial automorphisms. How are the students expected to know that the 6th degree polynomial splits when you adjoin one root? That seems that a real pain to work out.
@aidanhennessey5586 Жыл бұрын
Furthermore, because the galois group of a galois extension acts transitively on the roots of the polynomial, we should be able to immediately rule out S_3, so the only possible isomorphism class would be Z_6
@billkinneymath Жыл бұрын
Yes, you are correct, as I try to address (but don't do a very good job of doing) at 1:23:36. I should have said that as part of the problem statement. As far as f(x) actually splitting, I never checked that. I think I got this example out of a book somewhere. All in all, I should have chosen a different example or written the problem differently. Thanks for the feedback.