Thank you Dr. Salomone for your positive and humble contribution . Your videos are unique and excellent. I would like to know whether you may also have videos for the abstract algebra I. Thank you
Very nice, Dr Salomone. One question: Should 2^(2/3)[1+ζ] be 2^(1/3)[1+ζ^2]? I'm asking because the roots of t^3-2 are: 2^(1/3) 2^(1/3)ζ 2^(1/3)ζ^2 and since {t} => 2^(1/3) fixed, and {tr} => 2^(1/3) and 2^(1/3)ζ both fixed, then it seems logical that {tr^2} => 2^(1/3) and 2^(1/3)ζ^2 both fixed
Dear Matt Sir, Thanks for the excellent video! Around 8.37, you have mentioned that Z3 fixes zeta3 and zeta3 square. I'm not able to understand that concept. Could you please share your thoughts when you find time. Thanks !
`Z₃` is a cyclic group, and cyclic groups are the "building blocks" that other groups are made of, so they are always subgroups of other groups ;) (including the symmetric group `S₃`).
@@rockinroggenrola7277 I didn't say "NORMAL subgroups", I just said "subgroups". And I can't think of any group that wouldn't have cyclic groups as (not-necessarily-normal) subgroups, considering that you can always take some generator and calculate its powers until it wraps around, obtaining a cyclic group this way. I think that there even was some theorem about that, if I remember correctly.
He is rather fast paced but I would recommend utilizing the pause button to give yourself time to think on certain areas. Also I recommend watching videos by other people too. Galois Theory is like a language, and it can be hard to explain and get across, so learning it from multiple sources is a great way of getting your head around it.
5:15 the magenta-coloured card with the subfields has Q(alpha) and Q(zeta) mixed up, I think.
Thank you Dr. Salomone for your positive and humble contribution . Your videos are unique and excellent. I would like to know whether you may also have videos for the abstract algebra I. Thank you
Very nice, Dr Salomone. One question:
Should 2^(2/3)[1+ζ] be 2^(1/3)[1+ζ^2]?
I'm asking because the roots of t^3-2 are:
2^(1/3)
2^(1/3)ζ
2^(1/3)ζ^2
and since
{t} => 2^(1/3) fixed, and
{tr} => 2^(1/3) and 2^(1/3)ζ both fixed, then it seems logical that
{tr^2} => 2^(1/3) and 2^(1/3)ζ^2 both fixed
When Matt looks at the camera and tells you something is important, you better buckle up!
Excellent video.Thank you Dr. Salomone.
hey, are you working on Galois's theory?
Dear Matt Sir, Thanks for the excellent video! Around 8.37, you have mentioned that Z3 fixes zeta3 and zeta3 square. I'm not able to understand that concept. Could you please share your thoughts when you find time. Thanks !
On the other hand, at 14:20, he says that Q(zeta) fixes only Q - which makes much more sense, I think.
Do you record these videos at normal speed and then speed them up so that they'll be shorter? 😅 Who can process new material at 78 rpm?
sir any videos on solving polynomial by Galois theory
can z3 be a subgroup of S3?
derci ferreira Yep! It's worth thinking, though, about whether Z3 is a *normal* subgroup of S3 and why/why not.
Matthew Salomone if every group of n order is a subgroup of a simetric group of n order, then Z3 is a subgroup of S3.
`Z₃` is a cyclic group, and cyclic groups are the "building blocks" that other groups are made of, so they are always subgroups of other groups ;) (including the symmetric group `S₃`).
@@bonbonpony Well, not all groups have cyclic groups as normal subgroups. Take for example the alternating groups.
@@rockinroggenrola7277 I didn't say "NORMAL subgroups", I just said "subgroups". And I can't think of any group that wouldn't have cyclic groups as (not-necessarily-normal) subgroups, considering that you can always take some generator and calculate its powers until it wraps around, obtaining a cyclic group this way. I think that there even was some theorem about that, if I remember correctly.
This video is great, but I just can't keep up with its pace. I just started Galois theory
I agree! No one can process new material this fast.
Great video, thanks.
Respected Matthew Sir, I have a general query: Do You have notes on complex analysis as well? Please reply, it's a request...
I wish I did have complex analysis materials because I love that subject very much. But I don't at this time.
Subfields are dual to subgroups.
Rotations are dual to reflections.
Conjugate pairs = duality!
"Always two there are" -- Yoda.
You explained this way too fast without showing us how things work. I've watched the video several times, and I still understand nothing.
+Naveed Sayed watch the prevIous vIdeos you wIll understand
He is rather fast paced but I would recommend utilizing the pause button to give yourself time to think on certain areas.
Also I recommend watching videos by other people too.
Galois Theory is like a language, and it can be hard to explain and get across, so learning it from multiple sources is a great way of getting your head around it.
I actually put this up at 2x speed. Just Slow it down if u want
@@SuperYtc1- I listen at 0.75