How to Get Galois Groups Using Field Extensions

[@plranisch9509]

It was very clear and understandable and I hope it continues. Thank you very much!

[@cameronspalding9792]

@ 9:00 14 is a multiple of 2

[@dibeos]

@@cameronspalding9792 of 4, we put a card in to fix it :) should’ve popped up

[@victork8708]

The Einstein criteria s proof shall i see 💚

[@RickyMud]

A true genius. He even found a single shape that tiles the plane

[@dibeos]

It’s Eisenstein, sorry, read it weird! But we put a card there to correct it :)

[@victork8708]

@@dibeos ty for correction. Let it stay, u know what i meant :p

[@victork8708]

@@RickyMud Have you heared of neo-Riemanian theory? ;)

[@kpt123456]

14 is a multiple of 2.contrary what has been said around 9:00

[@dibeos]

of 4, we put a card in to fix it :) should’ve popped up

[@siddhadevapps]

I certainly like your content, but it could be even greater if you'd double check the statements in your videos, e.g. compare your version of the Eisenstein Criterium with the one from the book you cited.

[@davidhall6712]

Thanks for the great work on this and many of your other videos. I'd like to see an explanation like this using a field with slightly more complex symmetry than the simple conjugate reflections of Q(√2). Please take a look at the video titled "What is the square root of two? | The Fundamental Theorem of Galois Theory". The first 15 seconds of that video shows some graphic depictions of fields that I will describe below, but that video still leaves me with some unanswered questions about how to determine which symmetries exist in these only slightly more complex fields. For example, I'd like to see a discussion of the relationships between Q(√5) and Q(φ) where φ is the golden ratio (1+√5)/2. Also, what's the relationship from a Galois perspective between Q(φ) and the 5th complex roots of unity. All three of these fields (C, Q(√5) and Q(φ)) have 5-fold symmetry and can be used to generate the vertices of a regular pentagon, but the 3 fields are somehow similar but not identical. Similarly, consider Q(ρ,σ) where ρ and σ are the two internal diagonal lengths of a heptagon with a unit edge length. It can equivalently be defined as ρ^2 = 1+σ and σ^2 = 1+ρ+σ with 1 < ρ < σ. That field, like the complex 7th roots of unity has 7-fold symmetry. I happen to know that Q(√7) is a subfield of Q(ρ,σ). How can I determine if there are any other such subfields or symmetry groups or subgroups of such fields which are only slightly more complex than Q(√2)?

[@karl131058]

I'm sorry, but the fields Q(√5) and Q(φ) are identical...

[@davidhall6712]

@@karl131058 Thanks for your comment. Writing my response helped force me to clarify the question in my own mind, although I'm still not much closer to an answer. a) You are correct that the fields expressed as Q(√5) and Q(φ) are identical in the sense that any value of the form a+b√5 can also be expressed as c+dφ and vise-versa where a, b, c and d are all rational numbers, but if I understand it correctly, according to the video entitled "What is the square root of two? | The Fundamental Theorem of Galois Theory", the two expressions do NOT have the same **symmetries**. Specifically, √5 has a set of TWO algebraically interchangeable conjugate values that differ only in sign whereas φ has a set of FOUR algebraically interchangeable values, namely {φ^1, φ^2, φ^3, φ^4). My question is not about the equality of the two fields, but about the different symmetries of the permutations of these elements. Similarly, Q(√7) is a subset of Q(ρ, σ), but the permutations of the two expressions do not have the same symmetries. That's what I would like to see explained or explored. I'd also like to see the relationship between these extensions of Q and the Nth complex roots of unity which have the same symmetries geometrically. b) I corrected the inequality in my original post. You were correct that 1 < ρ < σ. I had originally posted the opposite inequality.

[@karl131058]

@@davidhall6712 please explain your statement about "FOUR algebraically interchangeable values, namely {φ^1, φ^2, φ^3, φ^4)". As far as I know, φ^2=φ+1, so either your "chain" ends here or goes on forever - there's no reason it ends at φ^4... The fields - you admit - are identical, and thus their automorphisms are identical! And they are: the defining polynomial for φ is x^2-x-1, the zeros are φ and 1-φ, which is, if you express it as 1/2*(1±√5) - just changeing the sign at √5 - same automorphism!

[@davidhall6712]

@@karl131058 I think I may be mixing up attributes of Q(φ) with Q(ζ) where, as you said, φ is the positive REAL root of x^2-x-1 and ζ is the COMPLEX 5th root of unity. Have you watched the video I mentioned? As shown in that video, there are permutations of {ζ^1, ζ^2, ζ^3, ζ^4) that can be made without invalidating equations involving them. I'm not sure, but I may be incorrectly trying to ascribe that same property to the permutations of {φ^1, φ^2, φ^3, φ^4) since they are geometrically identical AFAIK. Convoluting these two field extensions and trying to use them interchangeably may be the thing that's confusing me. In either case, I'm trying to wrap my head around the concept shown at the 5 second mark of that video. I appreciate your feedback.

[@karl131058]

@@davidhall6712 I watched that video, did so already half a year ago, as can be seen by the existence of a comment of mine that old concerning automorphisms NOT fixing Q (hint: they don't exist!). There are some other minor mistakes in there, but all in all I found it quite educational and very well done. And I'm quite sure you're definitely getting it wrong there, because, since φ is the solution to a quadratic equation, there can only be a pair(!) of conjugates, as I pointed out before: φ and 1-φ, which is just the same as switching √5 and -√5

[@siarya_math]

CORRECTION: 8:33 It's "Eisenstein" and not "Einstein". Those names look similar though. 8:52 1) should be "not a multiple of p" and not "not a multiple of p^2" 2) should be "the constant multiple is not a multiple of p^2" And you left out the other condition which is for all powers of x that are NOT the highest power of x, the coefficient is a multiple of p. So the constant term is also multiple of p. Also, at 9:00, 14 is not a multiple of 2^2. 14 is a multiple of 2

[@dibeos]

Thanks for your input! you’re right, don’t know how we missed that. we added cards as corrections :)

[@danielc1112]

@@dibeos I can't see these card corrections Edit: Oh, you put them in the summary. That's not really good enough, you can't edit the video with an overlay or something like that? If not, then delete the video and reupload it I think. It's very confusing and too many mistakes

[@adeniltonramos1853]

you are wrong in your videos in time 4:05 grafic 2, 1/2 is in point 2, 3/2.

[@dibeos]

The position of 2 + 1/2i on the graph is actually correct :) Try to take a closer look: The real part 2 places it 2 units along the x-axis, as you rightly pointed out, and the imaginary part 1/2 places it 1/2 unit above the x-axis in the positive imaginary direction. It’s was also represented exactly this way in the book Visual Group Theory, where we got the graph.

[@plranisch9509]

Lovely

[@HelloWorlds__JTS]

I love your passion for teaching this stuff, and I hope you continue and gain more momentum.... But you're losing a significant amount of value and credibility when you aren't careful with your explanations, e.g. already at 1:46 you call 0 the identity element, but in the same sentence as, and context of, the set S being an Abelian group under multiplication. This is a trivial mistake for anyone who knows what you mean, but who is your intended audience? And clearly the other commenters catching other such mistakes should present a huge clue. I hope you are taking the criticism constructively and are determined to make improvements!

[@dibeos]

@@HelloWorlds__JTS thank you for the constructive criticism. They are the consequence of small distractions. But we will fix these mistakes in future videos, because we will change a few things in our process of making videos.

[@deltalima6703]

Your next video was quite good. 👍

[@jursamaj]

I'm not sure what our original language is, but you miss some things in English. "Arithmetic" has pronunciations: with the accent on "-met-", it's an adjective relating to the noun form of arithmetic. This is how you pronounce it in this video. Instead, you should be using noun form, which is pronounced with the accent on "-rith-". You seem to get more common English words right, so it would probably help to consult an English dictionary for pronunciation of less common word.