Why you can't solve quintic equations (Galois theory approach) #SoME2

Mathemaniac July 3, 2022
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Mathemaniac

@mathemaniac

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This channel is meant to showcase interesting but underrated maths (and physics) topics and approaches, either with completely novel topics, or a well-known topic with a novel approach. If the novel approach resonates better with you, great! But the videos have never meant to be pedagogical - in fact, please please PLEASE do NOT use YouTube videos to learn a subject. While stylistically similar to 3Blue1Brown, I don't use Manim (Grant's mathematical animation Python library) to make my videos. I actually use PowerPoint, GeoGebra and (sometimes) Mathematica to produce my videos. Pull down the description of my most recent video to see how you can download the files.

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An entry to #SoME2. It is a famous theorem (called Abel-Ruffini theorem) that there is no quintic formula, or quintic equations are not solvable; but very likely you are not told the exact reason why. Here is how traditionally we knew that such a formula cannot exist, using Galois theory. Correction: At 08:09, I forgot to put ellipsis in between. Video chapters: 00:00 Introduction 00:23 Chapter 1: The setup 04:38 Chapter 2: Galois group 11:15 Chapter 3: Cyclotomic and Kummer extensions 19:43 Chapter 4: Tower of extensions 27:25 Chapter 5: Back to solving equations 35:23 Chapter 6: The final stretch (intuition) 43:25 Chapter 7: What have we done? Notes: I HAVE to simplify and not give every technical detail. This is made with the intent that everyone, regardless of their background in algebra, can take away the core message of the video. This can only be done if I cut out the parts that are not necessary for this purpose. As with my previous video series on “Average distance in a unit disc”, this is made to address the question I always had when I was small - treat this as a kind of a video message to my past self. For the “making everything Galois extension” bit, we will need to show that the only things fixed by ALL automorphisms over Q must be in Q itself. This is intuitive, but difficult to justify rigorously. All proofs I know involve “degree of field extension”, and the very satisfying result called the “tower law”, which I deliberately avoided throughout the video because it turns out not to be necessary for the core part of the video. For instance, this proof: https://math.stackexchange.com/questions/962898/on-a-proof-that-the-splitting-field-of-a-separable-polynomial-is-galois The reason we have this mess is that we defined Galois extension using the splitting field of a (separable, i.e. no repeated roots) polynomial. The usual definition given is exactly as above - only things fixed by ALL automorphisms over Q must be in Q itself. This typical definition will of course solve the problem above, but will now create the problem of why this definition implies the larger field is made by adjoining the roots of some polynomial. These two definitions are equivalent, but I just think that it makes much more sense to define it the way I did in the video, in the context of the video; and also I think this is an easier definition to accept. Resources on Tower law: https://en.wikipedia.org/wiki/Degree_of_a_field_extension#The_multiplicativity_formula_for_degrees https://artofproblemsolving.com/wiki/index.php/Tower_law Quotients of solvable groups are solvable (the elementary proof): https://math.stackexchange.com/questions/2203250/quotient-of-solvable-groups-is-solvable-whats-wrong-with-this-proof [The question is already the proof - it is a really elementary way to show the result that we want] More resources on proofs that A_n is not solvable: Sign of permutations: https://en.wikipedia.org/wiki/Parity_of_a_permutation Alternating groups: https://mathworld.wolfram.com/AlternatingGroup.html The proof that A_n is simple (i.e. no non-trivial normal subgroups): http://ramanujan.math.trinity.edu/rdaileda/teach/s19/m3362/alternating.pdf [You need to only go up to Page 5 towards the end of the proof of Theorem 2, but you definitely need group theory lingo] If you know a bit of group theory (orbit-stabiliser and Cauchy), then you can see that the polynomial x^5 - 6x + 3 has the full S5 Galois group, because it is (i) irreducible [this requires Eisenstein’s criterion, see link below], and (ii) exactly two complex roots [and hence the Galois group contains a transposition, i.e. complex conjugation]. Note that the Galois group is transitive. This again needs quite a bit of justification. For the proof assuming transitivity, see here: https://math.stackexchange.com/questions/3075225/f-irreducible-polynomial-with-p-2-real-roots-rightarrow-gal-mathbbq-f Eisenstein’s criterion: https://en.wikipedia.org/wiki/Eisenstein%27s_criterion Other than commenting on the video, you are very welcome to fill in a Google form linked below, which helps me make better videos by catering for your math levels: https://forms.gle/QJ29hocF9uQAyZyH6 If you want to know more interesting Mathematics, stay tuned for the next video! SUBSCRIBE and see you in the next video! If you are wondering how I made all these videos, even though it is stylistically similar to 3Blue1Brown, I don't use his animation engine Manim, but I will probably reveal how I did it in a potential subscriber milestone, so do subscribe! Social media: Facebook: https://www.facebook.com/mathemaniacyt Instagram: https://www.instagram.com/_mathemaniac_/ Twitter: https://twitter.com/mathemaniacyt Patreon: https://www.patreon.com/mathemaniac (support if you want to and can afford to!) Merch: https://mathemaniac.myspreadshop.co.uk Ko-fi: https://ko-fi.com/mathemaniac [for one-time support] For my contact email, check my About page on a PC. See you next time!

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