One of the few aspects of TH-cam that improves with time is the growing audience for content like this. It has always been a minuscule percentage of users, but with the growth of the user base there’s enough viewership to reward and motivate content creators like this. I’ll never take this for granted.
There is a broader viewership, consequently resulting in more motivation for any uploader but as you said, by force. If you think about all the crappy clickbaity content which is being uploaded over time just for the sake of some views, fame, (easy 'money'?) and it's only discouraging, so in terms of percentages I wouldn't risk it one bit. But all in all, I agree with you, good quality content can only enrichen the whole platform.
Anyone who was around when the internet first became available to the public knows that this kind of content was all you could get (in more rudementary, non-video form). The "average" person on the internet in the early-mid 90's was very educated and polite. It was a joy amd an honor to have experienced it.
I agree with the sentiment that perhaps too large of a jump was taken for the relationship between the tiling and the lack of formula. I followed everything before, and that step came seemingly out of nowhere. I couldn't believe the video was already over, made me feel like something was missing.
To my commentors: of course some books on the topic would cover this. That is not the point. I was saying that previous videos on this channel never made such leaps, and I could always follow them without being a proper mathematician. It was just some feedback towards Aleph 0. I still very much enjoy this video and the channel.
Yeah I agree, I could follow the general idea before the last section. But the last section was too much of a jump, took some time to see , that he is talking about tiling by subgroups and counting the numbers.
@@KemonoFren so this video is useless, it didnt showed the Insolvability of the quintic, it explained a bit of Galois group theory and then jump to conclusion. The same is as if it was a physics channel explaining a bit the atmosphere and by the end saying "bc the sky is blue the grass is green".
Nice explanation, although I'm slightly confused since you didn't explain everything. Where do the subgroup chains come from and what do they represent? I think you should show an example with concrete polynomials. And also, although this is probably a more advanced topic, I'd like to know why exactly do non-prime numbers mean that it's not solvable, since it seems that the entire proof hinges upon this single assumption.
Solvability of groups is defined in terms of some, well, let's call them "decomposition sequences". We take the (finite) group and break it apart into smaller and smaller pieces (the pieces being appropriately chosen subgroups contained in each other) until we hit rock-bottom, i.e. the trivial group. The group being solvable is then some conditions on how exactly this decomposition into smaller pieces takes place. As it turns out (and this is in fact not so trivial) if we have a solvable group we can always take such a "decomposition sequence" and refine it (by inserting some intermediate subgroups inbetween) until every "transition number" is prime. The converse is straightforward (if all "transition numbers" are already prime there's nothing to do) and hence a group is sovable if and only if no non-prime "transition numbers" appear in the "decomposition sequence". The details are bit more delicate (as every rigorous treatment of Galois Theory is) but I hope this conveys the basic idea.
@@Tsskyx No, only if can't get rid of them via interated refinements. So, you can start with, say, the whole group as "decomposition sequence" and then insert some suitable (not any kind of subgroup suffices here) subgroup to get a new refined sequence. If now all "transition numbers" are prime we're happy; otherwise repeat. But there are groups (notably the A_n for n>4 from the video) which just don't have any suitable subgroups to insert. Hence we get stuck at a non-prime "transition number". It's a bit hard to break down without going deep down the mathematical rabbit hole.
Bro, I hadn't checked your channel in a while and I notice there are two videos missing: -How to learn pure mathematics -Math isn't ready to solve this problem What happened to those videos:(? I really loved them.
Dude. I was reading the section about galois theory literally today in Modern Abstract Algebra and I didn't get it at all lol... Anyway, thank you so much for these videos, you truly are a hidden gem of youtube
8:00 "In general, if a polynomial is solvable by radicals the number of tiles is a prime number." But why is this the case? This is kind of a conceptual gap in the video that left me dissatisfied. Great video otherwise!
It has to do with the fact that in some sense every finite abelian group is a product of groups like Z/qZ for some q=p^k, and in some pretty specific sense this means that the number of tiles is a prime number.
There is alot of work that goes into the insolvability of the quintic if you want a complete proof. This is essentially a culminating theorem in a first semester of Galois Theory. I recommend Milne's free online notes on Galois Theory after you've taken/learned basic group and ring theory.
Maybe this is given as a fact because he has difficulty proving it concisely or intuitively in this video. In textbook it is stated that a finite group is solvable if and only if all its composition factors are groups of prime order. If he needs to explain the fact, more details have to be involved.
I wish they had your video during my Abstract Algebra days at Uni. The "roadmap" of a mathematics course at this level is generally obscured by the cramming of fundamental concepts into our heads, not how those concepts have built the map. Thank you for filling in a fundamental part of the map.
i see a lot of people in the comments here who find this explanation unsatisfying, and this is certainly a good instinct to have! the reason is however because to actually understand this result you need a courseful of context, and a 10 minute youtube video can only do so much. i don't think this video was ever supposed to be a comprehensive explanation, but just an introduction to the topic c: btw the way the music loops is killing me
Okay man I have to say that I've watched and read a fair amount of introductory content to Groups, Abstract Algebra and Field Extensions and this is by far the most comprehensive one. It is amazing how clearly you present the ideas, an statement of how organized you have the topic in your own mind, truly amazing! Thanks!
Actually, there is a very "algebrecity" of Sₙ for n ⩾ 5, that wraps this part out. I will copy an argument from my Galois course. Suposse that there are G₀, G₁, ..., Gₘ such that {e} = G₀ ◃G₁ ◃ ... ◃ Gₘ = Sₙ and every Gᵢ₊₁ --- Gᵢ have prime order, i.e., Gᵢ₊₁/Gᵢ is a cyclic group with prime order. Some algebra 1 theory tells us that these quotient group have the comutation property necessarily (it is abelian). Therefore, for any a,b ∈ Gᵢ₊₁, it would aGᵢ bGᵢ = bGᵢ aGᵢ and so a⁻¹b⁻¹ab ∈ Gᵢ. The element a⁻¹b⁻¹ab = [a,b] it is called the commutator, and we had seen that it transfers all the elements in Gᵢ₊₁ to Gᵢ. Okay, thats some generic bullsheet, and now we will see "why" the n ⩾ 5 its important. All those groups are subgroups of Sₙ, i.e., a set o permutations. Lets find out some dirty permutations. Chosse a,b,c,d,e ∈ {1,2,3,...,n} distinct (which can be done cause n ⩾ 5). After a lot good homework, the triplets factor as (a b c) = (d b a) ⁻¹ (a e c)⁻¹ (d b a) (a e c) = [ (d b a) , (a e c) ] Therefore, every triplet is an commutator (which we saw that trasnfers every element one step down. Using those "stepness" property on each group, every triplet in Sₙ should be in the trivial group G₀ = {e}, which is INSANE. WTF ALGEBRA DUDE?! IS THAT ALL ABOUT? SOME MIRACULOUS PROPERTIES OF AN RANDOM NUMBER? yep.
So, in the subgroup chain that was shown in the video, each of the subgroups creates a chain of "simple" factor groups. Factor groups are groups that you get when you "divide" successive groups in the subgroup change. Simple groups are groups that don't have any normal subgroups, a special type of subgroup. In these subgroup chains, each factor group must be simple. In the case of n = 2, S2 only has one subgroup, the subgroup of just the identity element. The factor group S2/{e} is a simple group. In the case of n = 3, S3 has a normal subgroup, namely C3, which is the cyclic group of order 3. When you take the factor group S3/C3, you get a simple group, and C3 can then be decomposed to the group with just the identity. However, at n >= 5, Sn can first be decomposed to the group An, which is the set of even permutations. However, An turns out to be simple for n >= 5, so the only subgroup which may come after it in the subgroup chain is the subgroup with just the identity. Therefore, you don't get a prime tiling from Sn to {e}. Instead, you get n!/2, which is not prime for n >= 5.
@@gustavopauznermezzovilla4833 Dude, that's crazy. It's well-known that solvability is equivalent to the commutator chain becoming trivial, but I didn't know you can use that to easily prove that S_5 is not solvable. However, there is a stronger statement than this, namely that A_n is simple for n\ge5, where that argument doesn't work. So that cool proof is sadly not as useful as I had hoped.
I think there was a video of yours in which you gave your recommendation of mathematics books. I can't find it on youtube, I've been trying to for a while. What happened to that video?
Although there may be a detail or two that you did not explain, still you break the discussion down to the crux of the matter and thus make the discussion accessible to the uninitiated. That is inspiring! Now I really want to learn the details. Thank you 😊
Many thanks. I did a course on Galois Theory in the third year of my maths degree, 40 years ago. I can't recall much of it, but this video has brought bits of it back to me. This has whetted my appetite to read up on it.
Two tiny nitpicks: First, at 2:31 I think it's misleading to write it as at every step a n-th roots is adjoined. This is not necessarily the case as one of your examples illustrates too. Might've been better to use n_i instead to indicate that the degree of the extension is not necessarily the same at each step. Second, I agree with the comment by @diribigal that the phrase "there is no general formula" is not exactly precise and might cause some confusion. For example, X^5-2 is a sovable polynomial of degree 5 as the Galois group is not all of S_5 but some proper solvable subgroup. However, at 8:25 you write "Gal(general degree n polynomial)=S_n" which, IMO, only adds to the confusion: the general polynomial meant here is in some variables t_1,...,t_n with no algebraic relation between them and not any polynomial over, say, Q. I know that it's hard to be this precise and still concise at the same time so I thought it might be worth adding a comment. Last but not least: I really enjoy your videos and I'm looking forward to more!
I loved Galois Theory. I wish I could continue this path, but the other mathematical subjects were too hard for me and could not pursue getting my Ph.D. I ended up taking 2 yeas self studying abstract algebra just to understand this proof. It was worth it.
It's my understanding that you basically need to be a genius to get a PhD. If this proof gives you so much trouble, then maybe a PhD isn't the right thing for you.
Love your videos! A recent problem set of mine involved proving that A_n is simple for n>4 and this is making me excited to learn Galois theory this spring
You should make a sister video called "The Insolubility of the Quintic" where you just drop a 3d-printed x^5 into a glass of water and watch as it doesn't dissolve for 10 minutes.
You used to have a great video for people who wanted to study pure maths with a syllabus suggested you tube videos and books where did it go I would love to see it again or at least have a list of the videos and books thanks
Great job!! I always wondered how that worked out. As I specialized in functional analysis and probability theory, I never got the chance to dive deeper into this stuff. Thank you!
A Cambridge University math course that comprehensively explains Galois theory so elegantly in 10 Mins. Wow thank you so much. I feel cheated by my university that could never really explain anything. We were simply following procedures.
You had a video on books to read for people who want to learn college mathematics and it had a link to an open letter for calculus, can you share it again please? Or anyone else who passes bu here and know what I am talking about.
I had a surprisingly difficult time finding someone bold enough to make the simple, explicit claim that nested radicals correspond to field extensions. You'd think that would be easier to find, but I had a heck of a time finding it. Didn't see it on math stack exchange, and darn sure didn't see it in grad level textbooks because they never explain anything clearly. Didn't find it until seeing it in your video. Thank you very much.
Congrats. This is the most understandable and intuitive presentation of the Galois Correspondence and quintic insolubility. Contents of books are a torture to assimilate. Other lectures (lecture series) can't see the forest for the trees. Thank you
Thank you for the video. But, to be honest, the last part of the video was not comprehensible for me. I hope someone would be able to clarify it to me.
Yes, when n\geq 5, you get a composition series S_n>A_n>{e}, and Jordan-Holder says that you’ll never find any other composition series, so S_n is not solvable.
I am really amazed by how mathematicians and engineers are incapable of giving conceptual explanations of some fundamental ideas in their areas, but you did it!! you made me understand what solvable by radicals means, in a simple and insightful way; many thanks! the most useful part for me was the beginning, when you explained we had to add radicals step by step; I know the Insolvability of the Quintic isn't in contradiction with the Fundamental Theorem of Algebra, but then what is a root when it is not given in radicals?! this is also a thing which I think is fundamental but I can't find a good insightful nutshell explanation of;
really nice Video! I quite like the long end card (like 20sec), because i gives time to rethink everything you said, before having to interrupt the autoplay
Great stuff as usual, but I still don't completely get it (logical, though - it's ten minutes, and I have only the most basic rudiments of Abstract Algebra).
Don't worry, it's just not possible to completely "get it" in 10 minutes. Galois theory is the culminating end result of what turned out to be a 300 year effort by various mathematicians to answer one question: Does there exist a "quintic formula" and above? The quadratic formula was known in ancient times (albeit in word form, as symbolism didn't exist yet!), the cubic formula was found around the 1530s, and the quartic formula shortly after around 1545. But what about polynomials with higher degree? Around the early 1800's, mathematics has advanced enough where the time was "just right". And in the 1820's, the matter had been settled by two young geniuses, independently of one another. Abel proved in particular the impossibility of solving the quintic in radicals, while Galois took it a step further, settling the matter once and for all by figuring out exactly when polynomials ARE solvable in radicals. What is truely remarkable is that not only did Galois manage to do that, but he did it when he was 19 or so, and he didn't even have the field concept yet! He was the first to recognize the importance of groups (his terminology, in fact!). He must have possesed immensely incredible insight, having figured out this new angle of attack into an old problem. It's one thing to try and understand something with all the modern cushioning and years of something being established, but a completely different thing to even invent it yourself, and forging the difficult path into that unknown. But even in its modern formulization, Galois theory rests on a bunch of abstract algebra (mostly group and field theory) And it's STILL quite a difficult thing to learn! A lot of texts don't even show you the historical way things were done in the early days of Galois theory, via symmetric functions and whatnot. That's how Galois himself has done it, and I feel a lot of intuition might be lost by not showing these origins.
@@Cardgames4children I believe the reason, that most people are unable to appreciate the beauty of mathematics, is that we are taught it in a way that makes it impossible for most people to see for ourselves the reasoning involved in producing the mathematical results that we are taught. We have no intuition for why something is true or why a certain technique is able to solve a problem. If the origins of mathematical concepts and how to make an attempt at a proof of mathematical statements were taught alongside the things which are required to solve problems, then most people would have a better appreciation for mathematics.
@@MrAlRats totally agree. While I appreciate the beauty of mathematics itself and the ideas / concepts it contains, I also really like the beauty of how those things fit together or why theorems are true. Seeing the thread of logic tie different ideas together at each step, then clearly reaching the final statement ie the theorem you're proving. It's fun to sit back and just marvel how closely-knit the concepts are. One of my favorites is showing why a field extension F(a) has a dimension equal to the degree of the minimal polynomial p(x) of a. Let it be n. F(a) is the set of all polynomials in a with coefficients in F: F(a) = {A(a) : A(x) € F[x]} By the division algorithm, we can write A(x) = q(x)*p(x) + r(x), where deg(r(x)) < n. So, A(a) = q(a)*p(a) + r(a) = r(a), since a is a root of p(x). Thus any element of F(a) can be written in the form r(a), where r(x) is a polynomial of degree n-1: f_0 + f_1 * a + ... + f_n-1 * a^(n-1). Is the set {1, a, a^2, ..., a^(n-1)} independent? If it were NOT, then some of the above f_i would be nonzero, and so a would be the root of a polynomial of degree n-1, this is less that it's minimal polynomial degree and hence not possible. Hence all the f_i are 0, and so {1, a, a^2, ..., a^(n-1)} are independent. Therefore {1, a, a^2, ..., a^(n-1)} forms a basis of length n, and so F(a) has dimension n. What I love here is when the any-old-normal polynomial f_0 + f_1 * a + ... + f_n-1 * a^(n-1), viewed as a single object, is now taken, not as a polynomial, but as a linear combination of F-elements with the vectors {1, a, a^2, ..., a^(n-1)}. It was such a slight of hand that I think this proof is amazing, and shows the beauty of mathematics!
I really like the style of your videos, with the handwritten illustrations. What do you use to edit and create your videos? Also, thanks for sharing your mathematical knowledge!
I am lost at “equations involving those roots.” Where do these equations come from? Do I just write down any equation to discuss roots’ interchangeability within it?
The most impressive thing is that Galois came up with most of his ideas and wrote everything just a couple of days before he died. At the time he was only 20.
I'll start with the pros of the video. Everything you explained was done very well through helpful visuals that made the content more digestible. Now for the cons: tldr, you missed a spot. For anyone in my position, with little to no knowledge of group theory nor Galois theory for that matter, some of the things in this video seem rather arbitrary, and therefore take away from the impact of this 'proof'. For example, why should the number of tiles be prime? Why should there be tiles in the table in the first place? What is a group table? (I know this one, but could've still done with an explanation) All in all, a good video, but missing bits and pieces that would make it a great one. Edit: Might as well leave a recommendation of sorts. This sort of thing would work much better if done as a series on group theory and galois theory, so that viewers can get to grips with the basics rather than tackling tough problems like this straight away. Then you could leave a video like this to be the grand finale, and viewers would actually understand it if they watched the other videos.
1. 5:10 - Sure, you explained what a algebraic symmetry is. You have equations that the roots satisfy and the solution is symmetric if you can swap those roots while still satisfying that equation, but where do the equations come from? I can come up with a infinite number of equations involving the roots and I'm sure that I can find at least one of them where swapping the roots around in any way would not satisfy the equation. So what defines what equations the roots must satisfy? Same goes for the equations on the extended fields, where do they come from? 2. 5:58 - Correct me if I'm wrong, that group table basically shows the resulting symmetry when 2 symmetries are applied? You never explained how the table is constructed in the first place. I only assumed it was that because I've seen similar tables in the context of symmetries, but who knows, I don't know if I'm right because it was never explained. 3. 7:16 - You can tile that group into sections of equal size, but you never proved that this is a general property of polynomials that are solvable by radicals. Also, you didn't connect the tiling of the group with the removed symmetries in any way, shape or form. It seems that when you extend the group a new restriction is applied that reduces the group to one of its tiles. I could, potentially, tile the group into 8 different tiles if I wanted, and that would break the whole thing since 8 is not prime. The key thing is that you CAN divide into a prime number of groups OR that this division is somehow related to the extensions, but the correlation is never explained. 4. 8:00 - That was just worded in a confusing way. "The number of tiles is a prime number." Wrong! What you mean to say was "The number of DIFFERENT tiles is a prime number." and each unique group is defined as containing a unique group of elementos (the green tile contains 0, 1, 2 and 3, while the blue on contains 4, 5, 6 and 7) with the positioning of elements not mattering, which, if true, is something that was not properly explained. I only understood what you mean on a second/third watch. Also, as others have pointed out, you never explained why it has to be a prime number. 5. 8:14 - The fact that the group is solvable implies that the function is also solvable? Also, we spend the whole video talking about "solvable by radicals", that leads me to think that a function could be solved in other ways. So proving polynomials of degree 5 or more can't be solved by radicals doesn't really prove that there is no solution, unless you also prove that to have a general solution it has to be solved by radicals, which you did not comment on. 6. 8:20 - Now, that whole last section explains nothing and just throws "facts" at the viewer. Like, what is Sn to begin with? By the wording, and according to my previous knowledge, the number of permutations of n things is just n!, so is Sn = n!? 7. 8:30 - Where do those subgroups come from and what are they? 8. 9:36 - Can you? At this point I stopped looking for proves, because I knew I wouldn't receive any. I can see that you put a lot of effort in this video and I'm glad that there are people making quality content for TH-cam in the field of mathematics, but you made a mistake that I've seen many others do, you believed you could properly explain a complicated subject in 10 minutes and, in the process, a lot of stuff that is needed to fully understand the proof was left behind. Sometimes you can get away with it, but particularly in mathematics people usually leave unsatisfied at the end because a lot of "magic" happened somewhere in the middle. This level of topics requires a longer video or even a whole series of videos with more examples, counter-examples and proofs, otherwise we leave knowing just as much as we did when we started watching. Anyway, consider the criticism as friendly advice for next time and keep making videos, I'm looking forward to them, you've just earned a new sub. 🥰
There seem to be some steps missing. And I don't mean the step of having taken Modern Abstract Algebra. Why is it S4--2->A4--3->V4--2->C2--2->e and not S4--2->A4--12->e? Or, alternatively, why is it not something like S5--2->A5--5->W4-->3->V4--2->C2---2->e? Why is the 60 of irreducible complexity when the 12 was not?
As A5 is simple whatever W4 denotes it won't be a normal subgroup of A5. However, solvability requires that every (sub)group in the composition series is a normal subgroup of the the (sub)group immediately preceding it and that the quotients are all abelian. Hence, S4→A4→e is not enough as A4 is non-abelian and S5→A5→W4→V4→C2→e doesn't work as W4 is not normal in A5.
@@mrtaurho8846 Hmm. I attempted a comment last night and it appears not to have gone through. I somehow didn't see the reply until then. I am largely self-taught and have not previously encountered the terminology of a "normal subgroup." However, I think I can surmise that it merely means that it is closed under the group operation. I have seen the term "abelian." But "commutative" is more commonly used, the definition of "abelian" is only that it is commutative, and we are hopefully trying to make this more accessible, not less. The question of why it was not S4 -2->A4-12->e asks why the quartic or biquadratic (I've seen both terms) does not fail the way the quintic does (12 not being prime or even a power of a prime.) And a big part of this gap is that the video does not explain what An is (the set of even permutations of a set with n elements, A2 being the identity permutation.) In preparing my reply to you (twice) I managed to work out what V4 (as mentioned in the video) is. It does help. And I note that your criterion does not match the author's -- at least in a technical sense. V4 is commutative (your criterion) so you would stop there. But it does not have a prime number of elements. (I need to be careful with my words here. I almost said it doesn't have a prime order. It does, 2.) The author breaks it down to C2, which is done wholly arbitrarily. When you tell me "whatever W4 denotes it won't be a normal subgroup of A5" you aren't really adding anything. I am not able to deduce the meaning when you call A5 "simple." I don't recognize the terminology and, perhaps more importantly, the context does not give me sufficient information.
@John Undefined You're missing some fundemental terminology commonly used (AFAIK) in any book on Abstract Algebra. This is fine as I only used it for brevity but let me expand. Normal Subgroup: A normal subgroup is not merely closed under the group operation (this is a subgroup) but satisfies a further technical condition. This condition states the the subgroup is mapped onto itself by conjugation (multiplying from the left by g and at the same time from the right by the inverse of g). Note that not every subgroup is normal in general, although this is the case for abelian groups. Abelian Groups: Indeed, abelian groups are simply commutative groups and this is actually the more common term in the literature. Commutative is more commonly used to describe a property which binary operations (roughly, something taking two inputs and producing one output) may or may not have. So, a group with a commutative group operation is called abelian by convention. Quartic≠Biquadratic: In general, an equation of the form ax⁴+bx³+cx²+dx+e=0 is called a quartic equation while a biquadratic equation is of the special form ax⁴+cx²+e=0, i.e. only the squares of squares appear (so bi-quadratics). V4 in A4: V4 is the Klein Four Group which is a (commutative) subgroup of A4, but not a normal subgroup. The precise definition of these decompositions, however, requires normal subgroups of which there are less. The fact that V4 is abelian is completely irrelevant for whether or not V4 a normal subgroup of A4. Moreover, I did not claimed the criterion being that V4 is abelian but rather that the quotient groups at each step have to be abelian. This is somethinh quite different! Also, breaking down V4 into C2 is the only possible way of further refining this decomposition to get prime order transitions (V4-e has a transition of order 4, which is not prime, while V4-C2-e has two transitions both of order 2, which is prime). And C2 is also the only non-trivial subgroup of V4. Simple Groups: A simple group is a group whose only normal subgroups are the trivial subgroup and the whole group considered as subgroup. If the original group is non-abelian and simple it won't be solvable. That's the crux about A5 (and hence S5). At this point I would advise you to read up the precise definition of solvability. Starting from there you might benefit from learning the necessary preliminaries for understanding the concepts properly. As long as we keep hand-waving some things I don't think it will get clear what I'm saying. If you mind, I can recommend the book by Judson (Abstract Algebra: Theory and Applications) which is freely avaiable online. Hope that helps :)
@@mrtaurho8846 "Quartic≠Biquadratic" I have seen the quartic called the biquadratic and seen the terms used interchangeably. But I not trying to get caught up in the terminology. I am trying to communicate.
Thank you very much, very informative video. Since your handle is Aleph 0, I'd love to hear your take on Cantor's countable/uncountable infinities (I'm not proponent of the idea).
I understand your confusion, but don't know how to clarify. :-) I think he just means the number of individual 'rectangular boxes' marked by colors. So if you can color them with prime number of colors, you're good. If not, either you can after some (complicated) tweaking, or you still need non-prime amount of colors, it's not solvable in radicals. Maybe.
Hello! I am a little mentally challenged and it is very hard for me to follow your wonderfui explanations with the background music (actually, it's only one percussive effect in the music that NaNs me :-) ). Would there be any way to maybe get a video without the music? Maybe on a second channel or something like that? Thank you so much for these productions and the wonderfully analog imagery!
What kind of equations should be preserved exactly when swapping the roots ? I see x_0 + x_1 = 0 and x_1*x_3+x_2*x_4 = 0 but where exactly do those come from ? Edit : Went back in the video and heard him say "every single polynomial relation involving these numbers (the roots) that has rational coefficients"
Great stuff, but you lost me at 8:30. What is e? How did you derive S(n), A(n), C(n), V(n), etc. and the blue or red numbers between them? Why exactly must they be non-prime? I definitely feel I'm missing something. Thanks.
Thanks for this explanation, I've learned quite a bit! It's just not clear to me why the additional equations at 5:406:106:40 must be satisfied. I understand that they are written in such a way that they would "produce" the adjoined element if rewritten slightly: (x_1 - x_2 + x_3 - x_4)/4 = sqrt(2) (x_1 - x_3)/2 = sqrt(3+sqrt(2)) -x_2 + 1 - sqrt(2) = sqrt(3 - sqrt(2)) What am I missing?
Hi.. you post great videos. I was wondering if u can suggest any book(s) on your previous video regarding Navier-Stokes Equation (how it is used in fluid dynamics)?
Now is there a way to generalize the notion of root so that a number that is a non-radical root of some polynome could be used to solve others of the same degree? Or at least do such non-radical numbers still have interesting properties?
i know n-tic formulas with n>5 don't exist in a generalized sense, but would it still be possible to make it on a localized sense? as in, a formula that specifically solves for the roots of the expression ax^5+bx^3+cx+d and also one for ax^5+bx^2+c, and maybe one for ax^5+bx^4+cx^3+dx^4+e (where e is a variable, not the constant)? i don't really know where i could start to prove this problem, and looking it up online didn't really help either, it just gave me an answer on how there isn't a generalized formula for quintics and so on, but does anyone in the comments have an idea? or is it just not possible because of another aspect of group theory or this exact principle covered in the video?
No. The roots of the polynomial x^n-x-1 are provable not expressible algebraically in terms of its coefficients. As this is of the simplest non-solvable type for quintic equation I think your endeavor is hopeless. There are, of course, specific polynomials for this is possible (for example, x^5-1) but in general not. Maybe the Bring radical together with the Tschirnhaus transformations is something of interest to you.
Very interesting video. I've always been curious about field extensions etc (liking abstract algebra), but have never really gotten the occasion to really study it. One question I have is, you say that there's no "general, single" way to solve such equations by radicals... but is there a "family" of specific solutions by radicals for higher degree polynomials ? I know some specific polynomials do have solutions. az^n = c can be solved using n-th real roots and roots of unity. Some polynomials in higher degrees can be reduced to smaller polynomial using X=x^n, and solved in multiple steps, generally with degree 2, 3 and 4 methods for X, then substituting X for x^n. Is there a way to "partition" S_n or A_n into subgroups to get direct sequences with only prime numbers, thus defining radical-based solutions for specific classes of higher degree polynomials (ie, building solutions for the whole of deree 5 polynomials through a set of multiple formulae) ? I suppose not, since somebody would probably have found it and categorized cases by now if it were possible; but I don't really understand why though.
I don't have an example on me, but there are degree 5 polynomials which can not be solved by radicals at all. There is a way to prove that no general equation exists AND there is not a specific equation for each polynomial. But, I have not learned it. Taking Galois Theory currently, and am unimpressed with how it is being taught online.
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One of the few aspects of TH-cam that improves with time is the growing audience for content like this. It has always been a minuscule percentage of users, but with the growth of the user base there’s enough viewership to reward and motivate content creators like this. I’ll never take this for granted.
Well said!
There is a broader viewership, consequently resulting in more motivation for any uploader but as you said, by force. If you think about all the crappy clickbaity content which is being uploaded over time just for the sake of some views, fame, (easy 'money'?) and it's only discouraging, so in terms of percentages I wouldn't risk it one bit. But all in all, I agree with you, good quality content can only enrichen the whole platform.
Anyone who was around when the internet first became available to the public knows that this kind of content was all you could get (in more rudementary, non-video form). The "average" person on the internet in the early-mid 90's was very educated and polite. It was a joy amd an honor to have experienced it.
Why have you removed your learning maths video, please reupload
plot twist: they are the same user base in every math channel
I agree with the sentiment that perhaps too large of a jump was taken for the relationship between the tiling and the lack of formula. I followed everything before, and that step came seemingly out of nowhere. I couldn't believe the video was already over, made me feel like something was missing.
There is a lot to go through to explain the specifics that a video like this can't cover. It would be fully covered in a book on Abstract Algebra.
Read about field extensions, any abstract algebra book covers this, of course maybe you lack the pre requisites
To my commentors: of course some books on the topic would cover this. That is not the point. I was saying that previous videos on this channel never made such leaps, and I could always follow them without being a proper mathematician. It was just some feedback towards Aleph 0. I still very much enjoy this video and the channel.
Yeah I agree, I could follow the general idea before the last section. But the last section was too much of a jump, took some time to see , that he is talking about tiling by subgroups and counting the numbers.
@@KemonoFren so this video is useless, it didnt showed the Insolvability of the quintic, it explained a bit of Galois group theory and then jump to conclusion.
The same is as if it was a physics channel explaining a bit the atmosphere and by the end saying "bc the sky is blue the grass is green".
Nice explanation, although I'm slightly confused since you didn't explain everything. Where do the subgroup chains come from and what do they represent? I think you should show an example with concrete polynomials. And also, although this is probably a more advanced topic, I'd like to know why exactly do non-prime numbers mean that it's not solvable, since it seems that the entire proof hinges upon this single assumption.
Solvability of groups is defined in terms of some, well, let's call them "decomposition sequences". We take the (finite) group and break it apart into smaller and smaller pieces (the pieces being appropriately chosen subgroups contained in each other) until we hit rock-bottom, i.e. the trivial group. The group being solvable is then some conditions on how exactly this decomposition into smaller pieces takes place.
As it turns out (and this is in fact not so trivial) if we have a solvable group we can always take such a "decomposition sequence" and refine it (by inserting some intermediate subgroups inbetween) until every "transition number" is prime. The converse is straightforward (if all "transition numbers" are already prime there's nothing to do) and hence a group is sovable if and only if no non-prime "transition numbers" appear in the "decomposition sequence".
The details are bit more delicate (as every rigorous treatment of Galois Theory is) but I hope this conveys the basic idea.
@@mrtaurho8846 So like, if non-primes still appear in the chain after the first decomposition, then it's not solvable?
@@Tsskyx No, only if can't get rid of them via interated refinements.
So, you can start with, say, the whole group as "decomposition sequence" and then insert some suitable (not any kind of subgroup suffices here) subgroup to get a new refined sequence. If now all "transition numbers" are prime we're happy; otherwise repeat. But there are groups (notably the A_n for n>4 from the video) which just don't have any suitable subgroups to insert. Hence we get stuck at a non-prime "transition number".
It's a bit hard to break down without going deep down the mathematical rabbit hole.
@@mrtaurho8846 I see
@@ikarienator you misunderstood the question
When this channel explains something I already know, somehow it makes me know it better
Bro, I hadn't checked your channel in a while and I notice there are two videos missing:
-How to learn pure mathematics
-Math isn't ready to solve this problem
What happened to those videos:(? I really loved them.
How to learn pure mathematics is here: th-cam.com/video/fo-alw2q-BU/w-d-xo.html It's unlisted.
@@nahblue Why did he unlist them?
Dude. I was reading the section about galois theory literally today in Modern Abstract Algebra and I didn't get it at all lol... Anyway, thank you so much for these videos, you truly are a hidden gem of youtube
@@AmourLearning I recommend richard borcherds videos.
8:00 "In general, if a polynomial is solvable by radicals the number of tiles is a prime number." But why is this the case? This is kind of a conceptual gap in the video that left me dissatisfied.
Great video otherwise!
It has to do with the fact that in some sense every finite abelian group is a product of groups like Z/qZ for some q=p^k, and in some pretty specific sense this means that the number of tiles is a prime number.
There is alot of work that goes into the insolvability of the quintic if you want a complete proof. This is essentially a culminating theorem in a first semester of Galois Theory. I recommend Milne's free online notes on Galois Theory after you've taken/learned basic group and ring theory.
Maybe this is given as a fact because he has difficulty proving it concisely or intuitively in this video. In textbook it is stated that a finite group is solvable if and only if all its composition factors are groups of prime order. If he needs to explain the fact, more details have to be involved.
I had the same issue.
This stuff has never been easy...
What happened to your video guide to self studying mathematics? It was super helpful.
I know bro I am searching for it
I wish they had your video during my Abstract Algebra days at Uni. The "roadmap" of a mathematics course at this level is generally obscured by the cramming of fundamental concepts into our heads, not how those concepts have built the map.
Thank you for filling in a fundamental part of the map.
there's nothing like the feeling of discovering a new maths channel. great content you're severely under-appreciated
i see a lot of people in the comments here who find this explanation unsatisfying, and this is certainly a good instinct to have! the reason is however because to actually understand this result you need a courseful of context, and a 10 minute youtube video can only do so much. i don't think this video was ever supposed to be a comprehensive explanation, but just an introduction to the topic c:
btw the way the music loops is killing me
I just want you to know that I was genuinely happy when I saw this in my abo feed.
Great job Aleph! Been waiting for this for a while
Hello there
I like what u did there. Treating Aleph like his first name and Noll like his second.😂
Okay man I have to say that I've watched and read a fair amount of introductory content to Groups, Abstract Algebra and Field Extensions and this is by far the most comprehensive one. It is amazing how clearly you present the ideas, an statement of how organized you have the topic in your own mind, truly amazing! Thanks!
Me too! I try to tackle these ideas for a while and this kind of 'quick but still precise' recap helps a lot.
Gonzalo Christobal Well said.
👏👏👏 just made sense of a full course on Galois Theory full of formalizations of automorphisms and company and empty of intuition. Great video.
The "why" you never get a prime tiling beyond 4 seems just as mysterious as to why there's no general formula for beyond quintics.
Actually, there is a very "algebrecity" of Sₙ for n ⩾ 5, that wraps this part out. I will copy an argument from my Galois course.
Suposse that there are G₀, G₁, ..., Gₘ such that
{e} = G₀ ◃G₁ ◃ ... ◃ Gₘ = Sₙ
and every Gᵢ₊₁ --- Gᵢ have prime order, i.e., Gᵢ₊₁/Gᵢ is a cyclic group with prime order. Some algebra 1 theory tells us that these quotient group have the comutation property necessarily (it is abelian). Therefore, for any a,b ∈ Gᵢ₊₁, it would
aGᵢ bGᵢ = bGᵢ aGᵢ and so a⁻¹b⁻¹ab ∈ Gᵢ.
The element a⁻¹b⁻¹ab = [a,b] it is called the commutator, and we had seen that it transfers all the elements in Gᵢ₊₁ to Gᵢ.
Okay, thats some generic bullsheet, and now we will see "why" the n ⩾ 5 its important. All those groups are subgroups of Sₙ, i.e., a set o permutations. Lets find out some dirty permutations.
Chosse a,b,c,d,e ∈ {1,2,3,...,n} distinct (which can be done cause n ⩾ 5). After a lot good homework, the triplets factor as
(a b c) = (d b a)
⁻¹ (a e c)⁻¹ (d b a) (a e c) = [ (d b a) , (a e c) ]
Therefore, every triplet is an commutator (which we saw that trasnfers every element one step down. Using those "stepness" property on each group, every triplet in Sₙ should be in the trivial group G₀ = {e}, which is INSANE. WTF ALGEBRA DUDE?! IS THAT ALL ABOUT? SOME MIRACULOUS PROPERTIES OF AN RANDOM NUMBER? yep.
So, in the subgroup chain that was shown in the video, each of the subgroups creates a chain of "simple" factor groups. Factor groups are groups that you get when you "divide" successive groups in the subgroup change. Simple groups are groups that don't have any normal subgroups, a special type of subgroup. In these subgroup chains, each factor group must be simple.
In the case of n = 2, S2 only has one subgroup, the subgroup of just the identity element. The factor group S2/{e} is a simple group.
In the case of n = 3, S3 has a normal subgroup, namely C3, which is the cyclic group of order 3. When you take the factor group S3/C3, you get a simple group, and C3 can then be decomposed to the group with just the identity.
However, at n >= 5, Sn can first be decomposed to the group An, which is the set of even permutations. However, An turns out to be simple for n >= 5, so the only subgroup which may come after it in the subgroup chain is the subgroup with just the identity. Therefore, you don't get a prime tiling from Sn to {e}. Instead, you get n!/2, which is not prime for n >= 5.
@@gustavopauznermezzovilla4833 so basically when n=5, Gn has multiple identities?
@@gustavopauznermezzovilla4833 Dude, that's crazy. It's well-known that solvability is equivalent to the commutator chain becoming trivial, but I didn't know you can use that to easily prove that S_5 is not solvable. However, there is a stronger statement than this, namely that A_n is simple for n\ge5, where that argument doesn't work. So that cool proof is sadly not as useful as I had hoped.
I think there was a video of yours in which you gave your recommendation of mathematics books. I can't find it on youtube, I've been trying to for a while. What happened to that video?
I've seen this topic a bunch of times but never going deep enough and without this level of clarity. Massive thanks.
Although there may be a detail or two that you did not explain, still you break the discussion down to the crux of the matter and thus make the discussion accessible to the uninitiated. That is inspiring! Now I really want to learn the details.
Thank you 😊
Why have you removed your learning undergraduate maths video, pleas reupload
I've not taken any analytic algebra courses and still able to follow along with the basics thanks to your great visuals and explanations.
Your channel rocks! Especially liked the self-study video. Thank you!
Where that video is I can't find it
Many thanks. I did a course on Galois Theory in the third year of my maths degree, 40 years ago. I can't recall much of it, but this video has brought bits of it back to me. This has whetted my appetite to read up on it.
Brillantissime vulgarisation ! En 10 minutes seulement les grandes ligne de la preuve et ses éléments de la théorie de Galois. Merci !
Tout à fait d accord. Seuls les anglo saxons arrivent à faire cela.
Two tiny nitpicks:
First, at 2:31 I think it's misleading to write it as at every step a n-th roots is adjoined. This is not necessarily the case as one of your examples illustrates too. Might've been better to use n_i instead to indicate that the degree of the extension is not necessarily the same at each step.
Second, I agree with the comment by @diribigal that the phrase "there is no general formula" is not exactly precise and might cause some confusion. For example, X^5-2 is a sovable polynomial of degree 5 as the Galois group is not all of S_5 but some proper solvable subgroup. However, at 8:25 you write "Gal(general degree n polynomial)=S_n" which, IMO, only adds to the confusion: the general polynomial meant here is in some variables t_1,...,t_n with no algebraic relation between them and not any polynomial over, say, Q. I know that it's hard to be this precise and still concise at the same time so I thought it might be worth adding a comment.
Last but not least: I really enjoy your videos and I'm looking forward to more!
good point
I loved Galois Theory. I wish I could continue this path, but the other mathematical subjects were too hard for me and could not pursue getting my Ph.D. I ended up taking 2 yeas self studying abstract algebra just to understand this proof. It was worth it.
It's my understanding that you basically need to be a genius to get a PhD. If this proof gives you so much trouble, then maybe a PhD isn't the right thing for you.
Love your videos! A recent problem set of mine involved proving that A_n is simple for n>4 and this is making me excited to learn Galois theory this spring
Just don't get into any duels ;-)
You should make a sister video called "The Insolubility of the Quintic" where you just drop a 3d-printed x^5 into a glass of water and watch as it doesn't dissolve for 10 minutes.
I'm looking forward for when you'll have millions of subscribers. Great presentation, unique aesthetics and great didactics ! perfect!
I love to explore the mathematics this way! Thank you so much.
You used to have a great video for people who wanted to study pure maths with a syllabus suggested you tube videos and books where did it go I would love to see it again or at least have a list of the videos and books thanks
Great job!! I always wondered how that worked out. As I specialized in functional analysis and probability theory, I never got the chance to dive deeper into this stuff. Thank you!
A Cambridge University math course that comprehensively explains Galois theory so elegantly in 10 Mins.
Wow thank you so much. I feel cheated by my university that could never really explain anything. We were simply following procedures.
I have no idea whats going on but, I just love ur video's.
Another totally amazing video. Huge thanks.
Finally, someone can explain this quintic mystery 👏. Thank you!
I'm going to be studying field extensions and Galois Theory next week so this video is extremely helpful.
You had a video on books to read for people who want to learn college mathematics and it had a link to an open letter for calculus, can you share it again please? Or anyone else who passes bu here and know what I am talking about.
looking for the same. please post here if you find.
@@thirdwave1777 I am still looking for it but yes, I ll post it here if I find it sure. Please do the same if you can.
Really enjoying your series of videos. Thank you.
Been waiting for your video for a long time!
You just read my mind wow I started learning about this topic since 2 days. thanks for the great explanation and quality.
5:41 where did that equation come from?
Very nice presentation of a difficult concept. Keep up good work.
Just wanna say, love your videos Aleph! This one was especially interesting and cool, really excited for when I get to group theory in Uni :)
The quality of your content is exquisite!
Keep going with this tempo!
Btw,why did you delete the video on Hodge Conjecture ?
great video as always, I'm looking forward to taking abstract algebra next year and this is an awesome preview thank you
Where are you getting the equations @5:41, @6:13, and @6:42?
RJ
I had a surprisingly difficult time finding someone bold enough to make the simple, explicit claim that nested radicals correspond to field extensions. You'd think that would be easier to find, but I had a heck of a time finding it. Didn't see it on math stack exchange, and darn sure didn't see it in grad level textbooks because they never explain anything clearly. Didn't find it until seeing it in your video. Thank you very much.
Congrats. This is the most understandable and intuitive presentation of the Galois Correspondence and quintic insolubility. Contents of books are a torture to assimilate. Other lectures (lecture series) can't see the forest for the trees. Thank you
Thanks a ton....scratching my head to understand this concept from last 4 years . Now i can make sense.
Thank you for the video. But, to be honest, the last part of the video was not comprehensible for me. I hope someone would be able to clarify it to me.
Does this have to do with the fact that the alternating groups with n \geq 5 are simple?
Yes, when n\geq 5, you get a composition series S_n>A_n>{e}, and Jordan-Holder says that you’ll never find any other composition series, so S_n is not solvable.
A very nice presentation! Thanks for referring to my book "Galois Theory for Beginners".
Man, your video is fascinating. What an age to live in
I am really amazed by how mathematicians and engineers are incapable of giving conceptual explanations of some fundamental ideas in their areas, but you did it!! you made me understand what solvable by radicals means, in a simple and insightful way; many thanks! the most useful part for me was the beginning, when you explained we had to add radicals step by step; I know the Insolvability of the Quintic isn't in contradiction with the Fundamental Theorem of Algebra, but then what is a root when it is not given in radicals?! this is also a thing which I think is fundamental but I can't find a good insightful nutshell explanation of;
I think he made a video on how to learn pure mathematics on your own , but I can't find his video . Does anyone remember it ?
Great intuitive video, taking my second semester abstract alg class rn and this is spot on
Awesome! (notifications ganggg)
really nice Video!
I quite like the long end card (like 20sec), because i gives time to rethink everything you said, before having to interrupt the autoplay
Marvellous work!
Another knockout! No fat but still clear and satisfying
I love this. The explanation was simple enough to understand as a Physics student.
Great stuff as usual, but I still don't completely get it (logical, though - it's ten minutes, and I have only the most basic rudiments of Abstract Algebra).
Don't worry, it's just not possible to completely "get it" in 10 minutes.
Galois theory is the culminating end result of what turned out to be a 300 year effort by various mathematicians to answer one question: Does there exist a "quintic formula" and above? The quadratic formula was known in ancient times (albeit in word form, as symbolism didn't exist yet!), the cubic formula was found around the 1530s, and the quartic formula shortly after around 1545. But what about polynomials with higher degree?
Around the early 1800's, mathematics has advanced enough where the time was "just right". And in the 1820's, the matter had been settled by two young geniuses, independently of one another. Abel proved in particular the impossibility of solving the quintic in radicals, while Galois took it a step further, settling the matter once and for all by figuring out exactly when polynomials ARE solvable in radicals.
What is truely remarkable is that not only did Galois manage to do that, but he did it when he was 19 or so, and he didn't even have the field concept yet! He was the first to recognize the importance of groups (his terminology, in fact!). He must have possesed immensely incredible insight, having figured out this new angle of attack into an old problem. It's one thing to try and understand something with all the modern cushioning and years of something being established, but a completely different thing to even invent it yourself, and forging the difficult path into that unknown. But even in its modern formulization, Galois theory rests on a bunch of abstract algebra (mostly group and field theory)
And it's STILL quite a difficult thing to learn! A lot of texts don't even show you the historical way things were done in the early days of Galois theory, via symmetric functions and whatnot. That's how Galois himself has done it, and I feel a lot of intuition might be lost by not showing these origins.
@@Cardgames4children I believe the reason, that most people are unable to appreciate the beauty of mathematics, is that we are taught it in a way that makes it impossible for most people to see for ourselves the reasoning involved in producing the mathematical results that we are taught. We have no intuition for why something is true or why a certain technique is able to solve a problem. If the origins of mathematical concepts and how to make an attempt at a proof of mathematical statements were taught alongside the things which are required to solve problems, then most people would have a better appreciation for mathematics.
@@MrAlRats totally agree. While I appreciate the beauty of mathematics itself and the ideas / concepts it contains, I also really like the beauty of how those things fit together or why theorems are true. Seeing the thread of logic tie different ideas together at each step, then clearly reaching the final statement ie the theorem you're proving. It's fun to sit back and just marvel how closely-knit the concepts are.
One of my favorites is showing why a field extension F(a) has a dimension equal to the degree of the minimal polynomial p(x) of a. Let it be n.
F(a) is the set of all polynomials in a with coefficients in F:
F(a) = {A(a) : A(x) € F[x]}
By the division algorithm, we can write
A(x) = q(x)*p(x) + r(x), where deg(r(x)) < n. So, A(a) = q(a)*p(a) + r(a) = r(a), since a is a root of p(x).
Thus any element of F(a) can be written in the form r(a), where r(x) is a polynomial of degree n-1:
f_0 + f_1 * a + ... + f_n-1 * a^(n-1).
Is the set {1, a, a^2, ..., a^(n-1)} independent? If it were NOT, then some of the above f_i would be nonzero, and so a would be the root of a polynomial of degree n-1, this is less that it's minimal polynomial degree and hence not possible. Hence all the f_i are 0, and so {1, a, a^2, ..., a^(n-1)} are independent.
Therefore {1, a, a^2, ..., a^(n-1)} forms a basis of length n, and so F(a) has dimension n.
What I love here is when the any-old-normal polynomial f_0 + f_1 * a + ... + f_n-1 * a^(n-1), viewed as a single object, is now taken, not as a polynomial, but as a linear combination of F-elements with the vectors {1, a, a^2, ..., a^(n-1)}. It was such a slight of hand that I think this proof is amazing, and shows the beauty of mathematics!
I really like the style of your videos, with the handwritten illustrations. What do you use to edit and create your videos? Also, thanks for sharing your mathematical knowledge!
Very nice explanation. Thank you for this excellent work!
I am lost at “equations involving those roots.” Where do these equations come from? Do I just write down any equation to discuss roots’ interchangeability within it?
The most impressive thing is that Galois came up with most of his ideas and wrote everything just a couple of days before he died. At the time he was only 20.
Well, Hardy used to say no good Mathematics got discovered by someone over 50, most likely 30 too. He got a point, somehow
That's the summary of Abel's, Galiois work. Brilliant
This is good enough.... if it's too long it'd be like a whole-semester course 😝
We want a quick overview and the interested reader can dig deeper.
Excellent. Thank you. Could not be explained better.
I'll start with the pros of the video. Everything you explained was done very well through helpful visuals that made the content more digestible. Now for the cons: tldr, you missed a spot. For anyone in my position, with little to no knowledge of group theory nor Galois theory for that matter, some of the things in this video seem rather arbitrary, and therefore take away from the impact of this 'proof'. For example, why should the number of tiles be prime? Why should there be tiles in the table in the first place? What is a group table? (I know this one, but could've still done with an explanation)
All in all, a good video, but missing bits and pieces that would make it a great one.
Edit: Might as well leave a recommendation of sorts. This sort of thing would work much better if done as a series on group theory and galois theory, so that viewers can get to grips with the basics rather than tackling tough problems like this straight away. Then you could leave a video like this to be the grand finale, and viewers would actually understand it if they watched the other videos.
Totally enjoyed this video!
1. 5:10 - Sure, you explained what a algebraic symmetry is. You have equations that the roots satisfy and the solution is symmetric if you can swap those roots while still satisfying that equation, but where do the equations come from? I can come up with a infinite number of equations involving the roots and I'm sure that I can find at least one of them where swapping the roots around in any way would not satisfy the equation. So what defines what equations the roots must satisfy? Same goes for the equations on the extended fields, where do they come from?
2. 5:58 - Correct me if I'm wrong, that group table basically shows the resulting symmetry when 2 symmetries are applied? You never explained how the table is constructed in the first place. I only assumed it was that because I've seen similar tables in the context of symmetries, but who knows, I don't know if I'm right because it was never explained.
3. 7:16 - You can tile that group into sections of equal size, but you never proved that this is a general property of polynomials that are solvable by radicals. Also, you didn't connect the tiling of the group with the removed symmetries in any way, shape or form. It seems that when you extend the group a new restriction is applied that reduces the group to one of its tiles. I could, potentially, tile the group into 8 different tiles if I wanted, and that would break the whole thing since 8 is not prime. The key thing is that you CAN divide into a prime number of groups OR that this division is somehow related to the extensions, but the correlation is never explained.
4. 8:00 - That was just worded in a confusing way. "The number of tiles is a prime number." Wrong! What you mean to say was "The number of DIFFERENT tiles is a prime number." and each unique group is defined as containing a unique group of elementos (the green tile contains 0, 1, 2 and 3, while the blue on contains 4, 5, 6 and 7) with the positioning of elements not mattering, which, if true, is something that was not properly explained. I only understood what you mean on a second/third watch. Also, as others have pointed out, you never explained why it has to be a prime number.
5. 8:14 - The fact that the group is solvable implies that the function is also solvable? Also, we spend the whole video talking about "solvable by radicals", that leads me to think that a function could be solved in other ways. So proving polynomials of degree 5 or more can't be solved by radicals doesn't really prove that there is no solution, unless you also prove that to have a general solution it has to be solved by radicals, which you did not comment on.
6. 8:20 - Now, that whole last section explains nothing and just throws "facts" at the viewer. Like, what is Sn to begin with? By the wording, and according to my previous knowledge, the number of permutations of n things is just n!, so is Sn = n!?
7. 8:30 - Where do those subgroups come from and what are they?
8. 9:36 - Can you? At this point I stopped looking for proves, because I knew I wouldn't receive any.
I can see that you put a lot of effort in this video and I'm glad that there are people making quality content for TH-cam in the field of mathematics, but you made a mistake that I've seen many others do, you believed you could properly explain a complicated subject in 10 minutes and, in the process, a lot of stuff that is needed to fully understand the proof was left behind. Sometimes you can get away with it, but particularly in mathematics people usually leave unsatisfied at the end because a lot of "magic" happened somewhere in the middle. This level of topics requires a longer video or even a whole series of videos with more examples, counter-examples and proofs, otherwise we leave knowing just as much as we did when we started watching. Anyway, consider the criticism as friendly advice for next time and keep making videos, I'm looking forward to them, you've just earned a new sub. 🥰
I hope you do a video on V.I. Arnold's proof of the insolvability of the quintic. It might be easier for viewers to follow.
There seem to be some steps missing. And I don't mean the step of having taken Modern Abstract Algebra.
Why is it S4--2->A4--3->V4--2->C2--2->e and not S4--2->A4--12->e? Or, alternatively, why is it not something like S5--2->A5--5->W4-->3->V4--2->C2---2->e? Why is the 60 of irreducible complexity when the 12 was not?
As A5 is simple whatever W4 denotes it won't be a normal subgroup of A5.
However, solvability requires that every (sub)group in the composition series is a normal subgroup of the the (sub)group immediately preceding it and that the quotients are all abelian.
Hence, S4→A4→e is not enough as A4 is non-abelian and S5→A5→W4→V4→C2→e doesn't work as W4 is not normal in A5.
@@mrtaurho8846
Hmm. I attempted a comment last night and it appears not to have gone through. I somehow didn't see the reply until then.
I am largely self-taught and have not previously encountered the terminology of a "normal subgroup." However, I think I can surmise that it merely means that it is closed under the group operation. I have seen the term "abelian." But "commutative" is more commonly used, the definition of "abelian" is only that it is commutative, and we are hopefully trying to make this more accessible, not less.
The question of why it was not S4 -2->A4-12->e asks why the quartic or biquadratic (I've seen both terms) does not fail the way the quintic does (12 not being prime or even a power of a prime.) And a big part of this gap is that the video does not explain what An is (the set of even permutations of a set with n elements, A2 being the identity permutation.)
In preparing my reply to you (twice) I managed to work out what V4 (as mentioned in the video) is. It does help. And I note that your criterion does not match the author's -- at least in a technical sense. V4 is commutative (your criterion) so you would stop there. But it does not have a prime number of elements. (I need to be careful with my words here. I almost said it doesn't have a prime order. It does, 2.) The author breaks it down to C2, which is done wholly arbitrarily.
When you tell me "whatever W4 denotes it won't be a normal subgroup of A5" you aren't really adding anything. I am not able to deduce the meaning when you call A5 "simple." I don't recognize the terminology and, perhaps more importantly, the context does not give me sufficient information.
@John Undefined You're missing some fundemental terminology commonly used (AFAIK) in any book on Abstract Algebra. This is fine as I only used it for brevity but let me expand.
Normal Subgroup: A normal subgroup is not merely closed under the group operation (this is a subgroup) but satisfies a further technical condition. This condition states the the subgroup is mapped onto itself by conjugation (multiplying from the left by g and at the same time from the right by the inverse of g). Note that not every subgroup is normal in general, although this is the case for abelian groups.
Abelian Groups: Indeed, abelian groups are simply commutative groups and this is actually the more common term in the literature. Commutative is more commonly used to describe a property which binary operations (roughly, something taking two inputs and producing one output) may or may not have. So, a group with a commutative group operation is called abelian by convention.
Quartic≠Biquadratic: In general, an equation of the form ax⁴+bx³+cx²+dx+e=0 is called a quartic equation while a biquadratic equation is of the special form ax⁴+cx²+e=0, i.e. only the squares of squares appear (so bi-quadratics).
V4 in A4: V4 is the Klein Four Group which is a (commutative) subgroup of A4, but not a normal subgroup. The precise definition of these decompositions, however, requires normal subgroups of which there are less. The fact that V4 is abelian is completely irrelevant for whether or not V4 a normal subgroup of A4. Moreover, I did not claimed the criterion being that V4 is abelian but rather that the quotient groups at each step have to be abelian. This is somethinh quite different!
Also, breaking down V4 into C2 is the only possible way of further refining this decomposition to get prime order transitions (V4-e has a transition of order 4, which is not prime, while V4-C2-e has two transitions both of order 2, which is prime). And C2 is also the only non-trivial subgroup of V4.
Simple Groups: A simple group is a group whose only normal subgroups are the trivial subgroup and the whole group considered as subgroup. If the original group is non-abelian and simple it won't be solvable. That's the crux about A5 (and hence S5).
At this point I would advise you to read up the precise definition of solvability. Starting from there you might benefit from learning the necessary preliminaries for understanding the concepts properly. As long as we keep hand-waving some things I don't think it will get clear what I'm saying.
If you mind, I can recommend the book by Judson (Abstract Algebra: Theory and Applications) which is freely avaiable online.
Hope that helps :)
@@mrtaurho8846
"Quartic≠Biquadratic"
I have seen the quartic called the biquadratic and seen the terms used interchangeably. But I not trying to get caught up in the terminology. I am trying to communicate.
Thank you very much, very informative video. Since your handle is Aleph 0, I'd love to hear your take on Cantor's countable/uncountable infinities (I'm not proponent of the idea).
7:56 what exactly do you mean by tiles? that part confused me a lot.
I understand your confusion, but don't know how to clarify. :-) I think he just means the number of individual 'rectangular boxes' marked by colors. So if you can color them with prime number of colors, you're good. If not, either you can after some (complicated) tweaking, or you still need non-prime amount of colors, it's not solvable in radicals.
Maybe.
Brilliant video!!
Good video! I am curious about the proof that for n >= 5 there is no general formula (you kind of skipped over that in the video)
Please provide a full course in calculus cause you really really explain well
Pretty informative and understandable for distilling a whole undergrad course into 10 minutes
Hello! I am a little mentally challenged and it is very hard for me to follow your wonderfui explanations with the background music (actually, it's only one percussive effect in the music that NaNs me :-) ). Would there be any way to maybe get a video without the music? Maybe on a second channel or something like that?
Thank you so much for these productions and the wonderfully analog imagery!
What kind of equations should be preserved exactly when swapping the roots ? I see x_0 + x_1 = 0 and x_1*x_3+x_2*x_4 = 0 but where exactly do those come from ?
Edit : Went back in the video and heard him say "every single polynomial relation involving these numbers (the roots) that has rational coefficients"
Pellucid again. Great job.
Great stuff, but you lost me at 8:30. What is e? How did you derive S(n), A(n), C(n), V(n), etc. and the blue or red numbers between them? Why exactly must they be non-prime? I definitely feel I'm missing something. Thanks.
Thanks for this explanation, I've learned quite a bit!
It's just not clear to me why the additional equations at 5:40 6:10 6:40 must be satisfied.
I understand that they are written in such a way that they would "produce" the adjoined element if rewritten slightly:
(x_1 - x_2 + x_3 - x_4)/4 = sqrt(2)
(x_1 - x_3)/2 = sqrt(3+sqrt(2))
-x_2 + 1 - sqrt(2) = sqrt(3 - sqrt(2))
What am I missing?
Hi.. you post great videos.
I was wondering if u can suggest any book(s) on your previous video regarding Navier-Stokes Equation (how it is used in fluid dynamics)?
Where is your guide to studying Pure Maths video?
You do a great job.
Now is there a way to generalize the notion of root so that a number that is a non-radical root of some polynome could be used to solve others of the same degree? Or at least do such non-radical numbers still have interesting properties?
The links are dead
Great video! Thanks
I love your videos !
Great video. I would like to ask how do we find the equation that the roots must verify
I hope you will make a video explaining the Yang-mills existence and mass gap.
And explore some possible solutions to it.
Anyway, that's a great video
i know n-tic formulas with n>5 don't exist in a generalized sense, but would it still be possible to make it on a localized sense?
as in, a formula that specifically solves for the roots of the expression ax^5+bx^3+cx+d and also one for ax^5+bx^2+c, and maybe one for ax^5+bx^4+cx^3+dx^4+e (where e is a variable, not the constant)?
i don't really know where i could start to prove this problem, and looking it up online didn't really help either, it just gave me an answer on how there isn't a generalized formula for quintics and so on, but does anyone in the comments have an idea?
or is it just not possible because of another aspect of group theory or this exact principle covered in the video?
No. The roots of the polynomial x^n-x-1 are provable not expressible algebraically in terms of its coefficients. As this is of the simplest non-solvable type for quintic equation I think your endeavor is hopeless.
There are, of course, specific polynomials for this is possible (for example, x^5-1) but in general not.
Maybe the Bring radical together with the Tschirnhaus transformations is something of interest to you.
@@mrtaurho8846 ok, interesting
Thanks!
Great video!
the list of "nonprime" number of tiles in the chain of subgroups seems peculiar...
60, 360, 2520, ....
wonder if there is pattern in this
Very interesting video. I've always been curious about field extensions etc (liking abstract algebra), but have never really gotten the occasion to really study it.
One question I have is, you say that there's no "general, single" way to solve such equations by radicals... but is there a "family" of specific solutions by radicals for higher degree polynomials ? I know some specific polynomials do have solutions. az^n = c can be solved using n-th real roots and roots of unity. Some polynomials in higher degrees can be reduced to smaller polynomial using X=x^n, and solved in multiple steps, generally with degree 2, 3 and 4 methods for X, then substituting X for x^n.
Is there a way to "partition" S_n or A_n into subgroups to get direct sequences with only prime numbers, thus defining radical-based solutions for specific classes of higher degree polynomials (ie, building solutions for the whole of deree 5 polynomials through a set of multiple formulae) ? I suppose not, since somebody would probably have found it and categorized cases by now if it were possible; but I don't really understand why though.
I don't have an example on me, but there are degree 5 polynomials which can not be solved by radicals at all. There is a way to prove that no general equation exists AND there is not a specific equation for each polynomial. But, I have not learned it. Taking Galois Theory currently, and am unimpressed with how it is being taught online.
So hyped for this one 💯💯💯