ฝัง
- เผยแพร่เมื่อ 5 มิ.ย. 2024
- "The best way to learn a new topic is to teach it" - Grant Sanderson aka 3blue1brown
Galois theory is a fascinating topic and I hope you learned as much as I did while making this video.
Please leave me any feedback, questions, etc. in the comments.
===Chapters===
0:00 Introduction
0:24 Groups
1:18 Fields
3:03 The Connection
5:30 Solving a polynomial
9:42 Conclusion
10:46 Why is there no quintic formula
11:38 Outro
Ignore the typo at 0:14. It should be "Group Theory"
At around 3:25 The last equation should equal -2, not 2. Still it has the same result even if you swap the roots.
===Further material===
link.springer.com/book/10.100... (Excellent book about Algebra and Galois Theory)
• Why you can't solve qu... (Very good video by @mathemaniac, more technical and in-depth)
===Made with===
Manim (www.manim.community/)
===Music===
Music by Vincent Rubinetti
Download the music on Bandcamp:
vincerubinetti.bandcamp.com/a...
Stream the music on Spotify:
open.spotify.com/playlist/3zN...
===Tags===
#galois #quintic #grouptheory #galoistheory #abelruffini
Hi, I hope you liked my video!
Please leave your feedback, questions, etc. in the comments.
Further material is in the description
Hi
Are quintics solvable if you introduce things like trig functions and exponentials?
I don't think so. WolframAlpha would know such a solution; for me it seems unlikely if you look at what we're trying to solve (unless something like e^⅔pi i of course)
@@mathkiwi Oh, quintics are solvable if you introduce this thing called a Bring radical
en.m.wikipedia.org/wiki/Bring_radical
Seems like it's just defined to solve this equation
It is clear that you understand what you are talking about and that your goal is to produce videos that highlight the essence of a certain mathematical topic. If you intend to address only those who already know it, to show how elegantly these could be conceptually summarized, the style is good and the graphics are clean. However, if you are also addressing "the rest of us", who do not know a certain topic, or never understood it properly, to put them on the right track, I would suggest a slower pace. It can also be helpful to check how TH-cam interprets your words. An example (4:54): "...color group of the polynomial of a q to the treble group the channel fields..." (= "Galois group.... trivial group... chain of fields... "). I would also suggest providing some information about your Channel. If you don't want to put any personal information in it, at least spell out your purposes. Clearly formulating what you want to do would be of benefit both for you and for your target audience. As you can see I'm interested in the kind of stuff you're doing and wish you all the best.
Thank you for your feedback!
I think it is fair for a producer to not be aiming at everybody. I have no idea what's going on, and yet the explanation itself managed to evoke some dim sense of "ah". That's definitely a success. There's something to this form of explanation and abstraction that other attempts at "more clarity" manage to somehow miss.
Like, the only thing I understood here was that you need to do field extensions (and you can't do that for quintics). What are field extensions, how do they happen, why would I want to do one, where did "Galois group" suddenly jump into the picture, and soo on. I basically understood close to nothing. But I come away with an understanding that you can't do the field extensions for a quintic equation. In a sense, that's all I need to know, although I could not pass the knowledge on too anyone else by an explanation.
And if you want to hear things more slowly, you can slow down the video :)
"Slow down" means "take smaller steps", not "do every same step in slow motion".
But anyway... Great video -- even though I couldn't exactly follow it all.
I've been watching several videos related to this topic (3Blue1Brown, Math Visualized, Aleph 0, etc.) and so far this video imo is the one that tries to explain the idea behind this concept in the cleanest way possible, the animations for this video are incredibly good and made me understand some concepts that I didn't get from other channels explaining the same topic
(found this video way more friendly in the given steps to build the group table, also the animations showing what do permutations do to the roots of a polynomial are so good) . Awesome video.
wait so there's no quintic formula because you can't always divide the permutations of a quintic ? Galois theory is the only thing I've seen like 3 videos on and am still lost lol
I would recommend reading a book about abstract algebra then. Frankly, these results are so deep into the study of abstract algebra that it's often the case that the "brief explanations" online are unsatisfying.
I would personally recommend "A Book of Abstract Algebra" by Pinter. It is like $15
If you only watched a few videos on Galois theory, obviously you're lost lol. For Galois theory you probably need a semester of group theory and another semester of ring and field theory to have enough prerequisite knowledge to fully understand it.
This is honestly so well explained. I haven't been able to understand this for a few weeks until I saw your video, and now I'm crystal clear about this elegant proof. Subscribed and thank you!
This is so u underrated, your editing and/or manim skills are definitely at 3b1b's level, really deserved more subs man!
Thanks for this. I have been trying to understand this topic for some time and have watched a number of videos about it. This one however, I find offers the best hope of gaining a grasp of the essential steps in the argument in a clear way without going into a lot of unnecessary detail.
DIDACTICALLY BEST VIDEO OF GALOIS THEORY. Nice your circle graphics !!
Relating this to the fundamental theorem of symmetric polynomials that relates coefficients to the roots will be fantastic. Why should roots obey the permutations in the first place for solvability? Because if they don't we will have the original polynomial altered.
Definitely one of the better videos on unsolvsbility of the quintic
Great video, nice explanation, thank you for this work
This was an awesome video, thank you. I haven’t taken a class in abstract, only seen videos to cure my curiosity, and this was a great learning. Subscribed
This is the video.
Finally I understand....
Every other video I've watched was way too overexplained, this was perfect.
Just a reminder if you are having trouble understanding all this:
Galois died at 20 years age.
And several months of his short life were imprisoned due political reason.
He is a genius.
good 3b1b style with a speedier narrative, i'm a fan
At 3.25, the last equation should equal -2. Excellent video!
You're right and thanks!
3:25
Great clip!!! Clear and short!
Does all of that mean that there is still hope that one day we find general solutions that includes trigonometry for polynomials of higher degrees? Somewhat like the "casus irreducibilis" for 3rd degree polynomials?
Look up ultra radicals.
it's worse, actually - there is no solution with any elementary functions (all trigonometry fuinctions included), meromorphic functions, their integrals and derivatives.
No. Using the proof by Vladimir Arnold it can be shown that not even trigs are enough
I have elementary background of abstract algebra from years ago. You made galois theory a bit easier to digest, I still can’t grasp it but I can at least understand a bit of the premise of why Quintics and above don’t have formulas
3:22 3rd equation should be equal to - 2.
3:34 1st equation should be equal to 2 * sqrt( 2 )
3rd equation should be equal to 2 * sqrt( 2 )
I would like if you recommend any sources for abstract algebra bcuz it was really a great video but was a lot to take if ur not familiar with the concepts which i was
Btw gr8 video but i advice to make ur own English captions so second language speakers of English can understand it more and slow the pace of the video
I once put a general form 10th order polynomial into Cayley in 1983 to see how it would handle it - it solved in under a second - but the formulea was 10 pages of print out! Amazing what a Group Theory program designed for infinite precision maths and to handle groups of up to 10 ^ 50 elements could do even back then!
Beautiful video!!
Thanks!
I want to study Group Theroy.
Why? You know, I have a cousin that has multiple PhDs in ancient languages. He studied languages we don't even know how to pronounce anymore. He makes minimum wage at a museum. That feels like studying group theory to me.
@@mrfarts5176, but why does she/he so poorly with a PhD?
@@erik-ic3tp What are you going to do with Phds like that?
@@mrfarts5176 you can become a professor, that's enough income (atleast in my country) that you can ask for
@@mrfarts5176 that's a weird thing to say
Where does the algebraic equation for each group comes from?
Ok, if they'res no quintic formula. How are we going write the root (or roots if you feel complex) of x⁵ - x - 1 without using decimals? My idea is to use this root function "R( f(x) , n ) where n is what specific constant it's talking about", so √2 will be R( x² - x , 1 ). Not only that but it could do non polynomials like this log2 (3) = R( 2^x - 3 , 1 ).
Actually, you can find a solution to the general quintic using a function called the Bring radical. You can use a quadratic Tschirnhaus transformation (change of variable) to reduce a general quintic to the form y^5+ay^2+by+c=0 (principal quintic form), then a quartic transformation to reduce it to the form v^5+dv+e=0. The solutions to this are (d^1/4)BR((d^-5/4)e) and its conjugates, where BR(a) is the Bring radical.
Love this!
Amazing video 🎉
At 6:00 how and from where we know that x1 x3+x3 x4 =0?
It is some invariant of this particular equation, but how we get it?
At 6:00, 8:56, and 9:21 he gives us equations that he uses to find the subgroup chains of the galois group of the original polynomial. He doesn't explain how he got these equations, but you can verify for yourself that these polynomials are true.
If you notice at 5:46, he gave us the 4 roots of the polynomial, so if you plug these values into the three equations, you'll see that they are true.
In general, if you want to compute the splitting field for some polynomial, you can either: (1) know the roots of the polynomial ahead of time and write down equations based on the roots (which is what he did in the video. Easier still, you can just write down the splitting field based on the roots directly, which he did in the first part at 1:48 - 2:28). Or (2) you can find normal subgroups of the original galois group via some clever trick. If you can't find any nontrivial normal subgroups, then the original polynomial can't be factored by roots. (This is essentially the proof of unsolvability of the quintic, which he briefly touched on but didn't explain in depth at 11:15)
One remark - unsolvable by radicals
Once we allow functions like hypergeometric functions or stuff like this
polynomials will be solvable
nice video for one who are new to galois theory like me!
First of all thank you a lot for your valuable explanation. Could you please suggest a book (in English) that discusses the topic?
awesome !!!
I don't get the 'symmetric table' part. What does it mean to be a symmetric table? And why can't you take the trivial group from S5 in the first step?
"Symmetric table" means the group is Abelian (commutative), i.e. ab=ba for all elements in the group.
That's very insightful, providing a rather precise explanation for why polynomial equations of degree five or higher cannot be solved using radicals.
If I start with a polynomal like x⁷ + x⁶ + ... + 1 how should I do any field extension (for getting this Galois group) without knowing - minimal - 1 root?
Of course I know S(7) but this helps me nothing for the above problem.
Or, is it easier to start with a S(n), and n not prime?
Or, in case x^n - 1 all roots are now well known and what are now these field extensions and for what is this Galois Group now good for?
Thanks to anybody who answers even only one of these questions!
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Or is Galois Theory useless in finding the roots of polynomials?
Disclaimer: I'm just an undergraduate and I don't think I'll specialize in field theory
For the particular polynomial you gave it's still possible to find the roots by radicals:
assuming you meant x^7+x^6+x^5+x^4+x^3+X^2+x+1, you can factor out a x+1 and get (x+1)(x^6+x^4+x^2+1). To factor the second part you can think of it as t^3+t^2+t+1 where you substitute t=x^2. By the formulas for degree three polynomials you can find the roots of that polynomial in t as radicals (say a_1, a_2 and a_3. If I'm not mistaken they should be -1, i and -i). So you factorized the polynomial into (x+1)(x^2+1)(x^2+i)(x^2-1). To finish factoring just take the square root of the remaining bits:
(x+1)(x+i)(x-i)(x+iz)(x-iz)(x+z)(x-z)
where z is the principal eighth root of unity. In retrospect the polynomial there is just x^8-1 divided by x-1, so it's not surprising that what's left are just the other eighth roots of unity (in fact Q(z) is splitting field). I'll keep the long derivation above just as an example of some techniques one might use to still find roots of polynomials of degree higher then 4.
Forgetting the specific structure of that degree 7 polynomial, in general you don't really have a way to tell which extensions to make out the gate I think. The first thing you should check is if the polynomial is irreducible over Q. You can do this in various ways but I don't think there's a general algorithm that find irreducible components. If the polynomial is reducible try to make separate extensions for each factor, this can help in calculating the Galois Group.
Suppose now that your polynomial is irreducible over Q, then if you divide by the leading term you have what is called the "minimal polynomial" of any of its roots. What you can do now is give any root of the polynomial a placeholder name like "a" or "alpha". You can then make Q(a) and it will necessarily be a degree 7 extension (in general the degree of the polynomial in question). Of course this doesn't tell you anything about who a is but it can sometimes help in finding other information like the order of the Galois group. It can also help in further factoring the polynomial you started with, perhaps leading to finding the remaining roots as an algebraic expression containing a.
Knowing the Galois group is far from useless. I don't know much Galois theory but for example you can use it to find conjugate roots really easily. For the polynomial you gave the Galois group is G(Q(z)/Q) = where g_1 sends z to -iz and g2 sends z to iz, all while keeping Q fixed.
Knowing that group and looking at the irreducible factors of the original polynomial (x+1)(x^2+1)(x^4+1) you can find all roots by making this group act on a single root of each irreducible component:
* -1 stays fixed because it's in Q
* g_1(i)=g_1(z^2)=(-iz)^2=-i, g_2(i)=g_2(z^2)=(iz)^2=-i
* g_1(z)=-iz, g_2(z)=iz, g_1(g_2(z))=g_1(iz)=g_1(z^3)=(-iz)^3=-z
(I omitted the other combinations because you'd just get back roots you already found). Notice also that the Galois group never mixes up roots of different irreducible factors. This can help in finding the specific irreducible factor containing a given root: just look at all images under the Galois group.
For the case of x^n-1, where, as you said, we know the roots, Galois theory isn't really used in finding the roots but it can be used to still prove some results about those types of polynomials. I'm not the most qualified person to talk about this but for example you could look at when other splitting fields contain roots of unity or you could try to decompose harder splitting field calculations in terms of the simpler and well understood ones (like x^n-1). For example, G(Q(z_n)/Q) is isomorphic to Z/nZ* (a fact I used before for n=8), so you can answer some questions in field theory by studying the subgroups of Z/nZ* (again, I'm not qualified to go into much more detail then this).
@@francescosorce5189 incredible! The G. group of my polynomial seems to be V4, generated by your g_1 and g_2.
But how did you find these two subgroups?
I don't understand your * notation, f.i. *-1 and also the following *'s.
#####
So I could start with a sequenz of irreducible factors and the unique Galois group foļlows, subgroup of the S(n), n is rank of polynomial.
But the procedure AND purpose is still a mystery.
#######
Thoughts and Prayers from
Dr. Anton Schober,
Mathematical Physicist and Alumnus of the Technical U/Berlin
3:34 the top and bottom equations should say 2*sqrt(2) on the right side, no?
exactly what I was thinking. First major confusion I got from this video
At 3:25, isn't there a mistake at last line?
There is mistake
There is actually Horner found it and we can use it for quintic sixtic....
There are now quintic polynomials with only A5 as admissable (Galois?) subgroup, perhaps x⁵ + x + 1?
Of course, such a polynomial has 5 root so it splits in a product like
(x - a)(x - b) ... (x - e).
Those a, b, ... are some complex numbers.
Question: does this Galöis Theory claim that the a, b, c, ...cannot be expressed as radicals of some integers?
Obiously, a quintic polynomial has a real root, say this a above.
What is the nature of this a between the irrational, transcendental and so on numbers of the real field?
No, none of these can be expressed as radicals of some integers, that's the point. If one could be expressed as such, the remaining polynomial would be of degree 4 and therefore solvable.
a is algebraic but irrational
really cool
There are constraint-specific quintic formulae if just 1 constraint is added to the co-efficients - e.g. one root is known, or all roots are rational, or ... and so on.
What does not exist is ONLY an entirely unconstrained (general purpose) quintic formula.
Didn't get it. Why permutations have anything to do with the solvability of equations?
I like math, but i just learn whatever the hell i come upon, since I’ve already learned a sufficient amount for the profession I want to do after college. But I want to continue being a student of life, which is why I learn more. I have no experience with any of the theoretical processing you did to prove, by means of these field extensions for each of the roots, that the quintic formula is impossible. So please bear with each of my questions, if you would be so kind. And if it appears to you that there is a certain foundation in any given area of your reasoning that I should learn first, please let me know. I’ll number my sections to make em easier to respond to if you get the chance 😁
[1] Firstly, from the video it seems like fields are just a way to group objects. Like natural numbers or rational numbers or irrational or complex, etc. Is this expansive enough of a definition? It appears that we “extend the field” to accommodate for our solution set for any particular polynomial. Why do we do this?
[2] I see how the addition of all x’s(the roots), or the multiplication of all them are the same no matter how they are interchanged, but after this, i see that how you form another equation before extending the field to include sqrt2: x1x3 + x2x4 = 0. Why do we choose this equation?
[3] Then, our next equation, when extending the field to sqrt 2: x1+x2 -(x3+x4). I’m not sure how we decide to use this equation either.
[4] I have the same question on how we deicide to use x3-x1 for extending the field to include sqrt(3+sqrt(2)), but I think seeing how the other two work should be enough to understand this.
[5] I did note that you said these were only some of the *possible* equations we could choose from(of which I’m not aware how to find the larger overall set of these either), so if that larger group of equations is important, please let me know how to construct them.
[6] Besides that, a big gap in my understanding is how these equations that we choose based on the current field we have included connects to the solvability of ax^n + bx^(n-1) … + k = 0? (This might be answered by addressing one of the next questions, I have no way to tell 😅)
[7] After the equation setup and finding valid interchangeabilities, you create the grid of permutations, first giving the sqrt 2 example. You say that the number of tiles(excluding repeats) is 2, since that is the degree of the root we are extending the field to. Why is this the case?
Additionally, in this sqrt2 extension, we notice how there are the two tiles which seem to be the “base” tiles if you might call it that, since we notice that the other two are an operation from our possible permutations applied to all its elements.
These “base” tiles are the only ones that contain the sigma(0) “do nothing” permutation. At the end there with the S4 case it seems like you just keep on choosing “base” tiles like this and zooming in on that until you get a 3x3 base tile.
[8] What is the purpose of these condensations and why do they operate like that, where we keep on choosing one of the ones with sigma(0)’s inside it?
[9] Why is symmetry necessary to guarantee solvability-would it interfere with the ability to split it up into the factors that constitute the splitting field?
3:22 3rd equation should be equal to - 2.
3:34 1st equation should be equal to 2 * sqrt( 2 )
3rd equation should be equal to 2 * sqrt( 2 )
That is what I thought too.
It is confusing.
5:53 How did you come about with these two equations ?
6:00 How did you come about with this equation ?
But it’s possible to solve x^5-x-1=0 using calculus. By taking derivative we could show the polynomial could have up to 3 real roots, then pick some values nearby extremes and use Newton’s method
So, extending the field with the differential operator we will be able to solve any polynomial. We could solve 5 grade polynomial because we could solve 4 grade in radicals. But as we could now solve 5 we could solve 6 because it’s derivative is 5 grade polynomial and so on. QED.
So Abel Ruffini theorem only applicable with arithmetic operators, powers and roots. Adding more operators makes possible to specify exact formula or algorithm.
Where do x1+x2+x3+x4 = 4 and the ones for -2 and 0 come from. It's clear to see the group structure, just not where they come from.
Can you give me a timestamp?
5:53 thanks for the fast response.
A few seconds earlier I show the 4 solutions of the equation. They are x1,x2,x3,x4 and they satisfy these equations
I get they satisfy those equations, but where do the x1x3-x2x4, x1x2x3x4 and x1+x2+x3+x4 expressions come from?
You can construct these from the group, and you can calculate the group, but that was out of the scope of this video. But it's easier to understand the other way around
For anybody wondering yes this music is from 3Blue1Brown😊
Subgroups (discrete, quantum) are dual to subfields (continuous, classical) -- the Galois correspondence.
The size of a Galois group measures the amount of symmetry of the roots of a polynomial -- symmetry breaking!
Randomness (entropy, lack of symmetry) is dual to order (syntropy, symmetry) -- Galois groups.
Symmetric polynomials have large Galois groups.
Symmetry wave functions (Bosons, waves) are dual to anti-symmetric wave functions (Fermions, particles) -- quantum duality.
"Always two there are" -- Yoda.
Symmetry is dual to conservation (invariance) -- the duality of Noether's theorem.
Patterns, symmetry = predictability, certainty or syntropy!
There is no quadratic formula...
with only +, −, ×, ÷
Would it be fair to claim that not all quintic polynomials got an axis of symmetry.
Correct. The guaranteed symmetry stops at cubics. All cubics have half-turn symmetry about the inflection, but quartics and beyond do not have a guaranteed axis of symmetry or point of rotational symmetry. Special cases of quartics, called biquadratics, have an axis of symmetry, but the general one doesnt.
This video could be called *To get a taste of Galois Theory in 12 minutes*.
Nice, but the tempo could be somewhat slower.
this video is a big ahhhh limbo reference
Theroy? Or, theory?
Interesting video. No clue what it is about.
It is the proof that the buck stops with quartic functions, when it comes to a generalized formula using arithmetic and roots, or even arithmetic, roots, and trigonometry. For quintics and higher, there is no general formula. Quartic means x^4 is the highest degree term, and qunitic means x^5 is the highest.
@@carultch Thanks. Yeah, that I got (and knew). But when he starts proving this I am lost right away.
👍
See buya method,hé resolve algebricly all quintic equation
0:58 this is wrong, since you show the elements arranged in a circle, the actual number of permutations is (n-1)! You can just say they are not in a circle here to avoid confusion.
quite complex wout background
I can do it for ya
5:07 pingnesnl
Thank-you for making this video. I recently completed an introductory course in Group Theory, and although I thoroughly understand the concept of a normal subgroup and a quotient group, your explanation of a "splitting field" still seems fuzzy in my mind. Also, no offense intended, but I find your accent difficult to understand for certain words; for example, I can't tell whether you're saying "can" or "can't" at [2:10]. I turned on the subtitles, but that didn't help much. Also, Chapter 2, on fields, doesn't actually *define* what a field is, or how it is related to a group. I feel like this video is very close to being the one that best explains why there is no quintic formula, but the pacing feels a bit rushed.
Correction for content around 0:18. Group Theroy should be Group Theory.
3:21 "All equations with just rational coefficients hold true even if you swap all instances of sqrt(2) and -sqrt(2)" then you immediately show a counter example in the next slide. x^3 has rational coefficients but does not hold...
That equation doesn't have rational coefficients, look at the constant term. Though there is still a mistake, since it should be -2sqrt(2), not -sqrt(2).
I can solve it with series.
Set theory is so cool.
completely difficult
well, list of errors does not stop at 03:25, obviously, what sqrt(2)^3=sqrt(2) or sqrt(2)*(-(sqrt(2)^2))=sqrt(2) at 03:34 is supposed to mean and why cubic power appeared in the context of what till that moment was a discussion of quadratic equation x^2 -2 = 0 ? Maybe you should polish your videos to a reasonable level before claiming to explain the reasons behind Galois theory...
(0:15) _Theroy_
what are you talking about?
I think the guy is either Asian or German. I can not even understand him.
Love the whisper talking.
Ah yes, Group Theroy
0:18 Group Theroy!
Your title card is misleading. There is a general formula if you include transendental functions
Either you speak too fast or I keep missing something. Either way I leave sadly, as unenlightned as I came.
If you asked me what value white people stood for, I’d probably say individuality
Fajny kontent. Ale gościu niewyraźnie mówi. Dykcja fatalna. Nagrać proszę jeszcze raz
group theroy
"Group Theroy" 💀
Way to slow 🤣
Please do improve your english pronunciation!
May be you are not right....
Honestly your explanations were too fast and im not sure ive followed the reason why all this group theory thing is mapped to finding expressions for n-degree polynomials, 5/10 at best, sorry.