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)
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.
Galois himself died before being able to explain it. his work was rejected during his lifetime, and only accepted about a decade after his death, in exactly the same form that was previously rejected, btw, because somebody bothered to go around proselytizing about the fact that it does work. there's no genuine evidence that anyone in history has ever actually understood Galois theory. now, that's not to dismiss the many people who are skilled in using it. but honestly, we all know it doesn't require deep knowledge about how and why something works simply to use it. I guarantee the other dude that commented here doesn't understand Galois theory. he can maybe use it, but he has absolutely no idea why there is no general quintic formula, only that there isn't and that it's somehow related to the predictions of Galois theory. this is why he's saying you need more time. that more time is needed to break down your resistance and inclination to ask questions so that you'll eventually just accept the doctrine that Galois theory does magically explain something. if I were wrong, then somebody would have composed a genuine explanation of Galois theory long ago, and it would be what we see, instead of this sideways talking nonsense. note that every time we begin finding a Galois Group we must assume that the roots of the polynomial are permutable... despite the fact that we already know this assumption is false in the general case. that's a strange place to begin if anyone understands what's happening, since the only actual correct knowledge we have going in must be rejected immediately. think about it.
@@sumdumbmick You have clearly never studied math past watching RU-vid videos. Galois theory is not some magical theory that nobody understands and just believes Galois on, because it would not be accepted in that case. Not to say that Galois wasn't a genius, but the difficult part is discovering his insights, not understanding it. Galois theory is completely derived from the axioms of ring theory and field theory and studies something that is pretty concrete: automorphisms of a field, leaving a certain subfield fixed. It easily follows from how automorphisms work that for some polynomial, the automorphism can only send roots of that polynomial to other roots or itself. This means that you can derive the group of automorphisms (the Galois group) from the roots of a certain polynomial (the minimal polynomial of the field extension, in fact). This is all not very difficult to prove, and I could prove it on my own. Not because I learned them by heart, but because I understand the theory. You mentioned that there exists no explanation of Galois theory. That is not true at all, look at dummit & Foote's abstract algebra. Chapter 14 is about Galois theory, and it proves every single statement that is made.
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.
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 RU-vid 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.
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.
Thank you. From one member of the rest of us. I do not understand you. You are speaking to people who already understand Galois theory and can follow you. Your accent makes it hard for me to understand your words.
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.
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.
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?
it's worse, actually - there is no solution with any elementary functions (all trigonometry fuinctions included), meromorphic functions, their integrals and derivatives.
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.
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.
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)
The quartic equation is the highest degree equation with a general solution that can be found with only addition, subtraction, multiplication, division, and root extraction.
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!
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
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...
0:58 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.
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
I don't understand. Do you need to cunstruct all possible equation using the roots? And check the permutations of those equations? If yes, why there was no note at the permutation tables indacating the corresponding equation?
And when you were constructing the equations why did you use only addition and multiplication? Why no division or subtractoin or nth root? What is the rule there?
the underlying premise is that the roots of the polynomial can be permuted, but that premise is false in the general case, so... why are we assuming that premise in the first place, and continuing to assume it every single time we apply Galois theory? literally every time you end up at anything less than S(n) being the Galois Group it's just a proof by contradiction. that's a pretty crap method for doing something that you claim to understand. it's one thing to show that Galois theory works, and to teach people how to find the Galois Group of any given polynomial, but I remain skeptical that you, or anyone else, truly understands what's really happening here.
it's also a problem that you're claiming to be explaining this but you never even hint at where your equations are coming from. why are we using (x1 +x3) -(x2 +x4) = 4v2 at 7:24? as far as anyone watching this without already knowing how to do this is concerned, you just pulled that equation out of your ass. which means you're not actually teaching anyone anything.
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.
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
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
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
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?
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.
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!
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).
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!
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! ######### 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
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.
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.
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.
Haven't watched the video yet... but I just wanted to say that putting "unsolvable" in the thumbnail kinda annoyed me because solving these numerically is trivial
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.
