I used to classify myself as an "analysis person" and later on an "algebra person". But eventually I learned that the most beautiful math happens when algebra and analysis mix together. Some subjects that show off their interplay include algebraic number theory, Lie groups, Hodge theory, elliptic curves, modular forms, and much much more! The proof of Fermat's last theorem uses LOTS of analysis and algebra for example, and all the subjects I listed above.
@@theflaggeddragon9472 Llevo años aprendiendo Math y cuando uno se atasca en algún momento, siempre hay personas maestras que saben enganchar a la Maravillosa Matemáticas
@@renatohugoviloriagonzalez4881 Que bien que continuas aprendiendo! Y tienes razon que somos muy afortunados teniendo estos videos de espertos gratis en TH-cam, para aprender y para desatascarnos. Aproposito, has visto los videos de Profesor Richard Borcherds? Disculpe mi Espanol, no he hablado en much rato.
This video is a thing of genuine beauty. You have a rare talent for illuminating these deeply technical subjects in a fascinating and accessible way. Many thanks.
Took group theory and ring and field theory but didn't get all the way to Galois theory by the end. What a clear and concise way to encapsulate the fundamental concept. Thanks for this.
Me too and I found abstract algebra the most difficult of all maths subjects I took. Watching this I felt like I was getting an insight and then it disappeared again, my brain obviously cant work in these terms.
Took Abstract Algebra and did much of what you did, with groups, rings, fields and the various domains (PID, ID) and ended after only after hitting a couple of the special topics (we hit finite simple groups, but missed intro to Galois theory)
Thank you for putting so much effort into making this. This is my first time hearing about Galois Theory, and this video was amazingly clear and a treat to watch. It's sad that so few people watch these compared to other channels of equal quality.
@@NonTwinBrothers I agree. It's not one of these channels that live from immediate hype. I wouldn't wonder if this video is still getting watched in 20 years, when 99.99999% of youtube content is long lost to irrelevance :-)
@@harriehausenman8623 Exactly. No one knows/cares who the Kim Kardashian equivalent of the 18th century was, but we all know who Euler or Bernoulli was.
When you said that the lines wouldn't go where I expected, I almost paused the video, because I was pretty sure I did see where they would go -- and I was right! My intuition was based on the understanding of multiplication by any complex number of magnitude 1 as rotation -- which of course wraps around after each full turn. So, ζ⟶ζ² applied twice is just the rotation by ζ² twice -- ζ² * ζ² = ζ⁴. Well, that's reasonably obvious, but the next step falls out of the wrapping nature of the rotation. ζ⁴ twice is just ζ⁸, but since ζ is the 5th root of unity, that means that ζ⁸ = ζ³ -- and so on. With this view, all of the graphs are immediately evident from the starting point of the given mapping of ζ.
This is the most perfect introduction to Galois Theory that I have seen over several decades. It gives us not just the bare bones of the theory, but also their subtlety, their power, and beauty, and every idea copiously illustrated by clear diagrams and algebraic formulae. However, there must be something wrong with my pc, or my old ears, as I can hardly hear the voice over the music. I wish I could turn the music down, down to zero, and then I would enjoy the video for its full worth!
@@johnnypiquel2295 interesting. I've heard that it's helpful if you want to program new languages and such but I'm not sure to what extent it might be applicable in ones software related job (although I suppose that depends on the type of role). Either way I'm planning on taking it in the future just out of curiosity. Extra bonus if I can apply it to one of my projects :D
@@FsimulatorX it’s probably not directly applicable to most software roles. Indirectly, though, there’s the fact that there’s a commutative ring over bit vectors of length n, with bit-wise XOR as the additive operation and bit-wise AND as the multiplicative operation, although that probably doesn’t affect most programmers. There is an algebraic structure that even the most junior of programmers has an intuitive understanding of, though, and that’s monoids. Integer addition is a trivial example, but the more instructive case is strings, with concatenation as the binary operation. It’s plainly associative, and the identity element is just the empty string. Any programming language with generics and interfaces (or traits, or protocols, whatever they end up calling it) is capable of representing monoids at the type level, although it’s really only in richly, statically-typed functional languages like Haskell where you see algebraic structures like monoids being actively modeled as interfaces that any function can be defined in terms of.
It's a testament to the complexity of groups and Galois Theory that simplified explanations still manage to fly over your head, but equally it is a testament to the beauty of these concepts that every time you want to go through it once again simply to understand more. This was a fantastic video - probably the most beginner friendly of all the videos I saw in this area!
I find it difficult to express just how GOOD this video was at explaining the general idea behind Galois theory. Genuinely, thank you. Thank you so much, you've given me another way to look at fields. Another tool that I didn't know even existed.
The connection to unsolvability of the general quintic: Suppose x is a root of an irreducible quintic polynomial, and x is expressible by radicals. Then we can "build up" to a field containing x with a sequence of fields like Q < Q(a) < Q(a,b) < Q(a,b,c), where each step we adjoin an nth root of some element of the previous field.* Each step's Galois group will be a cyclic group. Using the Galois correspondence, this means the Galois group of the last field will have the property that it has a sequence of (normal) subgroups where the quotient at each step is a cyclic group. This is what we call a "solvable group". However, the Galois group of a general quintic polynomial is the symmetric group S5, which does not have this property. When you try to form a sequence of subgroups, you run into the alternating group A5 which doesn't have any nontrivial normal subgroups. Hence we have a contradiction, so the original assumption that x was expressible by radicals is false. *e.g., say x = sqrt(2)+sqrt(3+sqrt(5)), then we'd do Q < Q(sqrt(2)) < Q(sqrt(2),sqrt(5)) < Q(sqrt(2),sqrt(5),sqrt(3+sqrt(5))) so x is contained in the last field in the sequence. You can do this for any radical expression for x.
A lot of this is not quite familiar enough for me to join the dots in your concise explanation, but I'm glad you added this, since it was presented as one of the motivating ideas of the video.
@@leif1075 if you have taken a course in group theory it is much easier to understand. I have no doubt the person who wrote this comment understands it
Very lovely addition by this person to the already amazing video. I had been meaning to study field theory, group theory etc and this is one of the first few videos that came up as a result. Even if we don’t fully understand the whole thing I think it’s fine. This universe has a whole plethora of things that are so beautiful but yet we understand nothing about them. But yet, even making the attempt and going through the motions is so rewarding just in and of itself! May be after repeated attempts - we will be able to develop the intuition for these higher order abstract ideas. I don’t think anyone can understand it fully on the first pass!
The thing that popped out at me when I finally understood the usefulness of this: It was not at all clear that Q(zeta) should necessarily contain Q(sqrt(5)). I can see why such a resemblance might exist given that zeta is the 5th root of unity, but this was not at all obvious. But, we'd already seen the subgroup of Q(zeta), it pops out very clearly in the table! and this alone is enough to prove the field contains some subfield.
I'm not convinced of this tbh. I don't see the connection between 1^(1/5) and sqrt(5). I don't think a field extension Q[zeta] contains sqrt(5), I don't see how you get there with algebraic operations
@@bobtheblob728 Indeed this is tricky to see and I’ve spent about the past hour trying to convince myself of the fact! For brevity, let’s let z denote the fifth root of unity mentioned in the video, namely the polar point (1,2pi/5). It turns out that the rectangular form of this point involves sqrt(5). In fact, you can check that 1+2(z+z^4)=sqrt(5). So, any a+bsqrt(5) in Q(sqrt(5)) can be written as a+b(1+2(z+z^4)) in Q(z), which is why Q(sqrt(5)) is indeed embedded in Q(z)! Hope this helps! If you’re curious, there is slightly more to be said! It is also true that sqrt(5)=-2(z^2+z^3)-1, which gives us another way to see the embedding. Moreover, the reason that Q(sqrt(5)) corresponds to the subgroup that it does is because the subgroup contains the permutation that swaps z^2 for z^3 and z for z^4, which just changes the order of addition in the two equivalent expressions for sqrt(5) given above. But, as addition commutes, these swaps preserve the expression being equivalent to sqrt(5). So, we see that that subgroup contains exactly those permutations which fix sqrt(5), which again explains the correspondence to Q(sqrt(5))!
@@bobtheblob728 You can easily check it in Wolfram alpha. If you do the fifth root of unity (e^(2πi/5)) the number you get is -1/4 + sqrt(5)/4 + i*(sqrt(2*(5 + sqrt(5))/4) If you want to add a root to a field you have to do a field extension. First you extend Q to Q(sqrt(5)) and then you extend again to get Q(sqrt(5), sqrt(2*(5 + sqrt(5)))) Luckily, the fifth roots of unity arrive if you extend Q by radicals (they are the solution of a quartic polynomial).
This is actually a really interesting comment. It's not at all obvious that Q(zeta5) is the only subfield in there- the only way (that I know of) to show that is via Galois theory!
Wonderful, clear videos! Great! So appreciated! At 19:47 there appears the incorrect equation "zeta + zeta^2 - zeta^3 -zeta^4 = SqRt[5]", which should read "zeta + zeta^4 - zeta^2 -zeta^3 = SqRt[5]". What follows in the video becomes correct after this revision. Small details.
Less than 5 minutes in watching and I have made more Google searches than required by an assignment. I love Maths and Engineering. Keep up the good work.
Never learned galois theory in school, just some basic group theory and field theory. I always imagined it was very daunting, but this 25 minute video was very easy to follow and gives me a sense of why people even care about this field. Thank you
Beautifully crafted content. How can one not love math or at least sense its underlying beauty? This video of yours really showcases how wonderful math can be. Thank you!
Very nice video, thank you! The only thing that confuses me a little, is when you say @15:20 that your mind can not make sense of what you're seeing. It is confusing me, because to me it is not just very logical and feels perfectly normal, but I've paused the video and could guess the rest of the permutations (after the first 4). Now I did a lot of group theory before, symmetry of groups, chemistry, knots, Rubik's cube, permutation matrices ... but it really is very natural to me, especially visually. Anyway, nice video and keep it going ;)
Yep, I think watching Mathologer helped with that. For example the 4th one along where z3 points to z5 and back. The z1 points to z7 so this is the 7th power permutation. z3 to 7th power is z21 and 21 is 5 mod 16 so it goes to 5. z5 to 7th power is z35 which is 3 mod 16 so that goes to 3 again. All the others work like that.
If there had been TH-cam in my teens, I would have studied math at college. Thanks for posting! I find this enormously interesting and satisfying to watch in my middle age.
It's always a treat to see a video from this channel. No other channel gets me as invested in modern mathematics like yours. I'm in my undergrad for physics, but I'll probably take my school's graduate algebra sequence starting next fall because of this. Keep it up.
this channel is extremely underrated, some of the best math content on youtube. no other vid has ever gave me as good of an intuition for this topic, and i've seen a lot of them
Great video. I might have to watch it a couple more times to grasp everything in it, but still, it's the best explanation of Galois Theory that I've found anywhere.
I'm going to watch this over and over again until I finally have an intuitive understanding of this theory. I did this in a Uni course but never got a good grasp on how we really arrive at the final result.
Trust me (as someone who doesn't live up to this advice nearly as much as he should): the way to really understand it isn't (just) to repeatedly watch things, but to find or (once you've begun to really grasp the idea through use) pose problems to play with; if you don't *use* the syntactic tools to navigate a constraint-wise consequential (i.e. well-defined) context in such a way that failure to *understand* the concept will more or less reliably ensure failure to resolve the problem - which sounds very negative, but what's important is the contrapositive supposition (which likewise seems to be the case in practice, at least up to fairly nuanced arguments about what really constitutes "understanding") that if one successfully resolves a nontrivial quantity and/or variety of examples, then one *must* possess some minimal degree of genuine understanding of the salient concepts (i.e. the ones indispensable for the problem's rigorous and persuasive resolution). In my experience, this is the hardest part about self-studying higher mathematics: not so much access to problems, but access to *feedback* with respect to solutions the consistency/coherence/general quality of which is non-trivial to determine lol but it's certainly easier now than ever before, at least. Hope none of what I said came off as condescending or pretentious or anything like that, I just feel for anyone who also yearns to understand these things and want to help them in any way I can frankly, so...godspeed my guy 🤟🏻 haha
@@BachelorChowFlavour hey man I wasn't trying to insinuate any such thing, like I said sorry if it came off that way I only say as much because it's a mistake I personally have made which has affected me
@@SliversRebuilt You hit the nail on the head as to a common stumbling block for students, ie, lack of "access to feedback with respect to solutions the consistency/coherence/general quality of which is non-trivial to determine". Your comments indicate you truly understand what it takes to learn a difficult subject effectively. Kudos!
Over at about 12:30, he skips a couple steps which confused me a bit. I'll use s for sigma and z for zeta here. According to his map, we assumed that s(z) = z^2 (since that's where the arrow points to). Also keep in mind that z^5 = 1. s(z^2) = s(z)^2 = (z^2)^2 = z^4 s(z^4) = s(z)^4 = (z^2)^4 - z^8 = z^5 * z^3 = 1 * z^3 = z^3 Those were his calculations.
@@akritworanithiphongBecause z is initially the 5th root of unity, and by definition the n-th root of unity raised to some integer the power n yields 1. See wiki for more.
Wow - I'd made several (admittedly rather half-hearted) attempts to figure out what Galois Theory was about over the years. Made no progress whatsoever !!! Anyway, this video actually made it all start to make sense. Truly remarkable how well you explained it all. Many thanks !!!
This is beautiful! I love how you started with the concept of the root-two conjugates. Such an elegant introduction to the deeper math. Fantastic presentation as always!
Wow. I'm at a loss for words. That simple question of which permutations preserve algebraic relations is such an interesting question and made everything click for me. Galois really was way ahead of his time. Thank you for an amazing video
This is an excellent video, with wonderful graphics that really make the ideas behind Galois theory come to life. There's only one thing that threw me. At around 7:42, you state that Aut F is "a little too big", so we restrict our attention to Aut F/Q. I spent a good hour trying to figure out what could possibly be in Aut F but not in Aut F/Q. At the end of it, I proved to myself that any field automorphism of any field containing Q must fix Q, so it ended up being an hour well spent. It doesn't detract from the main point you're making there, which is essentially, "this is the notation we use", and when we have Q
Actually this is not the error. The error occurs a few seconds later at 11:46. It is redundant to include this term, so it is correct at 11:43, incorrect (or at least redundant) at 11:46
Later, he doesn't say that that expansion is unique, does he? He just applies sigma, correctly. So I would not call it an "error" in either case. But good to note that, I wondered the same.
Could anyone please explain why there are only 8 roots of unity in the example with zeta = e^i2pi/16 ? I dont exactly understand what makes the other roots redundant :)
this video is genuinely amazing, please more content like this! This was just the right video length and buildup for the topic at hand, i could follow every step and im about to look up some more Galois Theory, bc im genuinely intrigued now!
I would like to take a moment to thank you, and all the incredible math explainers on youtube for making such clear, and well made content. For some reason, I decided I wanted to learn Galois Theory, and I have been doing nothing but trying to understand it for an entire week now. I have come further than any teacher could have brought me thanks to content like yours. Even though I am only a freshman, I am seriously considering studying this theory even further, in order to see all of it's power. I was not only intrdocued to group theory, but to all of algebra thanks to content like yours, mate, you're amazing, and it is thanks to you that algebra is beautiful. Thank you mate
Aleph, very much appreciated this very well done video. The starting point, that sqrt(2) and -sqrt(2) are interchangeable in any algebraic equation, not just polynomials, is great. could however use some clarifications, like how do you go about proving sqrt(2) and -sqrt(2) are always interchangeable and ditto for certain permutations of higher roots, and how are Zeta and sqrt(5) related and therefore the fields are related, and why only odd powers in the 16-th root of unity example, and the ending where "what is the square root of two" is answered by saying it depends on the field you're in needs an explanation, lastly I wouldn't go so far as to say this changes how I view fields unless I was convinced that all fields are extensions to the rationals (and I'd like to know in what sense are the reals an extension of the rationals). But I really like your style and very much hope you continue, I'd even go so far as to say I wish there were professors that were as good at explaining things as you are when I was studying math.
About the "how do the reals contain the rationals" question, one precise way to answer is this: there is a subfield of R that is canonically isomorphic (in the sense of field isomorphisms) to Q. The "canonical" part is just human convention, so really the only statement is "Q is isomorphic to a subfield of R" and that's it.
Let me take a shot answering the first question: why are conjugates interchangeable for certain permutations? I'll use the root 2 example. Let s be an automorphism of Q(root 2) fixing root 2. Let q be any number in Q. Then s(q) = q. Now express q as some algebraic combination of elements in in Q(root 2), say p1, p2,...p(n). Because s is an automorphism passing that algebraic combination through s will yield the same combination of the elements s(p1), s(p2),... s(pn) (because of the properties s(p1 + p2) = s(p1) + s(p2) & s(p1*p2) = s(p1) * s(p2)). because s(q) = q, the combination of elements after passing through s still equals q. Now for Q(root 2) the only possible permutations gave s(root 2) = root 2 or s(root 2) = - s(root 2), so applying those permutations won't change the value of expressions, which lie in Q.
I totally agree that this video is pretty good but, as has been pointed out, it is spoilt by the two issues raised by Peter relating to sqrt(5) and why just the odd powers of zeta are considered. Such a shame that what is an excellent step-by-step presentation should omit two crucial steps.
God I wish I could understand more of this, hope to come back one day and see the beauty of this explanation in the same way I could finally understand the beauty of the Stokes' theorem on manifolds. Keep the excellent job, you are inspiring!
If you enjoy this type of stuff, pick up "Infinity and the Mind" book by Rudy Rucker. One of my favorite books on the math subject. I can't recommend it enough!
I love algebraic number theory but I am still working on my intuition for Galois theory. This video is a very powerful intuition pump. Working through the examples is very enlightening. The arithmetic of roots of unity has a mesmerising beauty and things like Dedekind sums are among some of the deepest objects in mathematics. I am *so happy* to have discovered this video. I subscribed.
Nice work! One suggestion for improvement: when drawing roots of unity you state that they are equally spaced, but the drawings are sometimes very off. (2pi/5 sometimes looks like pi/2 and sometimes like pi/4). It could also help if you draw zeta^0=1.
I really like the music in this video. It's gentle and pensive but continuously progresses as if to say, "Consider the following: ..." Which is clearly just perfect; much like 3B1B's "Pause and ponder" music. I'm still quite a long way from grasping most of the deeper intuitions here. Still, I often find as I learn about a topic in math that my brain actually soaked up little bits and pieces of things like this as if by osmosis and then later on (sometimes much later) I'll suddenly realize I have enough pieces and enough context to understand the significance of what was being said. So thanks for making these videos and putting so much detail, care, and attention into them even though the number of people able to full understand everything in them on the first watch may be relatively small.
Amazing! Your best video so far! Just one note. I believe that Q has no non-trivial automorphisms. So every field extension of Q automatically fixes Q. That is, you could have saved on that technicality in the video I believe.
I had the same thought. Sigma(3/5) = sigma((1+1+1)/(1+1+1+1+1)) = (1+1+1)/(1+1+1+1+1) = 3/5. But also it's hard to believe someone who was freshly studying this stuff to make a video would put that in there as a mistake. So I'm scratching my head...
Wow...! I must have seen this video no less than 10 times through the years, the first while I didn't even know what a bijection was. Slowly but surely I learned everything to understand this. Thank you for being in my mathematical journey!
Nice video that reveals my own limitations. Galois theory is something almost understandable for me without ever really understanding. I wouldnt be able to go from this into why algebraic solutions of 5th order equations don't exist 😀 Hopefully some day I'll get there.
Say you have a general solution to a quintic, then you can express it in terms of the basic operations and taking a root. Everytime you take a root, you move to a bigger field where your new element lives in. The galois group of this new field over the field you just left is always very nice. Once you've arrived at your solution, the galois groups should then contain a chain of subgroups that are all nice, all the way to the trivial group. The thing is, you can just check this fact for any field you want, and if there isn't such a chain of nice groups, you know you can't construct that field by taking roots aka solving a general equation. So, we look at the splitting field for the general quintic, and we see that the galois group over Q must be isomorphic to S5, which _does not_ have a chain of nice subgroups. So we can't reach it by taking roots, and so there is no formula. Bit rushed, but QED.
@@ruinenlust_ "Galois group over Q must be isomorphic to S5, but S5 doesn't have a corresponding substructure" this puts into one concise sentence what I was so struggling to put into words. It's one thing to broadly understand, but better once you can articulate it clearly. Cheers.
19:31 "Phrasing this in a more down to Earth language: we first looked at permutations that preserved the algebraic relations over *Q.* Now we'll look at permutations that preserve algebraic relations over *Q* adjoin √5." That is exactly what an alien pretending to be a human would say.
Hey there, just finished watching the video and I had to congratulate you for the amazing work ! The way you explain and motivate all of the concepts as well as the editing make the video a joy to watch and kept me invested in wanting to fully understand the fundamental theorem. Some feedback : as someone who is absolutely not familiar with Galois theory and anything close, there were 2 or 3 times in the video where I was slightly confused, namely I don't understand/know how to compute the Q-conjugates of a given number and it is not clear to me why an automorphism on Q(zeta) has to send zeta to one of its conjugates (I guess these are exercises left to the watcher ?) Despite these minor confusions which shall be fixed upon rewatching the video or some reading/thinking on my side, I have to say that this video was pure bliss to watch, I learned a lot on a topic that is new to me, and I am very much looking forward to the next ones ! :)
I don't know if you've figured out your confusions on your own, but they gave me some thoughts that I wanted to share. So, first, there isn't any one formula you can apply to get the conjugates of a number, say, x, but the conjugates are the other solutions of the minimum-degree polynomial with rational coefficients defining x. So, for example, (sqrt(2)+sqrt(3))^2 = 5+2sqrt(6), so ((x^2-5)/2)^2-6=0, or x^4-10x^2+1=0, is a polynomial with a root of sqrt(2)+sqrt(3), and as the other roots are all +/-sqrt(2)+/-sqrt(3), which are all "related" to sqrt(2)+sqrt(3), the polynomial is probably minimum degree. You need the polynomial to have minimum degree to rule out polynomials like (x^2-2)(x-1) or (x^2-2)(x^2-3) making sqrt(2) and 1 or sqrt(2) and sqrt(3) conjugates under that definition, when they shouldn't be. With that characterization of conjugate numbers, it isn't too hard to show that automorphisms that fix rationals send numbers, again, say x, to their conjugates, as if p is a defining polynomial of x, having roots that are only conjugates of x, then if f is an automorphism fixing rationals f(p(x))=p(f(x)) but f(p(x))=f(0)=0, so p(f(x))=0 and f(x) is also a root of p, meaning f(x) must be a conjugate. f(p(x)) must equal p(f(x)) as, for example, f(x^4-10x^2+1)=f(x^4)-f(10)f(x^2)+f(1)=f(x)^4-10f(x)^2+1, just following from f(x+y)=f(x)+f(y), f(xy)=f(x)f(y), and f(q)=q for rational q. I don't know how easy any of that was to understand, or if you had figured any of that out yourself, but those were both good questions that prompted some enjoyable working out from me, so thanks for commenting them! Edit: Actually, the definition of conjugate is slightly more refined than that, for two numbers x and y to be conjugate they have to have the *same* minimum-degree defining polynomial, not just one being the root to the other's, which is why he mentions that even powers of the 16th root of unity aren't conjugate to odd powers, even powers are also powers of the 8th root of unity, giving a smaller defining polynomial. The proof that automorphisms map to conjugates is still essentially the same, just with the addition that, as f^(-1) is also an automorphism, we have p(x) = p(f^(-1)(f(x))=f^(-1)(p(f(x)), so if p(f(x))=0, p(x)=f^(-1)(0) is also 0, meaning that x and f(x) are always both roots or both not roots of any polynomial, so in particular, they must both be roots of the other's minimum-degree defining polynomial, so they must be the same.
@@carolinfardal Thank you sooooo much for writing this up, you cleared my confusion ! And no I didn't figure it out by myself yet, so this is definitely very helpful :) One last thing, how to show that a defining polynomial of x with rational coefficients has minimal degree ? Or alternatively, how to find one ?
This is an excellent video on group theory. I especially like the use of a unit circle in the complex plane to illustrate the automorphisms graphically. Unfortunately, there appear to be errors that can be very confusing to viewers unfamiliar with the material, including myself. At 19:51 the equation should read z - z^2 + x^3 - z^4 = sqrt(5). That’s necessary for the imaginary parts of the roots to cancel out leaving a real number result. But then there’s a second error that follows immediately in the animation of the automorphisms. The automorphism that preserves the equation, highlighted on the right, should swap z^2 with z^3 and swap z^1 with z^4. But the animation swaps z^1 with z^2 and swaps z^3 with z^4. The other two automorphisms, the ones that do not preserve the equation, are correctly animated. But the main problem here is that signs of the second and third terms of the equation are swapped in every case. What’s obscured by these errors is that the automorphism that swaps z^2 with Z^3 and swaps z^1 with z^4 preserves the equation because it also cancels out the imaginary parts of the roots. The other two automorphisms do not cancel out the imaginary parts of the roots and that’s why they are excluded from the group of Q adjoined to sqrt(5).
I’m not the best at math and I’m terrible at arithmetic but I absolutely adore your videos! please continue to make them. They’re like a glimpse into the fabric of reality and you make it so accessible. If my math teachers had been like you I’d be a far better mathematician.
I didn't understand more than 50% and I'm not even interested in the subject, but the wonderful music, your simple hand drawings and your repetition that this is beautiful made it the best video to watch before going to sleep.
Thank you for the video! The explanation is simple but illuminative. I just have a few (probably rookie) questions: 1. How did you define the conjugates of \zeta, the 16th root of unity? At first I thought they were defined by the algebraic equation \zeta^16 = 1, but that isn't right. In general, is it always defined by some algebraic equations? 2. Is it true that Q(\sqrt{5}) is a subfield of Q(\zeta), where zeta is 5th root of unity? Or is there a 1-to-1 mapping from Q(\sqrt{5}) to subfield of Q(\zeta)?
The conjugates of \zeta are the so-called primitive 16th roots of unity, ie. the solutions to the equation z^16 = 1 that are not solutions of z^k = 1 for k < 16. You can check that these conjugates are exactly the solution set of an algebraic equation: the sero set of the so-called minimal polynomial of \zeta. In this case, the conjugates are the solutions of the equation z^8 + 1 = 0.
Nice video, quick comment - doesn't every field automorphism fix Q? It fixes natural numbers n: f(1) = 1, so f(n) = f(1 + 1 + ... + 1) = f(1) + f(1) + ... + f(1) = n*f(1) = n*1 = n So it also fixes negative integers -n : f(0) = 0, but also f(0) = f(1+(-1)) = f(1)+f(-1) = 1+f(-1), so 0 = 1+f(-1), so f(-1) = -1 So it also fixes 1/n: f(1) = 1, but also f(1) = f(n*(1/n)) = f(n) * f(1/n) = n*f(1/n), so n*f(1/n) = 1, so f(1/n) = 1/n. Therefore f(a/b) = f(a * (1/b)) = f(a)*f(1/b) = a * (1/b) = a/b. I think it's only when you need the automorphism to fix bigger subfields that you need to add the "fixing" requirement explicitly.
Q is always fixed by the automorphisms of a field if the characteristic of the field is 0 (so you are right in this case). Otherwise, if the field has characteristic p (prime), the field contains a copy of F_p (field of p elements) that is fixed by its automorphism group. You can show that every field contains a isomorphic copy of Q or an isomorphic copy of F_p for some prime p. I guess he didn't want to get into technicalities and forgot to mention this fact
This is exactly my approach of GRH through rigs (not exactly fields, but the logic is the same) of L-functions: the automorphisms of the maximal such "L-rig" preserving the rig generated by a given L-function F are the symmetries of the multiset of non trivial zeros of F. In simple words, preserving F is preserving this multiset. I can send my pdf to whomever is interested.
You just made me, an analysis lover, watch 25 minutes of abstract algebra content. That is incredible
I used to classify myself as an "analysis person" and later on an "algebra person". But eventually I learned that the most beautiful math happens when algebra and analysis mix together. Some subjects that show off their interplay include algebraic number theory, Lie groups, Hodge theory, elliptic curves, modular forms, and much much more! The proof of Fermat's last theorem uses LOTS of analysis and algebra for example, and all the subjects I listed above.
exactly my thoughts.
Incredibly well done.
@@theflaggeddragon9472 Llevo años aprendiendo Math y cuando uno se atasca en algún momento, siempre hay personas maestras que saben enganchar a la Maravillosa Matemáticas
@@renatohugoviloriagonzalez4881 Que bien que continuas aprendiendo! Y tienes razon que somos muy afortunados teniendo estos videos de espertos gratis en TH-cam, para aprender y para desatascarnos. Aproposito, has visto los videos de Profesor Richard Borcherds?
Disculpe mi Espanol, no he hablado en much rato.
This video is a thing of genuine beauty. You have a rare talent for illuminating these deeply technical subjects in a fascinating and accessible way. Many thanks.
Thanks for the kind words, Jim! Appreciate you watching :)
Took group theory and ring and field theory but didn't get all the way to Galois theory by the end. What a clear and concise way to encapsulate the fundamental concept. Thanks for this.
Me too and I found abstract algebra the most difficult of all maths subjects I took. Watching this I felt like I was getting an insight and then it disappeared again, my brain obviously cant work in these terms.
Took Abstract Algebra and did much of what you did, with groups, rings, fields and the various domains (PID, ID) and ended after only after hitting a couple of the special topics (we hit finite simple groups, but missed intro to Galois theory)
holy shit you made the how fast is melee video, greetings fellow melee and math nerd lmao
Is there a second part where he actually explains why the 5th degree equation cannot be solved ?
@@sambtt thanks fellow nerd! :)
Thank you for putting so much effort into making this. This is my first time hearing about Galois Theory, and this video was amazingly clear and a treat to watch. It's sad that so few people watch these compared to other channels of equal quality.
Nah give it some time, these vids get a sizeable number of views
@@NonTwinBrothers yeah
@@NonTwinBrothers I agree. It's not one of these channels that live from immediate hype. I wouldn't wonder if this video is still getting watched in 20 years, when 99.99999% of youtube content is long lost to irrelevance :-)
@@harriehausenman8623 Exactly. No one knows/cares who the Kim Kardashian equivalent of the 18th century was, but we all know who Euler or Bernoulli was.
@@elidrissii 🤣
When you said that the lines wouldn't go where I expected, I almost paused the video, because I was pretty sure I did see where they would go -- and I was right! My intuition was based on the understanding of multiplication by any complex number of magnitude 1 as rotation -- which of course wraps around after each full turn. So, ζ⟶ζ² applied twice is just the rotation by ζ² twice -- ζ² * ζ² = ζ⁴. Well, that's reasonably obvious, but the next step falls out of the wrapping nature of the rotation. ζ⁴ twice is just ζ⁸, but since ζ is the 5th root of unity, that means that ζ⁸ = ζ³ -- and so on. With this view, all of the graphs are immediately evident from the starting point of the given mapping of ζ.
Same thing for me, except I was thinking in terms of modular arithmetic.
This is the most perfect introduction to Galois Theory that I have seen over several decades. It gives us not just the bare bones of the theory, but also their subtlety, their power, and beauty, and every idea copiously illustrated by clear diagrams and algebraic formulae. However, there must be something wrong with my pc, or my old ears, as I can hardly hear the voice over the music. I wish I could turn the music down, down to zero, and then I would enjoy the video for its full worth!
No it's not your pc or your old ears. The music overpowers the narration at times.
I loved this stuff so much when I was a young computer science student. Finite fields, coding theory, polynomials. Heck yes.
They taught abstract algebra in an undergrad CS program?
@@FsimulatorX abstract algebra is like 90% of math done in computing
@@johnnypiquel2295 interesting. I've heard that it's helpful if you want to program new languages and such but I'm not sure to what extent it might be applicable in ones software related job (although I suppose that depends on the type of role).
Either way I'm planning on taking it in the future just out of curiosity. Extra bonus if I can apply it to one of my projects :D
@@FsimulatorX it’s probably not directly applicable to most software roles. Indirectly, though, there’s the fact that there’s a commutative ring over bit vectors of length n, with bit-wise XOR as the additive operation and bit-wise AND as the multiplicative operation, although that probably doesn’t affect most programmers. There is an algebraic structure that even the most junior of programmers has an intuitive understanding of, though, and that’s monoids. Integer addition is a trivial example, but the more instructive case is strings, with concatenation as the binary operation. It’s plainly associative, and the identity element is just the empty string. Any programming language with generics and interfaces (or traits, or protocols, whatever they end up calling it) is capable of representing monoids at the type level, although it’s really only in richly, statically-typed functional languages like Haskell where you see algebraic structures like monoids being actively modeled as interfaces that any function can be defined in terms of.
It's a testament to the complexity of groups and Galois Theory that simplified explanations still manage to fly over your head, but equally it is a testament to the beauty of these concepts that every time you want to go through it once again simply to understand more. This was a fantastic video - probably the most beginner friendly of all the videos I saw in this area!
Huge fan of your explanation style and visuals! Can't wait to watch this.
I find it difficult to express just how GOOD this video was at explaining the general idea behind Galois theory. Genuinely, thank you. Thank you so much, you've given me another way to look at fields. Another tool that I didn't know even existed.
The connection to unsolvability of the general quintic:
Suppose x is a root of an irreducible quintic polynomial, and x is expressible by radicals. Then we can "build up" to a field containing x with a sequence of fields like Q < Q(a) < Q(a,b) < Q(a,b,c), where each step we adjoin an nth root of some element of the previous field.* Each step's Galois group will be a cyclic group. Using the Galois correspondence, this means the Galois group of the last field will have the property that it has a sequence of (normal) subgroups where the quotient at each step is a cyclic group. This is what we call a "solvable group".
However, the Galois group of a general quintic polynomial is the symmetric group S5, which does not have this property. When you try to form a sequence of subgroups, you run into the alternating group A5 which doesn't have any nontrivial normal subgroups. Hence we have a contradiction, so the original assumption that x was expressible by radicals is false.
*e.g., say x = sqrt(2)+sqrt(3+sqrt(5)), then we'd do Q < Q(sqrt(2)) < Q(sqrt(2),sqrt(5)) < Q(sqrt(2),sqrt(5),sqrt(3+sqrt(5))) so x is contained in the last field in the sequence. You can do this for any radical expression for x.
A lot of this is not quite familiar enough for me to join the dots in your concise explanation, but I'm glad you added this, since it was presented as one of the motivating ideas of the video.
Yea do you think anyone understands what you said there..do you even understand it honestly??
@@leif1075 if you have taken a course in group theory it is much easier to understand. I have no doubt the person who wrote this comment understands it
Very lovely addition by this person to the already amazing video. I had been meaning to study field theory, group theory etc and this is one of the first few videos that came up as a result. Even if we don’t fully understand the whole thing I think it’s fine. This universe has a whole plethora of things that are so beautiful but yet we understand nothing about them. But yet, even making the attempt and going through the motions is so rewarding just in and of itself!
May be after repeated attempts - we will be able to develop the intuition for these higher order abstract ideas.
I don’t think anyone can understand it fully on the first pass!
Is it fairly easy to explain why "Each step's Galois group will be a cyclic group"? This is where things get fuzzy for me.
Really missed a lot these videos. Thanks for coming back!
The thing that popped out at me when I finally understood the usefulness of this:
It was not at all clear that Q(zeta) should necessarily contain Q(sqrt(5)).
I can see why such a resemblance might exist given that zeta is the 5th root of unity, but this was not at all obvious.
But, we'd already seen the subgroup of Q(zeta), it pops out very clearly in the table! and this alone is enough to prove the field contains some subfield.
I'm not convinced of this tbh. I don't see the connection between 1^(1/5) and sqrt(5). I don't think a field extension Q[zeta] contains sqrt(5), I don't see how you get there with algebraic operations
@@bobtheblob728 Indeed this is tricky to see and I’ve spent about the past hour trying to convince myself of the fact! For brevity, let’s let z denote the fifth root of unity mentioned in the video, namely the polar point (1,2pi/5). It turns out that the rectangular form of this point involves sqrt(5). In fact, you can check that 1+2(z+z^4)=sqrt(5). So, any a+bsqrt(5) in Q(sqrt(5)) can be written as a+b(1+2(z+z^4)) in Q(z), which is why Q(sqrt(5)) is indeed embedded in Q(z)! Hope this helps!
If you’re curious, there is slightly more to be said! It is also true that sqrt(5)=-2(z^2+z^3)-1, which gives us another way to see the embedding. Moreover, the reason that Q(sqrt(5)) corresponds to the subgroup that it does is because the subgroup contains the permutation that swaps z^2 for z^3 and z for z^4, which just changes the order of addition in the two equivalent expressions for sqrt(5) given above. But, as addition commutes, these swaps preserve the expression being equivalent to sqrt(5). So, we see that that subgroup contains exactly those permutations which fix sqrt(5), which again explains the correspondence to Q(sqrt(5))!
@@ronaldhoagland9597 ohh that makes sense I didn't check to see what sin/cos (2*pi/5) were. super interesting that sqrt(5) shows up there
@@bobtheblob728 You can easily check it in Wolfram alpha.
If you do the fifth root of unity (e^(2πi/5)) the number you get is -1/4 + sqrt(5)/4 + i*(sqrt(2*(5 + sqrt(5))/4)
If you want to add a root to a field you have to do a field extension. First you extend Q to Q(sqrt(5)) and then you extend again to get Q(sqrt(5), sqrt(2*(5 + sqrt(5))))
Luckily, the fifth roots of unity arrive if you extend Q by radicals (they are the solution of a quartic polynomial).
This is actually a really interesting comment. It's not at all obvious that Q(zeta5) is the only subfield in there- the only way (that I know of) to show that is via Galois theory!
Finally a video after 9 months! Feels great man
Wonderful, clear videos! Great! So appreciated! At 19:47 there appears the incorrect equation "zeta + zeta^2 - zeta^3 -zeta^4 = SqRt[5]", which should read "zeta + zeta^4 - zeta^2 -zeta^3 = SqRt[5]". What follows in the video becomes correct after this revision. Small details.
Thanks! This confused me quite a bit. Also, it would help to explain in what sense and why Q(\sqrt 5) is between Q and Q(\zeta).
Less than 5 minutes in watching and I have made more Google searches than required by an assignment. I love Maths and Engineering.
Keep up the good work.
This is amazing! I have had a difficult time trying to find a good explanation of Galois theory, and this finally made it click. Thank you so much!
Thanks Jason! Glad you found it helpful :)
Never learned galois theory in school, just some basic group theory and field theory. I always imagined it was very daunting, but this 25 minute video was very easy to follow and gives me a sense of why people even care about this field. Thank you
As someone who has no experience in the more abstract side of math, this video was surprisingly clear!
oh wow, we have very similar profile picture
neato
Beautifully crafted content.
How can one not love math or at least sense its underlying beauty? This video of yours really showcases how wonderful math can be. Thank you!
Very nice video, thank you!
The only thing that confuses me a little, is when you say @15:20 that your mind can not make sense of what you're seeing. It is confusing me, because to me it is not just very logical and feels perfectly normal, but I've paused the video and could guess the rest of the permutations (after the first 4). Now I did a lot of group theory before, symmetry of groups, chemistry, knots, Rubik's cube, permutation matrices ... but it really is very natural to me, especially visually.
Anyway, nice video and keep it going ;)
Yep, I think watching Mathologer helped with that. For example the 4th one along where z3 points to z5 and back. The z1 points to z7 so this is the 7th power permutation. z3 to 7th power is z21 and 21 is 5 mod 16 so it goes to 5. z5 to 7th power is z35 which is 3 mod 16 so that goes to 3 again. All the others work like that.
If there had been TH-cam in my teens, I would have studied math at college. Thanks for posting! I find this enormously interesting and satisfying to watch in my middle age.
This is my 3rd watch through of this video. I still love it. I still only 80% grasp it.
Life goal: understand this video.
It's always a treat to see a video from this channel. No other channel gets me as invested in modern mathematics like yours. I'm in my undergrad for physics, but I'll probably take my school's graduate algebra sequence starting next fall because of this. Keep it up.
this channel is extremely underrated, some of the best math content on youtube. no other vid has ever gave me as good of an intuition for this topic, and i've seen a lot of them
One of the clearest and most elegant presentation of Galois theory I have seen!
One day this channel will get the recognition it deserves, keep at it!
I had nearly given up on learning Galois Theory, but your videos gave me the motivation to continue!
Never thought I would ever be able to comprehend anything in Galois Theory. Thank you for letting me prove myself wrong!
Great video. I might have to watch it a couple more times to grasp everything in it, but still, it's the best explanation of Galois Theory that I've found anywhere.
I'm going to watch this over and over again until I finally have an intuitive understanding of this theory. I did this in a Uni course but never got a good grasp on how we really arrive at the final result.
Trust me (as someone who doesn't live up to this advice nearly as much as he should): the way to really understand it isn't (just) to repeatedly watch things, but to find or (once you've begun to really grasp the idea through use) pose problems to play with; if you don't *use* the syntactic tools to navigate a constraint-wise consequential (i.e. well-defined) context in such a way that failure to *understand* the concept will more or less reliably ensure failure to resolve the problem - which sounds very negative, but what's important is the contrapositive supposition (which likewise seems to be the case in practice, at least up to fairly nuanced arguments about what really constitutes "understanding") that if one successfully resolves a nontrivial quantity and/or variety of examples, then one *must* possess some minimal degree of genuine understanding of the salient concepts (i.e. the ones indispensable for the problem's rigorous and persuasive resolution).
In my experience, this is the hardest part about self-studying higher mathematics: not so much access to problems, but access to *feedback* with respect to solutions the consistency/coherence/general quality of which is non-trivial to determine lol but it's certainly easier now than ever before, at least. Hope none of what I said came off as condescending or pretentious or anything like that, I just feel for anyone who also yearns to understand these things and want to help them in any way I can frankly, so...godspeed my guy 🤟🏻 haha
@@SliversRebuilt I didn't read all that but yes, I'm not only going to watch it I will play with the ideas myself as well. I'm not a media zombie
@@BachelorChowFlavour hey man I wasn't trying to insinuate any such thing, like I said sorry if it came off that way I only say as much because it's a mistake I personally have made which has affected me
@@SliversRebuilt You hit the nail on the head as to a common stumbling block for students, ie, lack of "access to feedback with respect to solutions the consistency/coherence/general quality of which is non-trivial to determine". Your comments indicate you truly understand what it takes to learn a difficult subject effectively. Kudos!
Wow I recently watched Borcherds’ Galois theory series and this elucidated so much in that. Incredible video!
Over at about 12:30, he skips a couple steps which confused me a bit. I'll use s for sigma and z for zeta here.
According to his map, we assumed that s(z) = z^2 (since that's where the arrow points to). Also keep in mind that z^5 = 1.
s(z^2) = s(z)^2 = (z^2)^2 = z^4
s(z^4) = s(z)^4 = (z^2)^4 - z^8 = z^5 * z^3 = 1 * z^3 = z^3
Those were his calculations.
Why does -z^8 appear?
@@3Triskellion3 Maybe it was a typo and was supposed to be '=z^8' instead of '-z^8'
Thank you so much, may I ask why does z^5 evaluate to 1?
@@akritworanithiphongBecause z is initially the 5th root of unity, and by definition the n-th root of unity raised to some integer the power n yields 1. See wiki for more.
A lot of new videos discuss Galois theory. This is by far the most profound and pedagogical discussion on Galois Theory out there.
Wow - I'd made several (admittedly rather half-hearted) attempts to figure out what Galois Theory was about over the years. Made no progress whatsoever !!! Anyway, this video actually made it all start to make sense. Truly remarkable how well you explained it all. Many thanks !!!
This is beautiful! I love how you started with the concept of the root-two conjugates. Such an elegant introduction to the deeper math. Fantastic presentation as always!
This will be a legendary video
Wow. I'm at a loss for words. That simple question of which permutations preserve algebraic relations is such an interesting question and made everything click for me. Galois really was way ahead of his time.
Thank you for an amazing video
This is an excellent video, with wonderful graphics that really make the ideas behind Galois theory come to life. There's only one thing that threw me. At around 7:42, you state that Aut F is "a little too big", so we restrict our attention to Aut F/Q. I spent a good hour trying to figure out what could possibly be in Aut F but not in Aut F/Q. At the end of it, I proved to myself that any field automorphism of any field containing Q must fix Q, so it ended up being an hour well spent. It doesn't detract from the main point you're making there, which is essentially, "this is the notation we use", and when we have Q
Minor error at 11:43. The elements of the field also have the term square 5-th of 1 elevated to 4, one term more
Actually this is not the error. The error occurs a few seconds later at 11:46. It is redundant to include this term, so it is correct at 11:43, incorrect (or at least redundant) at 11:46
Was curious if the 4th power was missed or redundant.
1 + z^1 + z^2 + z^3 + z^4 = 0
therefore z^4 = -1 -z -z^2 -z^3
so no multiple of z^4 required as a seperate term.
Later, he doesn't say that that expansion is unique, does he? He just applies sigma, correctly.
So I would not call it an "error" in either case.
But good to note that, I wondered the same.
AGHHHHHHHH I WAS WAITING FOR YOU TO UPLOAD SOMETHING 😭😭😭😭
It's a good day when this channel uploads
one of the clearest video introducing Galois group
Could anyone please explain why there are only 8 roots of unity in the example with zeta = e^i2pi/16 ?
I dont exactly understand what makes the other roots redundant :)
Good question, that was also my first thought: Why only take the uneven exponents?
@@markborz7000Yes, and why were the even exponents counted in the case with the fifth root of unity?
this video is genuinely amazing, please more content like this! This was just the right video length and buildup for the topic at hand, i could follow every step and im about to look up some more Galois Theory, bc im genuinely intrigued now!
I would like to take a moment to thank you, and all the incredible math explainers on youtube for making such clear, and well made content. For some reason, I decided I wanted to learn Galois Theory, and I have been doing nothing but trying to understand it for an entire week now. I have come further than any teacher could have brought me thanks to content like yours. Even though I am only a freshman, I am seriously considering studying this theory even further, in order to see all of it's power. I was not only intrdocued to group theory, but to all of algebra thanks to content like yours, mate, you're amazing, and it is thanks to you that algebra is beautiful.
Thank you mate
You got me! After reading four books of Galah theory and group theory, this was the best introduction ever thanks a lot😅
As a physicist this intution is really helpful for me.
Aleph, very much appreciated this very well done video. The starting point, that sqrt(2) and -sqrt(2) are interchangeable in any algebraic equation, not just polynomials, is great. could however use some clarifications, like how do you go about proving sqrt(2) and -sqrt(2) are always interchangeable and ditto for certain permutations of higher roots, and how are Zeta and sqrt(5) related and therefore the fields are related, and why only odd powers in the 16-th root of unity example, and the ending where "what is the square root of two" is answered by saying it depends on the field you're in needs an explanation, lastly I wouldn't go so far as to say this changes how I view fields unless I was convinced that all fields are extensions to the rationals (and I'd like to know in what sense are the reals an extension of the rationals). But I really like your style and very much hope you continue, I'd even go so far as to say I wish there were professors that were as good at explaining things as you are when I was studying math.
About the "how do the reals contain the rationals" question, one precise way to answer is this: there is a subfield of R that is canonically isomorphic (in the sense of field isomorphisms) to Q. The "canonical" part is just human convention, so really the only statement is "Q is isomorphic to a subfield of R" and that's it.
Let me take a shot answering the first question: why are conjugates interchangeable for certain permutations? I'll use the root 2 example.
Let s be an automorphism of Q(root 2) fixing root 2. Let q be any number in Q. Then s(q) = q.
Now express q as some algebraic combination of elements in in Q(root 2), say p1, p2,...p(n). Because s is an automorphism passing that algebraic combination through s will yield the same combination of the elements s(p1), s(p2),... s(pn) (because of the properties s(p1 + p2) = s(p1) + s(p2) & s(p1*p2) = s(p1) * s(p2)). because s(q) = q, the combination of elements after passing through s still equals q.
Now for Q(root 2) the only possible permutations gave s(root 2) = root 2 or s(root 2) = - s(root 2), so applying those permutations won't change the value of expressions, which lie in Q.
I totally agree that this video is pretty good but, as has been pointed out, it is spoilt by the two issues raised by Peter relating to sqrt(5) and why just the odd powers of zeta are considered. Such a shame that what is an excellent step-by-step presentation should omit two crucial steps.
15:25 "The pattern seems almost random"
Really? To me it seems very organized, symmetric!
P.S.: Brilliant video!
Yeah, where he sees random I just see modular exponentiation. But that's what you get from studying cryptography for as long as I have I guess. :)
Yep, I also was quickly looking at the order of each element and thinking most likely they were groups.
@@ingiford175 exactly! Pretty neat that it naturally emerges like that by studying solutions to polynomial equations.
Oh my... what a well put together video on Galois's theory. My textbook makes so much more sense now
This is better than other who had higher viewers..this is worth to watch..because there is no negative
God I wish I could understand more of this, hope to come back one day and see the beauty of this explanation in the same way I could finally understand the beauty of the Stokes' theorem on manifolds. Keep the excellent job, you are inspiring!
If you enjoy this type of stuff, pick up "Infinity and the Mind" book by Rudy Rucker. One of my favorite books on the math subject. I can't recommend it enough!
Wow I'm the opposite. I get Galois theory but wish to be good enough to understand Stoke's theorem!
I love algebraic number theory but I am still working on my intuition for Galois theory. This video is a very powerful intuition pump. Working through the examples is very enlightening.
The arithmetic of roots of unity has a mesmerising beauty and things like Dedekind sums are among some of the deepest objects in mathematics. I am *so happy* to have discovered this video. I subscribed.
Nice work!
One suggestion for improvement: when drawing roots of unity you state that they are equally spaced, but the drawings are sometimes very off. (2pi/5 sometimes looks like pi/2 and sometimes like pi/4). It could also help if you draw zeta^0=1.
When I watch your videos I understand close to nothing but I still love them
I really like the music in this video. It's gentle and pensive but continuously progresses as if to say, "Consider the following: ..." Which is clearly just perfect; much like 3B1B's "Pause and ponder" music.
I'm still quite a long way from grasping most of the deeper intuitions here. Still, I often find as I learn about a topic in math that my brain actually soaked up little bits and pieces of things like this as if by osmosis and then later on (sometimes much later) I'll suddenly realize I have enough pieces and enough context to understand the significance of what was being said.
So thanks for making these videos and putting so much detail, care, and attention into them even though the number of people able to full understand everything in them on the first watch may be relatively small.
Amazing! Your best video so far!
Just one note. I believe that Q has no non-trivial automorphisms. So every field extension of Q automatically fixes Q. That is, you could have saved on that technicality in the video I believe.
I had the same thought. Sigma(3/5) = sigma((1+1+1)/(1+1+1+1+1)) = (1+1+1)/(1+1+1+1+1) = 3/5. But also it's hard to believe someone who was freshly studying this stuff to make a video would put that in there as a mistake. So I'm scratching my head...
@@dstahlkeMonstrous moonshine group?
Outer automorphisms for S6 have unique properties.
Wow...! I must have seen this video no less than 10 times through the years, the first while I didn't even know what a bijection was. Slowly but surely I learned everything to understand this. Thank you for being in my mathematical journey!
I still don't understand everything in this video, but compared to most graduate level textbooks, this is a gift from God.
Man you are finally there again, Please don't let us wait too long...because without your videos
TH-cam math space looks lot less a better place!
Amazing refresher on Galois theory after learning last year!
This video is truly amazing ! I didn't imagine someone could explain so clearly and in only 25 minutes the roots of Galois Theory.
Amazing how deep these concepts can go
This is by far the best video on Galois Theory I have seen on youtube. Wish I had your videos back when I was in school 😅
I'm so looking forward to re-viewing this. Great job!
Ohh my god,I missed you a lot,you had a really amazing content it helped me a lot
Nice video that reveals my own limitations. Galois theory is something almost understandable for me without ever really understanding. I wouldnt be able to go from this into why algebraic solutions of 5th order equations don't exist 😀
Hopefully some day I'll get there.
Say you have a general solution to a quintic, then you can express it in terms of the basic operations and taking a root. Everytime you take a root, you move to a bigger field where your new element lives in. The galois group of this new field over the field you just left is always very nice. Once you've arrived at your solution, the galois groups should then contain a chain of subgroups that are all nice, all the way to the trivial group. The thing is, you can just check this fact for any field you want, and if there isn't such a chain of nice groups, you know you can't construct that field by taking roots aka solving a general equation. So, we look at the splitting field for the general quintic, and we see that the galois group over Q must be isomorphic to S5, which _does not_ have a chain of nice subgroups. So we can't reach it by taking roots, and so there is no formula. Bit rushed, but QED.
Also really don't recommend Borcherds, instead try Professor Macauley's series on Visual Group Theory. That's an genuinely good series.
@@ruinenlust_ "Galois group over Q must be isomorphic to S5, but S5 doesn't have a corresponding substructure" this puts into one concise sentence what I was so struggling to put into words. It's one thing to broadly understand, but better once you can articulate it clearly. Cheers.
19:31 "Phrasing this in a more down to Earth language: we first looked at permutations that preserved the algebraic relations over *Q.* Now we'll look at permutations that preserve algebraic relations over *Q* adjoin √5."
That is exactly what an alien pretending to be a human would say.
Been waiting for you to post a new video! Takes me back to my first encounter with this in college. Great content as usual!!
Just amazing. You truly moved my heart with this beautiful exposition. I wish sometime to have such understanding in any field.
Amazing job.
Finally after so many days
Hey there, just finished watching the video and I had to congratulate you for the amazing work ! The way you explain and motivate all of the concepts as well as the editing make the video a joy to watch and kept me invested in wanting to fully understand the fundamental theorem.
Some feedback : as someone who is absolutely not familiar with Galois theory and anything close, there were 2 or 3 times in the video where I was slightly confused, namely I don't understand/know how to compute the Q-conjugates of a given number and it is not clear to me why an automorphism on Q(zeta) has to send zeta to one of its conjugates (I guess these are exercises left to the watcher ?)
Despite these minor confusions which shall be fixed upon rewatching the video or some reading/thinking on my side, I have to say that this video was pure bliss to watch, I learned a lot on a topic that is new to me, and I am very much looking forward to the next ones ! :)
I don't know if you've figured out your confusions on your own, but they gave me some thoughts that I wanted to share. So, first, there isn't any one formula you can apply to get the conjugates of a number, say, x, but the conjugates are the other solutions of the minimum-degree polynomial with rational coefficients defining x. So, for example, (sqrt(2)+sqrt(3))^2 = 5+2sqrt(6), so ((x^2-5)/2)^2-6=0, or x^4-10x^2+1=0, is a polynomial with a root of sqrt(2)+sqrt(3), and as the other roots are all +/-sqrt(2)+/-sqrt(3), which are all "related" to sqrt(2)+sqrt(3), the polynomial is probably minimum degree. You need the polynomial to have minimum degree to rule out polynomials like (x^2-2)(x-1) or (x^2-2)(x^2-3) making sqrt(2) and 1 or sqrt(2) and sqrt(3) conjugates under that definition, when they shouldn't be. With that characterization of conjugate numbers, it isn't too hard to show that automorphisms that fix rationals send numbers, again, say x, to their conjugates, as if p is a defining polynomial of x, having roots that are only conjugates of x, then if f is an automorphism fixing rationals f(p(x))=p(f(x)) but f(p(x))=f(0)=0, so p(f(x))=0 and f(x) is also a root of p, meaning f(x) must be a conjugate. f(p(x)) must equal p(f(x)) as, for example, f(x^4-10x^2+1)=f(x^4)-f(10)f(x^2)+f(1)=f(x)^4-10f(x)^2+1, just following from f(x+y)=f(x)+f(y), f(xy)=f(x)f(y), and f(q)=q for rational q.
I don't know how easy any of that was to understand, or if you had figured any of that out yourself, but those were both good questions that prompted some enjoyable working out from me, so thanks for commenting them!
Edit: Actually, the definition of conjugate is slightly more refined than that, for two numbers x and y to be conjugate they have to have the *same* minimum-degree defining polynomial, not just one being the root to the other's, which is why he mentions that even powers of the 16th root of unity aren't conjugate to odd powers, even powers are also powers of the 8th root of unity, giving a smaller defining polynomial. The proof that automorphisms map to conjugates is still essentially the same, just with the addition that, as f^(-1) is also an automorphism, we have p(x) = p(f^(-1)(f(x))=f^(-1)(p(f(x)), so if p(f(x))=0, p(x)=f^(-1)(0) is also 0, meaning that x and f(x) are always both roots or both not roots of any polynomial, so in particular, they must both be roots of the other's minimum-degree defining polynomial, so they must be the same.
@@carolinfardal Thank you sooooo much for writing this up, you cleared my confusion ! And no I didn't figure it out by myself yet, so this is definitely very helpful :)
One last thing, how to show that a defining polynomial of x with rational coefficients has minimal degree ? Or alternatively, how to find one ?
@@carolinfardal Thanks for this! Really helped with some confusion :))
I'm here from Vince's Bandcamp. I'm intrigued by this explanation of the theorem of Galois as well as the background music. You have earned my sub.
beautiful and eloquent explanation, as always
you really are amazing to share advanced knowledge and boil it down to something way more understandable. Keep it up!
This is an excellent video on group theory. I especially like the use of a unit circle in the complex plane to illustrate the automorphisms graphically. Unfortunately, there appear to be errors that can be very confusing to viewers unfamiliar with the material, including myself. At 19:51 the equation should read z - z^2 + x^3 - z^4 = sqrt(5). That’s necessary for the imaginary parts of the roots to cancel out leaving a real number result. But then there’s a second error that follows immediately in the animation of the automorphisms. The automorphism that preserves the equation, highlighted on the right, should swap z^2 with z^3 and swap z^1 with z^4. But the animation swaps z^1 with z^2 and swaps z^3 with z^4. The other two automorphisms, the ones that do not preserve the equation, are correctly animated. But the main problem here is that signs of the second and third terms of the equation are swapped in every case. What’s obscured by these errors is that the automorphism that swaps z^2 with Z^3 and swaps z^1 with z^4 preserves the equation because it also cancels out the imaginary parts of the roots. The other two automorphisms do not cancel out the imaginary parts of the roots and that’s why they are excluded from the group of Q adjoined to sqrt(5).
Yes! I've been looking forward to this!
Omgggg I needed this so much for my exam last year! =''(
this is the most beautiful thing i've ever seen
I’m not the best at math and I’m terrible at arithmetic but I absolutely adore your videos! please continue to make them. They’re like a glimpse into the fabric of reality and you make it so accessible. If my math teachers had been like you I’d be a far better mathematician.
This is great i kept running into galois theory in my exploration of wolframs work and didnt really understand what it was until now.
I thought I knew some of this stuff, but the way you gave insight into it is brilliant. Thanks!
I forgot about these videos for a sec!! This upload was a surprise in a good way
I have never understood Galois Theory so well. Thanks for making this video. You are a great teacher.
I just feel confused watching this video since a lot seems wrong.
I didn't understand more than 50% and I'm not even interested in the subject, but the wonderful music, your simple hand drawings and your repetition that this is beautiful made it the best video to watch before going to sleep.
Thank you for the video! The explanation is simple but illuminative. I just have a few (probably rookie) questions:
1. How did you define the conjugates of \zeta, the 16th root of unity? At first I thought they were defined by the algebraic equation \zeta^16 = 1, but that isn't right. In general, is it always defined by some algebraic equations?
2. Is it true that Q(\sqrt{5}) is a subfield of Q(\zeta), where zeta is 5th root of unity? Or is there a 1-to-1 mapping from Q(\sqrt{5}) to subfield of Q(\zeta)?
Nvm question 2. The 5th root of unity is (-1+\sqrt{5})/4.
The conjugates of \zeta are the so-called primitive 16th roots of unity, ie. the solutions to the equation z^16 = 1 that are not solutions of z^k = 1 for k < 16. You can check that these conjugates are exactly the solution set of an algebraic equation: the sero set of the so-called minimal polynomial of \zeta.
In this case, the conjugates are the solutions of the equation z^8 + 1 = 0.
Your videos SLAP. This was one of the best explanations of the FTG. Thank you!
Thanks Michael! Glad you enjoyed the video :)
This video is a true jewel indeed! Great content Sir!
What? You uploaded!? 9 months later... but glad you finally upload!
Thank you for your masterful clarity!
Best explanation I’ve seen
Thank you for sharing your knowledge and for the outstanding way you do it
Nice video, quick comment - doesn't every field automorphism fix Q?
It fixes natural numbers n: f(1) = 1, so f(n) = f(1 + 1 + ... + 1) = f(1) + f(1) + ... + f(1) = n*f(1) = n*1 = n
So it also fixes negative integers -n : f(0) = 0, but also f(0) = f(1+(-1)) = f(1)+f(-1) = 1+f(-1), so 0 = 1+f(-1), so f(-1) = -1
So it also fixes 1/n: f(1) = 1, but also f(1) = f(n*(1/n)) = f(n) * f(1/n) = n*f(1/n), so n*f(1/n) = 1, so f(1/n) = 1/n.
Therefore f(a/b) = f(a * (1/b)) = f(a)*f(1/b) = a * (1/b) = a/b.
I think it's only when you need the automorphism to fix bigger subfields that you need to add the "fixing" requirement explicitly.
Q is always fixed by the automorphisms of a field if the characteristic of the field is 0 (so you are right in this case). Otherwise, if the field has characteristic p (prime), the field contains a copy of F_p (field of p elements) that is fixed by its automorphism group. You can show that every field contains a isomorphic copy of Q or an isomorphic copy of F_p for some prime p. I guess he didn't want to get into technicalities and forgot to mention this fact
This is exactly my approach of GRH through rigs (not exactly fields, but the logic is the same) of L-functions: the automorphisms of the maximal such "L-rig" preserving the rig generated by a given L-function F are the symmetries of the multiset of non trivial zeros of F. In simple words, preserving F is preserving this multiset. I can send my pdf to whomever is interested.
These are really heavy videos. And yet, I keep watching them
This video deserves way more views
very good video, loved watching it unfold and now i have a better understanding of what the heck galois theory is!
Just, too good. Really motivated to study abstract algebra now.
I really enjoy all your videos. Thanks for putting in the work