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This is the highest production video of yours I’ve seen so far, very nice

You are trying the Birch Swinnerton or RH I am sure. Had it been not RH, you might not have kept it a secret.

Thank you so much for your content, you have an incredibily captivating way of presenting things

4π^2 All prime products are 4π^2= [4π^2]^(8/8)= ⇔ [4π^2]^8 × [4π^2]^(1/8) E8 ÷ E8=1 (E8)^24 ×(E8)^(1/24) 3×8=24 The 3rd generation of quark・lepton And it gives a hint to the ABC problem❗️

That was awesome proof and very easy to understand thanks a lot

i \ didi 40 i \ 1.00 USD per higher dimension - lol - bb bb bb is a b0zo - limited - CPA - CFA - VBA - VMA - MDO i - do not fumble pling horse tongue - tuesday night lights i - viva marinanist central catholic secondary - SAtin - SeAttle - 40

If 2x+1 is not divisible by three then 2x+1 is a prime number

Great lecture, thank you.

Do you have an explanation of the key claim at 14:10 - that the string of p elements can’t repeat in <p rotations.

e^π +ie^πi +je^πj +ke^πk +le^πl =MC ^2 e^πi-1=0 jkl=0 Quarternion Octonion Principle of the constancy of the speed of light Law of conservation of energy Law of conservation of momentum ζ(s),η(s),Γ(s) The infinite sum of natural numbers is ∞, -1/12 Differential calculus, integral calculus

π= 3.141592653589793…, π≒√2+√3= 3.146264369941972…, This gives some hints. SU(2) → √1√2 U(1) → √1 Next, if we write it in this form √1√[√2+√3]√4 → 2√1√[√2+√3]→2√π √2+√3≡√5 If there are some special spaces where this relationship holds, √2+√3=√5 √5= √2+√3 √7=2×√2+√3=√1+2×√3 √11=√1+2×√5=2×√2+√7 √13=2×√5+√3 √17=√13+2×√2 √19=2×√7+√5=√9+√10 ...=... -∞<...<-4<-3<-2<-1<0<1<2<3<4<5<7<8<9<10<11<12<13<14<15<16<17<18<19<...<+∞ These are true in a space with many one-dimensional lines, which at first glance appear to form triangles and intersect at three-dimensional angles. Each line has an origin 0, and the distance from the origin 0 is 1 unit. The axis of the one-dimensional real part that constitutes the Cartesian coordinate system and the oblique coordinate system has an origin 0 and a unit 1. When you look up at the night sky, the stars shine like the moon, galaxies, and constellations in the sky, and numbers are scattered like shooting stars that suddenly shine brightly.

Can't believe you slandered potatoes. Shameful.

6:40 all groups of order 6 are solvable, but not simple, so I assume it's S₃ 8:00 the eigenvalue gives the left column, not the bottom row

solved. P=peneka. NP=nau-peneka. next...

Absolute life saver!!

7:10 In the example of the construction of "a" splitting field of p(x) = x^3-2 over Q, the field of rational numbers, we factor p(x) = x^3-2 = (x-2^{1/3})(x^2 + 2^{1/3} x + 2^{2/3}) = p_1(x)p_2(x), so that p_2(x) is just irreducible over the field L = Q(2^{1/3}) not over the Q. this irreduciblity over L = Q(2^{1/3}) leads us to the new field M := L [x]/p_2(x) 8:12. 13:33 In the general construction of the fields, i.e., the proof of the existence theorem of the splittiing field of any polynomial p(x) over the field K 13:33, we factor p(x) = p_1(x)p_2(x) .... so that each p_i(x) is irreducible over K but ... as in the example 7:10 should we assume that each p_i(x) is irreducible over K_i, not just K, where K_i is defined by the recursive manner K_{i+1} = K_i[x]/(p_i(x)), and each K_i would be a field if we assumed that each p_i(x) was irreducible over K_i, not K. Or I am not sure but is each p_i(x) automatically irreducible over K_i, if we assume that each p_i(x) irreducible over K???

Thank you very much for these very insightful videos. I appreciate the clarity and lucidness of your presentation. You never sacrifice those in favor of "formal rigor". I am wondering if you could cover the subgroups of GL(2), namely the maximally compact subgroups and the groups corresponding to simple Lie algebras? I understand that the topic is highly dependent on the field in question, but your insight and explanations are extremely valuable nonetheless.

2:58 why is k2 an extension of k1?

18:45 why the three notations for a field extension

The general statement here is that a domain has unique factorisation if and only if numbers can't indefinitely be subdivided under multiplication, and there exists a gcd-operation (up to units), with the following three properties: 1. If d|a, d|b, then d|(a,b) 2. (a,b) divides both a and b If there exist two of these operations, they divide each other, so they are equal (up to units). It also has some nice algebraic properties, such as (ad,bd)=(a,b)d. If p is prime, and p|ab, then (a,p)(b,p)=(ab,ap,bp,p²), and p divides the right side, so p divides the left side, so (a,p)(b,p)≠1, so p|a or p|b.

Can anyone please explain how he replaced A by the matrix with reA and imA ? And thank you

Beside all the interesting and insping maths, I love your humor, which is legendedry: My rabbits are spherical. Made my day :D:D:D Thank you

2(362310209.122785/(1008.185959)(.7071087516))^3 = 2(1008.185959(.7071087516)(508222.29668339))^2 where 1008.185959(.7071087516) = 362310209.122785/508222.29668339

(1.904631362/1.903613583) + (1.903613583/1.904631362) = 2.000000285

The comments at 7:40 really improved my personal experience with this.

Constructions using ruler and compass can also be made using only a compass.

Galois theory only describes the relation between fields up to their perfect closure, just as homological algebra only describes modules up to their injective envelope. So you might as well take Galois theory over perfect fields, since any other properties have to established seperately anyways. The advantage is that this gets rid of the concept of non-seperable extensions, so normal and Galois extensions are now the same thing, similar to the Galois correspondence in covering maps.

Idk how many videos Id been warching in a row but this is very entertaining and enlightning.....I was literally reading category theory a while ago and watched a video yesterday on the fundamental group so all this connections are so OP and beautiful

20:40 - You have also said before that Spec(k^n) has n points, but isn't that wrong? Take R^2. It has 3 prime ideals. The 0 ideal as well as the ones generated by (1,0) and (0,1), respectively. Those ideals are maximal as R^2/((1,0)) is isomorphic to R. In general, k^n should have Krull dimension n-1 and 2^n ideals, 2^n-1 of which prime, since the whole ring is not a prime ideal.

Great lectures! thank you. I didn't understand at 9:58, how is fact that abs(Z1*Z2)=abs(Z1)*abs(Z2) related to sqrt(x^2+y^2)=sqrt((x+iy)*(x-iy))

the best

-rep

You keep talking of "closures of points" and "closures of singletons" as in 22:30, but Spec(R) is always Tychonoff, or T1 space, so all singletons are closed sets.

If anyone knows: at 10:56, how do we know L even exists by only doing a countable number of iterations?

3:40 - Shouldn't Spec(k[x]/(x^2)) have two points? (0) and (x) := (x+(x^2))?

Did we just skip over why x is transcendental at 6:00 but spent an entire 3 minutes explaining the trivial case of cos 2pi/7? :D

at 1:27 Mr Borcherds says "...1 is a unit..." Can someone expand on this idea?

The best

Would have been a great talk, but the slides are out of focus. It was in focus for the first few seconds and then the focus suddenly changes and nothing can be read.

Borcherds concludes the proof of Hilbert 90 at about 15:17 by saying that θ + α•σ(θ) + α^2•σ^2(θ) + … = β. But this looks incorrect to me. The terms of β are (α•σ)^k(θ), which are not equal to (α^k)•σ^k(θ). E.g. (α•σ)^n = N(α)•σ^n = id ≠ (α^n)•σ^n unless α^n = 1. The proof still works, but we need to show that 1 + α•σ + (α•σ)^2 + … + (α•σ)^(n-1) is not identically = 0. It is a linear combination of the characters σ^k, but with coefficients 1, α, α•σ(α), etc.

The circle of the Fano plane shouldn't be outside the triangles, it is the line containing the three midpoints of the sides of the triangle.

My general philosophy here is that definition is irrelevant, you can always define yourself right, but this is meaningless when you can't use this in a mathematical proof. The only right thing to do mathematically is to find the places where you would intuitively use the dimension in your argument, and construct a "measure" with the properties you would want here. In order to construct the Krull dimension, or any dimension with ordinals, you would already want to know everything you can proof with it, making them completely useless.

Great 👍

3:57 Anyone can explain more about the proof for K is uncountable using Aoc?

I dunno y im watching this but learning this interesting Number Theory even though i passed my graduation from College lol.

I don't understand what you mean when you say "the closure of a point" or "this point is not closed", since Spec(R) is always a T1-space, so singletons are always closed.

Hi, professor, I wonder if there is a book which your lectures are mainly based on. I just want to find such a book so that I can keep watching your lectures and learn well. THANK YOU!

took abstract algebra with him many years ago at Berkeley.

First lecture goes in Riemann hypothesis lol

i was thinking about this lately (the permutation of letters in the alphabet) turns out its a real established concept