I like that there are like 24k views and only 270 likes which means that this is like some of the most helpful content out there. I've got my final today LFG!
Is Joe Brawley still at Clemson? I went to N. C. State with him in the 60s. A long time ago! We tried to learn Galois theory there from Nathan Jacobson's books. I learned it incompletely at U of Illinois. Thanks for your lessons. The theory is now revealed. Bravo.
2 questions: 1) In the previous video at the very end, it determined that zeta*cube root 2 has minimal polynomial (x^3)-2 over Q and that [Q(zeta*cube root 2):Q] = 3. Here we find [Q(alpha):Q]=6. This feels contradictory to me. 2) In the final exercise, what is the minimal polynomial? alpha^4 should give us an element in Q, but I calculate it as 49 + 20*sqrt 6
Answer to question 1: Q(zeta, cbrt(2)) = Q(i*sqrt(3), cbrt(2)), but Q(zeta*cbrt(2)) does not need to equal Q(i*sqrt(3)*cbrt(2)). Answer to question 2: a minimal polynomial to alpha is x^4-10x^2+1=0 (up to a constant factor). It doesn't have to be in the form of x^n. It just needs to be a minimal degree polynomial that alpha satisfies. It can be proven that the minimal degree polynomial must also be irreducible, and unique up to a constant factor (here over \Q, you can search online for a proof). How did I find this x^4-10x^2+1? Specific to this problem, I just wrote out four powers of alpha and played around with them. Not sure if a general method exists.
when you say the only automorphism of Q(sqrt(2)) is isomorohic to C2 aren't you assuming the only automorphism of Q itself is the identity map ?, isn't that a pretty big assumption worthy of explanation ?
Yes, the identity is the only automorphism. Suppose that we have an automorphism f such that f(1) = a. Then f(1^n) = (f(1))^n = a^n therefore the only possibility is f(1) = 1 (it can’t be 0 because then f(x) = f(x*1) = f(x)*f(1) = 0 and f wouldn’t be bijective). From this you can see that f fixes all the whole numbers because if n is natural f(n) = f(1+1+1+…+1) = f(1)+f(1)+…+f(1) = n (and the same but with minus signs for the negative numbers). But any rational r is of the form r = q/p where q and p are whole numbers (p not being 0). Therefore f(r) = f(q)/f(p) = q/p = r. So the only automorphism is the identity.
Great lecture series! Deep understanding of this complex topic and nice visual examples. But... So many arithmetic errors... 25:55 alpha^4 should be "+18*2^(1/3)" not "-18*2^(1/3)" . Earlier : [254 =256-2 = 16^2-2 =] (16+2^0.5)(16-2^0.5) = (??) 194
Conjugate root theorem implies duality, roots come in pairs! Conjugate = Duality! "Sith lords come in pairs (duals)" -- Obi Wan Kenobi. "Always two there are" -- Yoda. SINE is dual to COSINE -- the word "co" means mutual and implies duality. SINH is dual to COSH -- hyperbolic functions. Injective is dual to surjective synthesizes bijective or isomorphism. Conjugate pairs = automorphism! Sub groups are dual to sub fields -- the Galois correspondence.
Very nice visual proof of the tower law!
I like that there are like 24k views and only 270 likes which means that this is like some of the most helpful content out there. I've got my final today LFG!
Very helpful video, thank you!
1/(sqrt(2) + sqrt(3)) did the trick for the last question. You can derive sqrt(2) and sqrt(3) easily with that primitive element.
Great video !!! Thanks.
Is Joe Brawley still at Clemson? I went to N. C. State with him in the 60s. A long time ago! We tried to learn Galois theory there from Nathan Jacobson's books. I learned it incompletely at U of Illinois. Thanks for your lessons. The theory is now revealed. Bravo.
May I give a small reminder that alpha^4 = 18*2^(1/3) and not -18*2^(1/3)
2 questions:
1) In the previous video at the very end, it determined that zeta*cube root 2 has minimal polynomial (x^3)-2 over Q and that [Q(zeta*cube root 2):Q] = 3. Here we find [Q(alpha):Q]=6. This feels contradictory to me.
2) In the final exercise, what is the minimal polynomial? alpha^4 should give us an element in Q, but I calculate it as 49 + 20*sqrt 6
Answer to question 1: Q(zeta, cbrt(2)) = Q(i*sqrt(3), cbrt(2)), but Q(zeta*cbrt(2)) does not need to equal Q(i*sqrt(3)*cbrt(2)).
Answer to question 2: a minimal polynomial to alpha is x^4-10x^2+1=0 (up to a constant factor). It doesn't have to be in the form of x^n. It just needs to be a minimal degree polynomial that alpha satisfies. It can be proven that the minimal degree polynomial must also be irreducible, and unique up to a constant factor (here over \Q, you can search online for a proof). How did I find this x^4-10x^2+1? Specific to this problem, I just wrote out four powers of alpha and played around with them. Not sure if a general method exists.
when you say the only automorphism of Q(sqrt(2)) is isomorohic to C2 aren't you assuming the only automorphism of Q itself is the identity map ?, isn't that a pretty big assumption worthy of explanation ?
Yes, the identity is the only automorphism. Suppose that we have an automorphism f such that f(1) = a. Then f(1^n) = (f(1))^n = a^n therefore the only possibility is f(1) = 1 (it can’t be 0 because then f(x) = f(x*1) = f(x)*f(1) = 0 and f wouldn’t be bijective). From this you can see that f fixes all the whole numbers because if n is natural f(n) = f(1+1+1+…+1) = f(1)+f(1)+…+f(1) = n (and the same but with minus signs for the negative numbers). But any rational r is of the form r = q/p where q and p are whole numbers (p not being 0). Therefore f(r) = f(q)/f(p) = q/p = r. So the only automorphism is the identity.
Great lecture series! Deep understanding of this complex topic and nice visual examples.
But... So many arithmetic errors... 25:55 alpha^4 should be "+18*2^(1/3)" not "-18*2^(1/3)" .
Earlier : [254 =256-2 = 16^2-2 =] (16+2^0.5)(16-2^0.5) = (??) 194
Conjugate root theorem implies duality, roots come in pairs!
Conjugate = Duality!
"Sith lords come in pairs (duals)" -- Obi Wan Kenobi.
"Always two there are" -- Yoda.
SINE is dual to COSINE -- the word "co" means mutual and implies duality.
SINH is dual to COSH -- hyperbolic functions.
Injective is dual to surjective synthesizes bijective or isomorphism.
Conjugate pairs = automorphism!
Sub groups are dual to sub fields -- the Galois correspondence.
😭