This is by FAR the cleanest and concise explanation I've ever seen explaining Galois theory. THANK YOU XMILLION
@sil1235
5 жыл бұрын
Slight mistake at about 6 minute, but q=q^2 cannot be for q=-1, and you showed q!=0, so q=1 follows. No need to consider q=q^3. Also the proof ends by showing phi(1)=1, but it should be mentioned why this shows that actually phi(q)=q.
@kyleellefsen
27 күн бұрын
I agree, the proof wasn't quite complete. Once phi(1)=1 is proven, then you can use the additive property of the homomorphism to prove that all the natural numbers are fixed. For all natural numbers n, phi(n) = phi(1+...+1) = phi(1) + ... + phi(1) = n phi(1) = n. Then any rational number q can be written as a/b for natural numbers a, b. Rearranging gives q*b = a. Then phi(qb) = phi(a) = a = phi(q) * phi(b) = phi(q) * b. Rearranging gives phi(q) = a/b = q, which completes the proof that phi fixes all rational numbers.
@VietTran-du4ip
8 жыл бұрын
I think you are one of the best professors who can explain these abstract concepts very clearly. Thank you Dr. Macauley. I really enjoy and learn a lot from your lecture videos.
@redlander55
5 жыл бұрын
When you construct Q(root 2) you use the fact that root 2 is irrational, so it's too late to question this now :))
@yanghwanlim4499
3 жыл бұрын
I agree! The professor is using circular logic which is logical false! You are quite sharp not falling into circular logic!
@Brien831
3 жыл бұрын
@@yanghwanlim4499 No, since the construction of Q(sqrt(2)) can be done by the fact, that X^2-2 is irreducible over Q, by the Eisenstein criterion. Now it is true, that you can follow, that sqrt(2) is irrational by the irreducibility of X^2-2 but this does not mean, that the proof given is invalid.
@NielsPetterLiset
Жыл бұрын
This is a very good introduction to field automorphisms and (start of) Galois theory! Thank you for making this video and making it public!
@MathProofsable
4 жыл бұрын
Great videos! A minor quibble I have is that I think it would be better to get students to be able to understand "morphism" of algebraic objects in terms of preservation of structure. This would mean that a field morphism is just a ring morphism (multiplicative inversion is not an algebraic structure but a property). In this case, the morphism must preserve addition, additive inverses, multiplication and the multiplicative and additive identities. Then you wouldn't need to prove q(1)=1. A better structural result is that all Field morphisms are monomorphisms. How a field is expressed as a subfield via different monomorphisms is of interest to Galois theory.
@brownarjun
7 жыл бұрын
Am a little bit skeptical of the proof on root 2 is irrational. By defining the automorphism as flipping elements which use root 2 in combination with the base field and then saying that rationales are fixed so root 2 is irrational as it's flipped is a bit tautological. Don't you need to prove from the start that root 2 isn't in the rationals so that the field extension is non trivial? Otherwise one could make the same claims on root 4 etc which of course is not true.
@JeanIbarz
7 жыл бұрын
Agree : I prooved that root 4 is not in the rational that way and that totally doesn't make sense !
@zhengkuang6025
6 жыл бұрын
Root 4 stays unmoved since you've moved both root 2 to -root 2, and their product is still root 4.
@antoniolewis1016
6 жыл бұрын
The proof was that phi(1) = 1, but from that you can say phi (integers) = integers (using the additive property of phi), and then from that you can say phi(fractions) = fractions (using the multiplication/division property of phi) so phi(q) = q for all rational numbers, no matter what that automorphism does to other numbers. What was done earlier with the root 2 example already assumed that root 2 is irrational, so this proof doesn't apply. You're right!
@sirfurness
5 жыл бұрын
If one makes a similar argument for root 4, it fails. Say you make an automorphism phi which takes root 4 to negative root 4 and fixes all other elements. So, phi(root 4) = -root 4, -root 4 = -2, therefore, phi(root 4) = -2. However, phi(root 4) = phi(2) = 2. Therefore, phi is not a bijection and is therefore not an automorphism.
@partialorder5596
4 жыл бұрын
Yeah, basically the "proof" relies on the assumption that phi is well-defined, which in turn relies on the assumption that sqrt(2) is not in Q. (Suppose sqrt(N) is in Q. Then we have multiple ways of writing each element in Q(sqrt(N)) in the format a*1+b*sqrt(N), where a & b are in Q. For example, (-N)*1+ sqrt(N)*sqrt(N) and 0*1+0*sqrt(N) are two ways of writing the element 0. Since phi ((-N)*1+ sqrt(N)*sqrt(N))=(-N)*1 - sqrt(N)*sqrt(N)= -2N and phi(0*1+0*sqrt(N))=0*1 - 0*sqrt(N))=0, phi is not well-defined.)
@josenaeliton5509
5 ай бұрын
In the "proof" that √2 is not a rational número using Galois theory you need to know that {1,√2} is a basis for Q(√2). Otherwise the automorphism isn't well defined. But it's the same as knowing that √2 isn't rational.
@fsaldan1
4 жыл бұрын
In the proof that phi(q) = q you already assume that phi(1) is rational. This is easy to prove for field extensions like Q(√2) that are generated by adding a finite number of irrational numbers to Q. I don't know if it can be proved for all field extensions.
@ChuanChihChou
Ай бұрын
Certainly a strange part of the lecture. phi(0) = 0 and phi(1) = 1 should be automatic given that the field automorphism is an automorphism of the additive group & multiplicative group & 0 & 1 are their identities respectively. Then you can build up the mapping of all q in Q from there.
@erds6579
6 жыл бұрын
Great lecture. Thanks for uploading.
@dmj94044
Жыл бұрын
I got lost for a while when you show that Q adjoin (cube root of 2) takes three terms (alpha, beta, and gamma). I guess this is because if there were only an alpha and beta term, then (cube root of two) squared would need to be in the field in order for the field to be closed under multiplication. But cube root of 4 would not be in that field (you can't construct cube root of four by a rational number times the cube root of two)--so you need the three terms.
@Gipsy4u
7 жыл бұрын
Thanks a lot, very helpful videos.
@adykaadyka
8 жыл бұрын
Very good and helpful video as well as explanation, thanks for this! :)
@sabrinapark8473
5 жыл бұрын
Thanks so much!! These videos are so helpful
@dashnarayana
Жыл бұрын
Very enlighting
@rasraster
7 жыл бұрын
I do not recall an earlier lecture where you covered subgroup lattices. Which one is that, please?
@shacharh5470
6 жыл бұрын
@6:33 that's not the end of the proof.. You showed that phi(1) = 1, how does it follow that phi(q) = q for all q in Q ?
@Hyebze
6 жыл бұрын
I was confused by this too :(
@srenladegaardkristensen1076
5 жыл бұрын
Shachar H No, that’s not the end of the proof. But since phi(1)=1, phi must also fix all intergers. And since phi(1/p)=1/phi(p) for any nonzero integer p (infact for any nonzero element), phi must also fix the rationals.
@chetedoherty
5 жыл бұрын
I came here to ask the same question but I've just figured it out..... phi(2) = phi(1 + 1) = phi(1) + phi(1) = 1 + 1 = 2 and by induction phi(z) = z (z an integer). Remains only to show that phi(p/q) = phi(p)/phi(q). Seems likely, but not sure if it follows directly from the definition of field automorphism.....
@willjohnston2959
2 жыл бұрын
No wonder he "forgot" to put the end of proof box there. :-)
@sambhusharma1436
Жыл бұрын
Thanku thanku so much 🙏
@lucasnobre6255
4 жыл бұрын
Boa boa macaulei calquin tu é pica dms
@howmathematicianscreatemat9226
4 жыл бұрын
I am sorry to intervene with your wonderful lectures but I don't think it's clear for many students why it was proofed that automorphisms fix rational numbers. The proof just showed that the neutral element of multiplication in a field is fixed under automorphisms. Not everybody of your students can conclude out of this that this already means that it fixes ANY rational number. For example the implication phi(1)=1 => phi(257/7)=(257/7) isn't clear for every watcher of your video.
@dabenxiongusa
Жыл бұрын
The professor skipped the part he thinks obvious. Hope this can help. assuming a is integer, it should be straightforward to prove that phi(a) = a as a can be treated as sum of a number of one; then considering phi(1) = phi(a / a) = phi(a) * phi(1/a) = a* phi(1/a) = 1, the phi(1/a) = 1/a.
@LibertyAzad
6 жыл бұрын
I thought that that the domain of the automorphism if a Galois Group is the field OF WHICH F is an extension, not necessarily the entire extension field itself. Or is there more than one version of the theorem perhaps?
@prfontaine5387
3 жыл бұрын
I think that some parts of the presentation are quite misleading. One can believe that Q(v) is always equal to {1.a + b.v | a,b in Q}. This is not the case in general. The true definition of Q(v) is {P(v) | P[X] is in k[X] } i.e. it is the set of all values taken by all polynoms with factors in Q and evaluated on value v. This set is a Q-vector space of finite dimension if v is algebraic over Q. The algebricity of v is a necessary condition for Q(v) = {P(v) | P[X] is in k[X] } to be a finite Q-vs. Consider Q(2^(1/3)) given as an example. This set is also a finite Q-vector space because 2^(1/3) is algebraic over Q (as beeing a root of X^3 -2 ) ; its basis is not {1,2^(1/3)} but {1, 2^(1/3), 4^(1/3) } . This can be proven as so. Take a value y in Q(2^(1/3)) it is P(2^(1/3)) for some P in Q[X]. Consider a monom a_k.X^k of P. With the euclidian division by 3, k = 3.q + r where r < 3. So a_k.X^k is equal to a_k.(X^3)^q + X^r . Evaluated on 2^(1/3) this latter monom is equal to: a_k.((2^(1/3))^3)^q +(2^(1/3)) ^r = a_k.2^q . (2^(1/3)) ^r. Yet a_k.2^q is in Q. As r is 0,1 or 2, we find that this monom is a product of an element of Q with 1, 2^(1/3) or (2^(1/3))^2 = 4^(1/3). That's it.
@markborz7000
7 ай бұрын
Why so complicated? The simple argument is that the extended field must be closed under miltiplication like any field. That's where this cubed sqrt(4) comes from. The same argument with Zetas. For minimal extension take one Zeta root and than complete. You get 6 basis elements for the extended field. Check then that this extended field is indeed complete by multiplication of all the 6 possible basis elements with each other and themselves. This must be the minimal completion because you started with just the three roots, while one root can be expressed by the other two and so can be "killed".
@travelgalaxy8291
5 жыл бұрын
Very Nice
@mymath1313
Жыл бұрын
Hey could someone explain the proof towards the beginning : If phi is an automorphism of an extension field then phi fixes q for every q being a rational number
@karthikmaths4175
5 жыл бұрын
Nice video
@mathiasbjerregard445
4 жыл бұрын
SPLENDID
@raybroomall8383
3 жыл бұрын
@ 2:30 ф(a+b sqrt(2)) + (c+d sqrt(2) cannot be equal to ф(a+b sqrt(2)) +
@justingreenough4296
3 жыл бұрын
Pardon, but the proposition on slide 3 was not proved. You proved that, if \phi is a field automorphism of an extension, then \phi(1) = 1. But that follows trivially from the assumption that \phi is automorphism.
@rasraster
7 жыл бұрын
Why does cube root of 4 show up in the basis, when you can easily generate it algebraically from cube root of 2?
@antoniolewis1016
6 жыл бұрын
Because the coefficients have to be rational
@djkroeger
2 жыл бұрын
I think it's because that equation is generated by using Z1, Z2, Z3, and then combining like terms . The equation is formed from all of the roots, plus all of the multiplication permutations of all of the roots, it seems. for example ∛(4) = Z1*Z3 (because zeta^3 = 1) (i.e. making all possible unique combinations of zeta and ∛(2); remembering zeta^2 = -zeta -1 ∈ [Q, zeta] so not unique; and (∛(2))^3 = 2 ∈ [Q] so also not unique)
@shalvagang951
2 жыл бұрын
i dont understand that at 12 :01 what is that v4 can someome explain me pls
@shrutianand2259
3 жыл бұрын
Translation in English in written disturbing to watch the concept plz write in bottom not overlap plz
@samtux762
7 жыл бұрын
Square root of minus one is like tobacco. Allowed to adults, but not to kids.
@bonbonpony
6 жыл бұрын
Not true. I teach complex numbers to kids and they get it.
@roshanshihab8515
Жыл бұрын
Still didn't understand why ζ² wasn't included in the basis of Q(ζ,∛2) Edit: okay I got it. So basically, ζ² = -(ζ+1)
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