Visual Group Theory, Lecture 6.2: Field automorphisms

Visual Group Theory, Lecture 6.2: Field automorphisms

A field automorphism is a structure preserving map from a field F to itself. This means that it must be both a homomorphism of both the addtive group (F,+) and the multiplicative group (F-{0},*). We show that any automorphism of an extension of the rationals Q must fix Q element-wise. The set of automorphisms of a field F forms a group called the Galois group, denoted Gal(F). We see several examples of this, and look into an intriguing connection between the subfield lattice of F and the subgroup lattice of Gal(F).

[CatherineBerrouet]

This is by FAR the cleanest and concise explanation I've ever seen explaining Galois theory. THANK YOU XMILLION

[sil]

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.

[VietTran-duip]

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.

[NielsPetterLiset]

This is a very good introduction to field automorphisms and (start of) Galois theory! Thank you for making this video and making it public!

[redlander]

When you construct Q(root 2) you use the fact that root 2 is irrational, so it's too late to question this now :))

[brownarjun]

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.

[MathProofsable]

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.

[rasraster]

I do not recall an earlier lecture where you covered subgroup lattices. Which one is that, please?

[fsaldan]

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.

[josenaeliton]

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.

[adykaadyka]

Very good and helpful video as well as explanation, thanks for this! :)

[shacharh]

@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 ?

[sabrinapark]

Thanks so much!! These videos are so helpful

[LibertyAzad]

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?

[dmj]

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.

[kaydenlimpert]

according to this video, -1 times -1 equals -1, and that would mean a negative number times another negative number equals a negative number, how could that make sense 5:42

[howmathematicianscreatemat]

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.

[mymath]

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

[justingreenough]

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.

[prfontaine]

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.