I must deeply thank you for this!! This is really what I have missed so much throughout my studying for Bachelor-mathematics!! You help us so much and really enable us to understand these thrilling things intuitively! Thanks!! :-))
this is by far the Best introduction to Fields, & group theory as a whole, with many Takeaways to learn from Thank you so much Professor, Field F Abelian Group F \{0} (without 0 ) closed under multiplication 3. Distributive law holds: a*(b+c) = a*b +a*c Field {is a Ring} where you can do basic Arithmetic sets; not fields: N Z, Zn (n composite) Fields: Q R C , Zp (prime) Extension coeffs of linear equationa a__+b___ are the Extension Q(,) i.e. a*sqrt(2) + b* 19:00 supporse have a Vector Space, Spanned by V1, V2, V3, V4 any element in little x written as a*v1 + bv2 cv3 + dv4 which forms a 4D vector space like in group & subgroup, we arrange Fields & subfields.. graph: label each extension as Degree Q. Abelian Group does not have Division (/) [only in Hilbert's] it has only + * - } as 3 operations , 1-to-1 & onto , correct?
Excellent. So nice and explicitly was presented. If the thump up did not have a limit, I would have stay all day and all night long up to click and click on it and still not enough. Thank you
The format and methods along with the sound explanation of these lectures will work well for an Online degree program in applied and computational mathematics. Thank you for your hard work and sharing it online.
As I recall, a field are given a set F with any to arbitrary binary operators + and * (they are just symbols representing some binary operator), with e_+, e_* are identity elements satisfying 1. (F,+) is an abelian group and 2. (F\{e_+},*) is an abelian group. 3. For a,b,c in F, a*(b+c) = a*b+a*c.
mistake on 16^2...nevertheless it doesn't take anything away from your lesson. Very helpful..been trying to understand field extensions and no one made it as clear as you did. Thanks!
Hey friend. It's probably a not insignificant amount of work to do so, but could you organize your videos into playlists? I like your visual group theory lectures and I would like to be able to binge watch them easily or find them quickly by saving the playlist. It would be a really helpful study tool, as well as make sharing an entire series of lectures with friends easier.
Some parts of the presentation are factually correct but potentially misleading I think. When one adds (adjoins) an element x to a field k, forming k(x),we obtain not only {a + b.x | a,b E k } but the bigger set {P(x) | P E k[X] } where k[X] denotes the set of all polynoms of variable X with factors in k. Notice that k(x) is not a field in general (this should be said explicitly). On revenge what can be proved is that is that k(x) is a vector space over k of finite dimension if x is algebraic over k. This is why your examples work: you always adjoin to Q values that are algebraic over Q and this is why you can obtain a finite basis for the vector space k(x). I think that the definition of k(x) should be separeted from the extension field notion as k(x) is not always a field.
Thank you! My text book does not explain "adjoin" anywhere near the way you did - perfect! To me it seems like many authors of textbooks dont write for their audience - they write them like we already have a phd in math!
Ah I understand the link between the last two questions: Yeah, it has structure V_4; not just a subgroup lattice V_4. The second last qn.'s answer has that adjoining x^2-3 or x^2-2 works to make Q(sqrt(6)) into Q(sqrt(2),sqrt(3)), but moreover, as in V_4, you can adjoin x^2-6 to either Q(sqrt(2)) or Q(sqrt(3)) to make either Q(sqrt(2),sqrt(3)), similar to V_4.
Q seems larger than we need... In a naive sense that does not respect the requirements to create a field a subset of Q with only all prime numbers would be enough to generate the a and b parameters, right? This is probably obvious, but reason I am asking is because it seems that we only need square roots of prime numbers too, which if correct makes things a bit more interesting.
why is the basis of Q(sqrt(2),sqrt(3)) equal to {1,sqrt(2),sqrt(3),sqrt(6)}? Isn't sqrt(6) a linear combination of sqrt(2) and sqrt(3) under multiplication? Ie.: sqrt(2)sqrt(3)=sqrt(6) therefor sqrt(6) should not be a basis?
We're only working with _rationals_ as scalars in this example. All four of those numbers are linearly dependent if we are allowed to multiply by reals.
This used to confuse me also. I had to always remind myself that there is no multiplication between elements in a vector space. It was confusing to me because Q(sqrt(2),sqrt(3)) is also an algebra over Q.
@@MathProofsable you're right, this is algebra over Q so sqrt(6) can only be constructed from Q(sqrt(2)) extended with sqrt(3) or Q(sqrt(3)) with sqrt(2)
These splitting field extensions of Q are at the same time vector spaces over Q, and under the same addition as in Q. Hence, neither we can apply multiplication from the field as addition in vector space, nor can scalars be irrational. IOW, no way to get sqrt(6) from 1, sqrt(2) and sqrt(3) as linear combination in that vector space.
Good introduction, definitions of terminology used. The rest of the presentation exercise is an example of (pure-math relative-timing ratio-rates motion and tangency rates), learning by doing-> Actual Intelligence Intuition. If the First Principle Observation of WYSIWYG QM-TIME resonance is just such interrelated definitions of terminology, floating superposition identification above the basic elements of 1-0 e-Pi-i sync-duration probability of integrated coherence-cohesion resonances.., ie quantization terminology used, then there's the "Measurement Problem" inherent in definitions other than self-defining elemental e-Pi-i sync-duration relative-timing ratio-rates in Perspective Principle.., and this is the Singularity-Duality proof-disproof situation in "solid" inclusive focus on the 1-0 axis of continuous creation connection Origin, and the Quantum Operator Fields Modulation Mechanism tangency-dualistic density-intensity real-numberness condensation spectrum of pure-math relative-timing rates. Fluxion-Integral Temporal superposition->numerical foci, Calculus. Fields of n‐² and i are incident, ie line-of-sight, and reverse direction functional process in sequence.., putting it all in pure-math perspective.
Many. You can adjoin sqrt(2), or sqrt(3), or 2*sqrt(2), or 2*sqrt(3), etc... Note that if you have two of sqrt(2), sqrt(3), sqrt(6), you have all three of them.
I think it's even clearer when you look at it from a "group" perspective, and you're pretty much giving it away by saying it's the Klein group lattice. Indeed, V4 has three element plus the identity (h, v and hv) and you only need two to generate the third (for example hv*v = h). The same way, if I already have sqrt(2) and sqrt(6) in my field, then by multiplying them I get sqrt(12)=2*sqrt(3), see how sqrt(2)*sqrt(2) vanish because it's order 2, exactly like V4. Same thing if you have sqrt(3) and sqrt(6).
Is it cheating to say the answer to the last problem is h(x)/(x^2 - 6) ? Algebraically it appears correct, but I have a gut feeling it’s not the most elegant/simple answer
If anyone finds this and wants to chime in: I’ve been able to work it out to (x - sqrt(6))/(x + sqrt(6)), but I am unsure how to proceed to the final step
This was far from trivial, at least not in my hands. I had to resort to long polynomial division to get the answer. First, (long) division of (x^4 - 5x^2 + 6)/(x^2 - 6) yielded x^2 + 1 plus a residual term of 12/(x^2 - 6). Confirming by multiplying back (x^2 - 6) with (x^2 + 1) + 12/(x^2 - 6) gave x^4 - 5x^2 - 6 + 12 = x^4 - 5x^2 + 6. So the answer is x^2 + 1+ 12/(x^2 - 6)... maybe there is a simpler way to write that...
New time traveller fantasy: Go back in time and snipe Galois' opponent of the duel to let him live longer. What would happen to the field (ha!) of abstract algebra? 🤔
I think so, looks like @Matthew Macauley and @Baptiste Portenard above explained problem. In general field needs to be extended with anything what change a+b sqrt(6) into a+b sqrt(2)+c sqrt(3)+d sqrt(6) and sqrt(1/2) looks like homomorphic with sqrt(2)
What kind of life those people must be living? Solving equations and fighting duels..? Surely absence of modern technology, social media drove people to extreme research at the same time extreme games.
Wait you've got more to cover in group theory first! Commutators and derived group, composition series, Jordan-Holder thm, solvable groups.. why did you skip all that?
I was having problem in understanding this concept, but after watching this lecture, I truly appreciate your effort.
Best channel for Galois theory. Thank God I found this!
I must deeply thank you for this!! This is really what I have missed so much throughout my studying for Bachelor-mathematics!! You help us so much and really enable us to understand these thrilling things intuitively! Thanks!! :-))
I’ve been pushing and pulling my brain through a book in Galois Theory by Emile Artin. Your video is so much more concise and helps a lot. Thanks.
I have to say your lectures are very clear. This is greatly appreciated.
This video is very helpful! Thank you for being very detailed and the transition from the simplest to the more advanced examples is very enlightening
this is by far the Best introduction to Fields, & group theory as a whole, with many Takeaways to learn from
Thank you so much Professor,
Field
F Abelian Group
F \{0} (without 0 ) closed under multiplication
3. Distributive law holds: a*(b+c) = a*b +a*c
Field {is a Ring} where you can do basic Arithmetic
sets; not fields: N Z, Zn (n composite)
Fields: Q R C , Zp (prime)
Extension
coeffs of linear equationa a__+b___ are the Extension Q(,)
i.e. a*sqrt(2) + b*
19:00 supporse have a Vector Space, Spanned by
V1, V2, V3, V4
any element in little x written as
a*v1 + bv2 cv3 + dv4
which forms a 4D vector space
like in group & subgroup, we arrange Fields & subfields..
graph: label each extension as Degree
Q. Abelian Group does not have Division (/) [only in Hilbert's]
it has only + * - } as 3 operations , 1-to-1 & onto , correct?
Excellent. So nice and explicitly was presented. If the thump up did not have a limit, I would have stay all day and all night long up to click and click on it and still not enough. Thank you
Oh man, fantastic course. It all starts coming together.
For the homework question near the end, I believe you can adjoin the roots of either x^2-3 or x^2-2 and get the same field extension.
The format and methods along with the sound explanation of these lectures will work well for an Online degree program in applied and computational mathematics. Thank you for your hard work and sharing it online.
Honestly, the first field theory lectures I came across that helped me get good intuitive understnading
Thank you Professor. Please post more. These are great to review concepts.
I love your explanation! You did it good! Keep up the good work! Hope you make more videos about visual group theory!
Really awesome.Nicely explained.Thank you sir
As I recall, a field are given a set F with any to arbitrary binary operators + and * (they are just symbols representing some binary operator), with e_+, e_* are identity elements satisfying
1. (F,+) is an abelian group and
2. (F\{e_+},*) is an abelian group.
3. For a,b,c in F, a*(b+c) = a*b+a*c.
Thank you sooo much, you saved a life in Turkey with sharing this video!!!
Awsome! Thankyou! I celebrate the existence of this chanel
mistake on 16^2...nevertheless it doesn't take anything away from your lesson. Very helpful..been trying to understand field extensions and no one made it as clear as you did. Thanks!
10:42 256 - 2 = 254
Many thanks. Very clear.
@1:26 It was a DUEL, not a "dual".
OMG! It's mindblowing!
Hey friend. It's probably a not insignificant amount of work to do so, but could you organize your videos into playlists? I like your visual group theory lectures and I would like to be able to binge watch them easily or find them quickly by saving the playlist. It would be a really helpful study tool, as well as make sharing an entire series of lectures with friends easier.
Nice
Some parts of the presentation are factually correct but potentially misleading I think. When one adds (adjoins) an element x to a field k, forming k(x),we obtain not only {a + b.x | a,b E k } but the bigger set {P(x) | P E k[X] } where k[X] denotes the set of all polynoms of variable X with factors in k. Notice that k(x) is not a field in general (this should be said explicitly). On revenge what can be proved is that is that k(x) is a vector space over k of finite dimension if x is algebraic over k. This is why your examples work: you always adjoin to Q values that are algebraic over Q and this is why you can obtain a finite basis for the vector space k(x).
I think that the definition of k(x) should be separeted from the extension field notion as k(x) is not always a field.
You are a really good teacher kudos
Great lecture, thank you so much!
great explanation
Brilliant explanation!
Thank you! My text book does not explain "adjoin" anywhere near the way you did - perfect! To me it seems like many authors of textbooks dont write for their audience - they write them like we already have a phd in math!
great video. Very good motivation of the subject.
Very informative
Explained in a great manner.
Very clearly put👌👍except for the little error 16 square
16 squared is 256. And 14 squared is 196.
Numbers were incorrect.
Still I prefer the theory to be correct than the other way round.
Yeah mistakes involving powers of 2 really stand out for people with CS background 😅
Very clear explanation, thank you
Ah I understand the link between the last two questions:
Yeah, it has structure V_4; not just a subgroup lattice V_4. The second last qn.'s answer has that adjoining x^2-3 or x^2-2 works to make Q(sqrt(6)) into Q(sqrt(2),sqrt(3)), but moreover, as in V_4, you can adjoin x^2-6 to either Q(sqrt(2)) or Q(sqrt(3)) to make either Q(sqrt(2),sqrt(3)), similar to V_4.
This was good sir! thank u
Explained very nicely
great explanation thank you.
25:19 I think either one will work. namly, the roots of the polynomial x^2-3 or x^2-2.
this helps me so much! thank you
At 16:43 I think you meant to say a + b(square root of 2) not a + bi.
Both are correct.
Q seems larger than we need... In a naive sense that does not respect the requirements to create a field a subset of Q with only all prime numbers would be enough to generate the a and b parameters, right? This is probably obvious, but reason I am asking is because it seems that we only need square roots of prime numbers too, which if correct makes things a bit more interesting.
10:55 proof?
why is the basis of Q(sqrt(2),sqrt(3)) equal to {1,sqrt(2),sqrt(3),sqrt(6)}?
Isn't sqrt(6) a linear combination of sqrt(2) and sqrt(3) under multiplication?
Ie.: sqrt(2)sqrt(3)=sqrt(6) therefor sqrt(6) should not be a basis?
We're only working with _rationals_ as scalars in this example. All four of those numbers are linearly dependent if we are allowed to multiply by reals.
This used to confuse me also. I had to always remind myself that there is no multiplication between elements in a vector space. It was confusing to me because Q(sqrt(2),sqrt(3)) is also an algebra over Q.
@@MathProofsable you're right, this is algebra over Q so sqrt(6) can only be constructed from Q(sqrt(2)) extended with sqrt(3) or Q(sqrt(3)) with sqrt(2)
These splitting field extensions of Q are at the same time vector spaces over Q, and under the same addition as in Q. Hence, neither we can apply multiplication from the field as addition in vector space, nor can scalars be irrational. IOW, no way to get sqrt(6) from 1, sqrt(2) and sqrt(3) as linear combination in that vector space.
10:45 194 ??
16*16 - 4 = 252 :) а не 194 ... с арифметикой проблемы, а в поля Галуа полез, бедный.
Good introduction, definitions of terminology used. The rest of the presentation exercise is an example of (pure-math relative-timing ratio-rates motion and tangency rates), learning by doing-> Actual Intelligence Intuition.
If the First Principle Observation of WYSIWYG QM-TIME resonance is just such interrelated definitions of terminology, floating superposition identification above the basic elements of 1-0 e-Pi-i sync-duration probability of integrated coherence-cohesion resonances.., ie quantization terminology used, then there's the "Measurement Problem" inherent in definitions other than self-defining elemental e-Pi-i sync-duration relative-timing ratio-rates in Perspective Principle.., and this is the Singularity-Duality proof-disproof situation in "solid" inclusive focus on the 1-0 axis of continuous creation connection Origin, and the Quantum Operator Fields Modulation Mechanism tangency-dualistic density-intensity real-numberness condensation spectrum of pure-math relative-timing rates. Fluxion-Integral Temporal superposition->numerical foci, Calculus.
Fields of n‐² and i are incident, ie line-of-sight, and reverse direction functional process in sequence.., putting it all in pure-math perspective.
Thank you!
So which polynomial extends the field Q(root 6) to Q(root 2, root 3)?
Many. You can adjoin sqrt(2), or sqrt(3), or 2*sqrt(2), or 2*sqrt(3), etc...
Note that if you have two of sqrt(2), sqrt(3), sqrt(6), you have all three of them.
I think it's even clearer when you look at it from a "group" perspective, and you're pretty much giving it away by saying it's the Klein group lattice.
Indeed, V4 has three element plus the identity (h, v and hv) and you only need two to generate the third (for example hv*v = h).
The same way, if I already have sqrt(2) and sqrt(6) in my field, then by multiplying them I get sqrt(12)=2*sqrt(3), see how sqrt(2)*sqrt(2) vanish because it's order 2, exactly like V4.
Same thing if you have sqrt(3) and sqrt(6).
@@baptisteportenard3246 Great explanation. Thank you.
Thank you
Thanks
Is it cheating to say the answer to the last problem is h(x)/(x^2 - 6) ? Algebraically it appears correct, but I have a gut feeling it’s not the most elegant/simple answer
If anyone finds this and wants to chime in:
I’ve been able to work it out to (x - sqrt(6))/(x + sqrt(6)), but I am unsure how to proceed to the final step
This was far from trivial, at least not in my hands. I had to resort to long polynomial division to get the answer. First, (long) division of (x^4 - 5x^2 + 6)/(x^2 - 6) yielded x^2 + 1 plus a residual term of 12/(x^2 - 6). Confirming by multiplying back (x^2 - 6) with (x^2 + 1) + 12/(x^2 - 6) gave x^4 - 5x^2 - 6 + 12 = x^4 - 5x^2 + 6. So the answer is x^2 + 1+ 12/(x^2 - 6)... maybe there is a simpler way to write that...
23;54 LIKE DID WITH DID .... spelling mistake
Z2*Z2?
New time traveller fantasy: Go back in time and snipe Galois' opponent of the duel to let him live longer. What would happen to the field (ha!) of abstract algebra? 🤔
Video stuck at 15 seconds. This did not happen with previous lectures (RUclips app Android)
❤️❤️❤️
uttam nice
nice
Q(sqrt (1/2))(sqrt(6)) is the answer?
I think so, looks like @Matthew Macauley and @Baptiste Portenard above explained problem. In general field needs to be extended with anything what change a+b sqrt(6) into a+b sqrt(2)+c sqrt(3)+d sqrt(6) and sqrt(1/2) looks like homomorphic with sqrt(2)
"duel" not "dual"
👌
What kind of life those people must be living? Solving equations and fighting duels..? Surely absence of modern technology, social media drove people to extreme research at the same time extreme games.
was saying same thing watching a movie about Jesus's Crucifixion. like wtf
16 x 16 = 256
duel* 🤦🏻♂️
Wait you've got more to cover in group theory first! Commutators and derived group, composition series, Jordan-Holder thm, solvable groups.. why did you skip all that?
your history is wrong
I need alcohol. But I dont drink because of the Muslim cast at work. Whiskey and beer. Beer filled my fridge at one point. Now I'm stuck.
This lecture is a bs. First non trivial question arises and the guy says I leave to you as homework :)