You’re right that’s it’s not obvious via the definition of linear independence. This is where the field theory and minimal polynomials come in handy. Since x^3-2 is the minimal polynomial of 2^(1/3) over Q, the extension K=Q(2^(1/3)) over Q is degree 3, meaning Q(2^(1/3)) is a 3 dimensional vector space over Q. The minimal polynomial of 3^(1/3) over K=Q(2^(1/3)) is x^3-3, so the extension L=Q(3^(1/3), 2^(1/3)) over K=Q(2^(1/3)) is also degree 3. We’ve just factored the extension L over Q, its degree is 3*3=9. So this tells us this spanning set is maximal, so it’s a basis, thus linearly independent over Q.
Oh my god you mentioned Brahmagupta! Finally professors have stopped teaching that the Greeks and Persians came up with everything! Mathematics has had such a rich history and india was a big part of it, its so sad that this history has been almost totally omitted up until relatively recently. Lots of love, Gaurahari Das
Wow, that was awesome. The theorem is beautiful, I wonder how did I not know about it before! Also, those animation in Manim are so great. I also use it for my videos, so could I kindly ask you for some sources you used to learn Manim? Or did you do it completely on your own? Thanks :)
Thank you! Sure here are some Manim resources I typically refer to: youtube.com/@behackl?si=1jizEU32Gg7FwM2B docs.manim.community/en/stable/examples.html www.devtaoism.com
5:55 You don't need to check whether powers with natutal exponents give new elements. Just the basic multiplication table from a few seconds earlier and a *very* short induction argument. The reciprocals are there because this is a field, after all (if you're being pedantic you have to show that this field exists at all first), and then a very similar induction covers the rest of the negative integer exponents. Completely unrelated: It is very often the case that K(a, b)=K(a+b), but not always. It is the case whenever [K(a,b):K]=[K(a):K][K(b):K]. And the simplest counterexample I've seen is an extension of K(s, t) with K having order p for some odd prime p, and s, t being transcendental. If a is a root of x^(p-1)-s and b is a root of x^p-sx-t, then K(a, b) has degree p(p-1) but K(a+b) only has degree p, as it is a root of x^p-sx-t. K(a, b) is still simple, but a+b isn't a generator.
Related to the unrelated remark: If L|K is any finite field extension of degree d, let's say with K-linear basis x_1,...,x_d, then an element u=a_1x_1+...a_dx_d is primitive for L if and only if 1,u,...,u^{d-1} are linearly independent. But this is the same as saying that their coefficient vectors wrt x_1,...,x_d are linearly independent, i.e. that the matrix formed by them has nonzero determinant. But now, this determinant is just a (degree d*(d-1)) polynomial in the variables a_1,...,a_d. Now, from analysis/algebraic geometry, we know that this polynomial is either identical to zero (which means that no primitive element exists) or the set of roots is "small" (a zero set). This second case however implies that "most" elements in the field extension are actually primitive (but writing down such a random element would require all powers, so it is probably not very efficient)
Super cool video! Seeing (gamma = alpha + beta) at the end was a bit jarring though: I think a common reaction to this result would inspire the question "is that just always true?". A quick statement like "in this case, it happens to be..." would help contextualize that answer since the result is presented almost as trivial without explanation. @MasterHigure has a nice comment below that added some important information to this point, though I cannot understand the notation 🙂
Yeah the proof of the theorem shows almost all expressions alpha+t*beta work, but I didn’t want to get into that in the video, rather just demonstrate a particular element works.
So what does the theorem say? If you add any finite number of extensions, that you can always represent then with a single element extension that is often just their sum?
Ok I got it... In field theory, the primitive element theorem states that every finite separable field extension is simple, i.e. generated by a single element. This theorem implies in particular that all algebraic number fields over the rational numbers, and all extensions in which both fields are finite, are simple.
The theorem statement is that all finite, separable extensions are simple. The statement of the theorem does not give a recipe for how to construct a primitive element. The proof of the theorem gives such a recipe. planetmath.org/proofofprimitiveelementtheorem
Good video. I had no audio issues (saying that because i see other people saying the video is quiet. Listening with headphones, have the volume 3/4 of max and can hear quite clearly.)
@@DrMcCrady grab a compressor, dial the threshold down to the volume of when your voice is quietest, and then dial the ratio up to decrease the distance between the highest volume and lowest volume parts. Adjust attack to be faster if you feel your voice is popping to loudly initially, and increase your release on the compressor based on how “smooth” you want the audio to sound (but bear in mind, it can sound unnatural and like an overall flat volume decrease if the release is too low). Throw a de-easer on top and pick out the 3k+ range to cut. Those two things generally are enough.
Here's a fun problem. Let a,b,c be formal variables. Let L be the field Q(a,b,c), and let K be the field Q(a+b+c,ab+bc+ca,abc). Clearly K is a subset of L. (In fact it is precisely the set of _symmetric_ rational functions in a,b,c.) The problem is to prove that L=K(a-b). Note that this is an instance of the primitive root theorem, in the sense that the theorem guarantees us that L=K(gamma) for _some_ gamma in L.
Not necessarily, for example if you adjoined sqrt(p) to the rationals for each prime p, then that extension would not have a primitive element. Does that help?
the proof is what's crazy (imo), because it turns out almost all gamma = alpha+x*beta can do the job. But I don't remember if there is some clever choice of x that will always work.
I have a math major and I was lost within 75 seconds :( what is Q? What is Q(i)? Heck I don't remember if R stands for "reals" or something else! Maybe I'm not your target audience :(
ℝ is the real numbers, ℚ the rationals, ℂ the complex numbers, and ℚ(i) (for example) consists of the rational numbers adjoint with i, so you just add i to the rational numbers, and make it closed under addition. This is now a new field.
Oh no! You got R right as the reals! I can see how abstract algebra could be a forgettable topic, and how the symbols could be vague if you haven’t seen them or it’s a been a while. Thank you for your comment.
Cannot believe your clickbait title worked. Super nice video
Me neither, thank you!
6:38 It’s clear to me that the elements shown here span the field, but how do we know that they are Q-linearly independent?
You’re right that’s it’s not obvious via the definition of linear independence. This is where the field theory and minimal polynomials come in handy. Since x^3-2 is the minimal polynomial of 2^(1/3) over Q, the extension K=Q(2^(1/3)) over Q is degree 3, meaning Q(2^(1/3)) is a 3 dimensional vector space over Q. The minimal polynomial of 3^(1/3) over K=Q(2^(1/3)) is x^3-3, so the extension L=Q(3^(1/3), 2^(1/3)) over K=Q(2^(1/3)) is also degree 3. We’ve just factored the extension L over Q, its degree is 3*3=9. So this tells us this spanning set is maximal, so it’s a basis, thus linearly independent over Q.
@@DrMcCrady Thanks for the explanation!
Oh my god you mentioned Brahmagupta! Finally professors have stopped teaching that the Greeks and Persians came up with everything!
Mathematics has had such a rich history and india was a big part of it, its so sad that this history has been almost totally omitted up until relatively recently.
Lots of love,
Gaurahari Das
Great video, I took an abstract algebra class as a comp sci major, and loved learning more about this type of thing. Keep it up!
Thank you!
The clickbait situation is crazy, but awesome video 👍🏻
Thank you!
Wow, that was awesome. The theorem is beautiful, I wonder how did I not know about it before!
Also, those animation in Manim are so great. I also use it for my videos, so could I kindly ask you for some sources you used to learn Manim? Or did you do it completely on your own? Thanks :)
Thank you! Sure here are some Manim resources I typically refer to:
youtube.com/@behackl?si=1jizEU32Gg7FwM2B
docs.manim.community/en/stable/examples.html
www.devtaoism.com
@@DrMcCrady Thank you!)
5:55 You don't need to check whether powers with natutal exponents give new elements. Just the basic multiplication table from a few seconds earlier and a *very* short induction argument.
The reciprocals are there because this is a field, after all (if you're being pedantic you have to show that this field exists at all first), and then a very similar induction covers the rest of the negative integer exponents.
Completely unrelated: It is very often the case that K(a, b)=K(a+b), but not always. It is the case whenever [K(a,b):K]=[K(a):K][K(b):K]. And the simplest counterexample I've seen is an extension of K(s, t) with K having order p for some odd prime p, and s, t being transcendental. If a is a root of x^(p-1)-s and b is a root of x^p-sx-t, then K(a, b) has degree p(p-1) but K(a+b) only has degree p, as it is a root of x^p-sx-t. K(a, b) is still simple, but a+b isn't a generator.
Sounds very efficient, thank you.
Related to the unrelated remark: If L|K is any finite field extension of degree d, let's say with K-linear basis x_1,...,x_d, then an element u=a_1x_1+...a_dx_d is primitive for L if and only if 1,u,...,u^{d-1} are linearly independent. But this is the same as saying that their coefficient vectors wrt x_1,...,x_d are linearly independent, i.e. that the matrix formed by them has nonzero determinant. But now, this determinant is just a (degree d*(d-1)) polynomial in the variables a_1,...,a_d. Now, from analysis/algebraic geometry, we know that this polynomial is either identical to zero (which means that no primitive element exists) or the set of roots is "small" (a zero set). This second case however implies that "most" elements in the field extension are actually primitive (but writing down such a random element would require all powers, so it is probably not very efficient)
the stupid title was why i clicked on it btw, good vid
Welcome aboard, thank you!
What sort of software did you use for the visualizations for the video? 3B1B python library?
Yep.
love this video! I do hope the video is a bit louder though
Super cool video! Seeing (gamma = alpha + beta) at the end was a bit jarring though: I think a common reaction to this result would inspire the question "is that just always true?". A quick statement like "in this case, it happens to be..." would help contextualize that answer since the result is presented almost as trivial without explanation. @MasterHigure has a nice comment below that added some important information to this point, though I cannot understand the notation 🙂
Yeah the proof of the theorem shows almost all expressions alpha+t*beta work, but I didn’t want to get into that in the video, rather just demonstrate a particular element works.
This video has a volume that I can’t barely hear with hardware volume maxed out, and suddenly an ad showed up, That was ear blowing.😂
Sorry about that!
I can hear it fine bro
The volume is fine to me?
@@bunnyben5607 have the same issue though 🥲
I think something might be wrong with whatever you’re using for sound output. Volume is completely fine for me.
Pretty dope theorem
So what does the theorem say? If you add any finite number of extensions, that you can always represent then with a single element extension that is often just their sum?
Ok I got it... In field theory, the primitive element theorem states that every finite separable field extension is simple, i.e. generated by a single element. This theorem implies in particular that all algebraic number fields over the rational numbers, and all extensions in which both fields are finite, are simple.
The theorem statement is that all finite, separable extensions are simple. The statement of the theorem does not give a recipe for how to construct a primitive element. The proof of the theorem gives such a recipe. planetmath.org/proofofprimitiveelementtheorem
Extra credit: Give an explicit, understandable process for finding the primitive element from a list of algebraic extension generators.
Yeah that’s even crazier that there are only finitely many linear combos alpha+c*beta that aren’t primitive elements!
Thanks! You math Guys are great!
Thank you for your comment!
Good video. I had no audio issues (saying that because i see other people saying the video is quiet. Listening with headphones, have the volume 3/4 of max and can hear quite clearly.)
Thank you!
You need to do a little work on your audio editing skills. Other than that, great video!
Thanks, I’ll figure it out!
@@DrMcCrady grab a compressor, dial the threshold down to the volume of when your voice is quietest, and then dial the ratio up to decrease the distance between the highest volume and lowest volume parts. Adjust attack to be faster if you feel your voice is popping to loudly initially, and increase your release on the compressor based on how “smooth” you want the audio to sound (but bear in mind, it can sound unnatural and like an overall flat volume decrease if the release is too low).
Throw a de-easer on top and pick out the 3k+ range to cut.
Those two things generally are enough.
Very nice video!
Thank you!
Here's a fun problem.
Let a,b,c be formal variables. Let L be the field Q(a,b,c), and let K be the field
Q(a+b+c,ab+bc+ca,abc).
Clearly K is a subset of L. (In fact it is precisely the set of _symmetric_ rational functions in a,b,c.)
The problem is to prove that L=K(a-b).
Note that this is an instance of the primitive root theorem, in the sense that the theorem guarantees us that L=K(gamma) for _some_ gamma in L.
That does sound cool!
Would this also work for extending a field with a countable infinity of elements?
Not necessarily, for example if you adjoined sqrt(p) to the rationals for each prime p, then that extension would not have a primitive element. Does that help?
Great video :)
Thank you!
the proof is what's crazy (imo), because it turns out almost all gamma = alpha+x*beta can do the job. But I don't remember if there is some clever choice of x that will always work.
Oh yeah I agree. I’d been watching a lot of Penguinz0 and thought I’d try one of his titles 😁
I thought the title meant too many primitive elements are being discovered or sth🤣
commenting for the algorithm gods
Thank you for your support as I stand at their altar awaiting judgement.
Cool video!
Thank you!
Good video. A little rushed. Volume is too low!
2:30 It is del Ferro, not 'Farro'
But really good video, very well explained
Thank you!
the clickbait worked 😂
I can’t believe it either, but I’m happy about it.
Liked for AD
very deep result. Galois couldn't demonstrate it.
It’s interesting that he knew some extent of it.
moistcritical physics
Definitely the inspiration for the title. I Just started getting into his channel this year.
I have a math major and I was lost within 75 seconds :( what is Q? What is Q(i)? Heck I don't remember if R stands for "reals" or something else!
Maybe I'm not your target audience :(
ℝ is the real numbers, ℚ the rationals, ℂ the complex numbers, and ℚ(i) (for example) consists of the rational numbers adjoint with i, so you just add i to the rational numbers, and make it closed under addition. This is now a new field.
Oh no! You got R right as the reals! I can see how abstract algebra could be a forgettable topic, and how the symbols could be vague if you haven’t seen them or it’s a been a while. Thank you for your comment.
Pi is equal to 3
You’re so close!
You are talking too fast. I can't follow at that speed.
Playback speed -> 0.5 speed 😁