no he really meant commutes because you can do a*b then conjugate conj(a*b) and it ends up being the same as conjugating and then multiplying conj(a)*conj(b)
Could you should start a series of lessons in mathematics please (such as:Lifting the exponent lemma; ord and primitive Roots;ect. It would be of great help
I have a whole primitive roots playlist: ruclips.net/p/PL22w63XsKjqx1bbzG_-BYd0P6Gbe__lHr In fact, I have over 100 videos covering a full course in number theory: ruclips.net/p/PL22w63XsKjqwAgBzVFVqZNMcVKpOOAA7c
I think the domain of N should be D\{0} instead of D. Because the degree of zero polynomial (i.e. f(x) = 0) SHOULD BE minus infinity. But in your definition, it can't be smaller than zero, which is not good.
Your last video has had me stumped on trying to find a direct method towards computing a solution that differential equation, I think it was f(f'(x)) = x. I wrote it in that form since the inverse notation doesn't look so nice in these comments (f'(x) = f^(-1)(x), was the actual problem). I was trying so many different ways to formulate solutions, and I wanted to run something by you. Do you think a power series solution could for work for this problem? The solution you found actually has a pretty nice power series representation if you expand about x =1, and I was wondering if that could work for a problem like this. We can formulate a compositional power series out of f(f'(x)) and try and match coefficients to both sides. The matching wouldn't be that difficult, its the computation of the coefficients that I think would prove tricky.
Does the norm function need to be defined at zero? From what I can tell, the degree of the zero polynomial is either undefined or negative infinity, so I'm not sure how the degree function works as a norm on all of F[x]. It doesn't seem like you used the norm of zero anywhere in the video.
Many texts specify that the norm function isn't defined at 0. However, with the way the definition is set up, it's irrelevant. The value of the norm of 0 (if one is even defined) is totally irrelevant to the definition of Euclidean domain here.
Sir, I think the given domain and range of the valuation function needs a little inclusion.. According to the definition ,what can we say about the valuation image of the zero polynomial of K[x]? And the image of 0 in the domain of Gaussian Integers?? If the domain be non-zero and the range be non-negative Integers ,then everything fits.. btw loved the way you interpret Mathematics..
Yes, we typically want to do something different with 0. There are two approaches people usually take. Either, you can choose to have N(0) undefined, or you can choose to have N(0) = −∞. From an algebraic perspective, I'm quite fond of the N(0) = −∞ approach.
I should point out, though, with the way he defined the norm function, it's completely okay for N(0) to be defined - as anything. All of the relevant conditions for the norm function are based on the assumption that the inputs are nonzero. If a and b are _nonzero,_ then N(a) ≤ N(ab): This condition says nothing about the norm of 0, so no restrictions here. If a and b are in D with b _nonzero,_ then there exist q and r so that a = qb+r where _r = 0_ or N(r) < N(b): Again, this puts no restriction on the norm of 0 - because it only considers nonzero divisors and allows r = 0 as a valid option, regardless of the norm of 0. How one chooses to handle all of this depends on what is most convenient. If you want typical absolute value/modulus/field norm to be Euclidean norms, then those things can quite naturally give a definition for N(0). So it's convenient to just say "absolute value is a Euclidean norm" rather than to say "absolute value, when restricted to nonzero integers, is a Euclidean norm". So if you use norms that are typically used in topological and/or analytic contexts, allowing N(0) to be defined but irrelevant might be the most convenient choice of definition. From a purely algebraic standpoint, it might be convenient to have N(0) undefined. However, I find it more convenient to have N(0) = −∞ because you can then get rid of most of the "nonzero" requirements in the definition - we still need to keep b nonzero in all of the statements, but we can allow a to be zero in N(a) ≤ N(ab), and we can replace "r = 0 or N(r) < N(b)" with "N(r) < N(b)". Of course, the downside then is that the standard absolute value on Z is not a Euclidean norm unless you redefine |0| to be −∞, so it's harder to state examples. It's a trade off on what you care more about.
I have a video already: ruclips.net/video/E_mU_fCmcxc/видео.html In fact, I prove it over polynomial rings as well: ruclips.net/video/83mIBDaUhDQ/видео.html
![Abstract Algebra | If D is a UFD then D[x] is a UFD.](http://i.ytimg.com/vi/YyDXxe-Itfo/mqdefault.jpg)
![Abstract Algebra | If D is a UFD then D[x] is a UFD.](/img/tr.png)
Thank you so so so much!
It's a great explanation sir.Thank you
nicely explained...crisp and clear.
but inverse of beta might not be in Z[i], how is it okay to start with alpha*beta^(-1) ?
What about the uniqueness of (q,r)?
3:40 you say conjugation and multiplication commutes. Don't you mean conjugation distributes over multiplication? conj(a*b) = conj(a)*conj(b)
no he really meant commutes because you can do a*b then conjugate conj(a*b) and it ends up being the same as conjugating and then multiplying conj(a)*conj(b)
Could you should start a series of lessons in mathematics please (such as:Lifting the exponent lemma; ord and primitive Roots;ect. It would be of great help
I have a whole primitive roots playlist: ruclips.net/p/PL22w63XsKjqx1bbzG_-BYd0P6Gbe__lHr
In fact, I have over 100 videos covering a full course in number theory:
ruclips.net/p/PL22w63XsKjqwAgBzVFVqZNMcVKpOOAA7c
@@MichaelPennMath thank you so much it is such a great help
I think the domain of N should be D\{0} instead of D.
Because the degree of zero polynomial (i.e. f(x) = 0) SHOULD BE minus infinity. But in your definition, it can't be smaller than zero, which is not good.
Your last video has had me stumped on trying to find a direct method towards computing a solution that differential equation, I think it was f(f'(x)) = x. I wrote it in that form since the inverse notation doesn't look so nice in these comments (f'(x) = f^(-1)(x), was the actual problem). I was trying so many different ways to formulate solutions, and I wanted to run something by you. Do you think a power series solution could for work for this problem? The solution you found actually has a pretty nice power series representation if you expand about x =1, and I was wondering if that could work for a problem like this. We can formulate a compositional power series out of f(f'(x)) and try and match coefficients to both sides. The matching wouldn't be that difficult, its the computation of the coefficients that I think would prove tricky.
Does the norm function need to be defined at zero? From what I can tell, the degree of the zero polynomial is either undefined or negative infinity, so I'm not sure how the degree function works as a norm on all of F[x]. It doesn't seem like you used the norm of zero anywhere in the video.
Many texts specify that the norm function isn't defined at 0. However, with the way the definition is set up, it's irrelevant. The value of the norm of 0 (if one is even defined) is totally irrelevant to the definition of Euclidean domain here.
Sir, I think the given domain and range of the valuation function needs a little inclusion.. According to the definition ,what can we say about the valuation image of the zero polynomial of K[x]? And the image of 0 in the domain of Gaussian Integers?? If the domain be non-zero and the range be non-negative Integers ,then everything fits.. btw loved the way you interpret Mathematics..
Yes, we typically want to do something different with 0. There are two approaches people usually take. Either, you can choose to have N(0) undefined, or you can choose to have N(0) = −∞.
From an algebraic perspective, I'm quite fond of the N(0) = −∞ approach.
I should point out, though, with the way he defined the norm function, it's completely okay for N(0) to be defined - as anything. All of the relevant conditions for the norm function are based on the assumption that the inputs are nonzero.
If a and b are _nonzero,_ then N(a) ≤ N(ab): This condition says nothing about the norm of 0, so no restrictions here.
If a and b are in D with b _nonzero,_ then there exist q and r so that a = qb+r where _r = 0_ or N(r) < N(b): Again, this puts no restriction on the norm of 0 - because it only considers nonzero divisors and allows r = 0 as a valid option, regardless of the norm of 0.
How one chooses to handle all of this depends on what is most convenient. If you want typical absolute value/modulus/field norm to be Euclidean norms, then those things can quite naturally give a definition for N(0). So it's convenient to just say "absolute value is a Euclidean norm" rather than to say "absolute value, when restricted to nonzero integers, is a Euclidean norm". So if you use norms that are typically used in topological and/or analytic contexts, allowing N(0) to be defined but irrelevant might be the most convenient choice of definition.
From a purely algebraic standpoint, it might be convenient to have N(0) undefined. However, I find it more convenient to have N(0) = −∞ because you can then get rid of most of the "nonzero" requirements in the definition - we still need to keep b nonzero in all of the statements, but we can allow a to be zero in N(a) ≤ N(ab), and we can replace "r = 0 or N(r) < N(b)" with "N(r) < N(b)". Of course, the downside then is that the standard absolute value on Z is not a Euclidean norm unless you redefine |0| to be −∞, so it's harder to state examples. It's a trade off on what you care more about.
sir can you please prove bezout's theorem as I am preparing for PRMO. I t will help me a lot.
PLEASE SIR!!!!!!!!!!!!!!!!!!! 🙏🙏
I have a video already:
ruclips.net/video/E_mU_fCmcxc/видео.html
In fact, I prove it over polynomial rings as well:
ruclips.net/video/83mIBDaUhDQ/видео.html
@@MichaelPennMath ok thanks
@@MichaelPennMath sir can you please make videos on detailed solutions of PRMO 2019 , RMO 2019 and also of INMO 2018.
❤️😊😊😊😊❤️