It seems that you accidentally put a tilde too much in the definition of universal side divisor. As it's written on the board every ring has one as you can take u=1 and z=0 for all x. It's clearly supposed to be $u \in D \setminus \tilde D$ such that you get nontrivial examples like $\pm 2$, $\pm 3$ for the rational integers and $x-a \in k[x]$ for any field $k$.
At 22:50 surely at c=5 1/4+19/25 is 101/100 and greater than 1? Does he mean the other way around that it's greater than 1 if c is less than or equal to ?
The backwards direction of your 2nd box at 6:28 needs to assume b is not 0. But then it runs into the problem that sa-tb could be 0. That's where I got stuck
It seems like you're switching between i*sqrt(19) and (1+i*sqrt(19))/2. For example, what you wrote at 25:19 seems like it works if you suppose w=i*sqrt(19), but if it's what you defined it as originally, there's no reason a and b need have opposite parities. If a=b=1, then (1+1/2(1+i*sqrt(19)))/c=1/4(3+i*sqrt(19)) isn't in the integer ring.
And I'm looking at Dummit and Foote now, and they write alpha/beta=(a+bsqrt(-19))/c, so I think that's probably what you meant to write, and we can replace that there from 17:16 to 31:52. Every other symbol on the board is correct, I think.
We're proving that an integral domain is a principal ideal domain if and only if it has some Dedekind-Hasse norm (in general a PID could have many different Dedekind-Hasse norms), which we later use to show that the domain Z[(1 + i sqrt(19))/2] is a PID, by exhibiting one (it doesn't matter that the norm exhibited for Z[(1 + i sqrt(19))/2] doesn't look like the norm in the abstract proof at 10:25; we just need to exhibit _some_ norm).
The point is that 1 is not a useful side divisor, so we exclude it. We showed that Euclidean domains have non-unit side divisors, and that’s a more restrictive thing, so we just restrict to them. Basically we make things harder on ourselves because we know the theory can take it.
To elaborate on what noahtaul said, I think Michael meant to say u came from D-(D tilde), rather than D tilde. Obviously any unit divides everything, but it's the nonunits we are interested in. For example, in Z[i], a universal side-divisor is 1+i. This is a nonunit, but dividing by it always gives remainder 0 or a unit.
Great video and content. One question: you mention that when a1, a2, a3,..., an in Z are comprime in then Bezout’s lemma applies I.e there are x1, x2,..., xn in Z such that the sum of ai.xi is equal to the gcd so 1 in case the ai s are coprime and you mention... elementary arithméticiens proof. Could you kindly show how this would be derived as this is not so obvious to me? Thank you
Some of your content is off-topic such that they do not follow one another. For instance, polynomial rings are generally discussed if you have exhausted PID, ED, UFD and their properties.
![Abstract Algebra | If D is a UFD then D[x] is a UFD.](http://i.ytimg.com/vi/YyDXxe-Itfo/mqdefault.jpg)
![[안방1열 풀캠4K] 베이비몬스터 'DRIP' (BABYMONSTER FullCam)│@SBS Inkigayo 241110](http://i.ytimg.com/vi/AQ4RIghEvUs/mqdefault.jpg)
It seems that you accidentally put a tilde too much in the definition of universal side divisor. As it's written on the board every ring has one as you can take u=1 and z=0 for all x. It's clearly supposed to be $u \in D \setminus \tilde D$ such that you get nontrivial examples like $\pm 2$, $\pm 3$ for the rational integers and $x-a \in k[x]$ for any field $k$.
I've been meaning to look up this argument. Thanks for going over it.
congratulation on reaching 20k! Amazing content on your channel
This is just an exercise problem in "Algebra" (GTM73) by Thomas W. Hungerford.
Could anyone explain why we don't need to check c=5? I'm confused since 1/4+19/5^2=0.25+0.76=1.01, which means c=5 does not satisfy 1/4+19/c^2
It is indeed confusing
Since r = ay - 19bx - cq is an integer, given c = 5, |r|
At 22:50 surely at c=5 1/4+19/25 is 101/100 and greater than 1? Does he mean the other way around that it's greater than 1 if c is less than or equal to ?
The backwards direction of your 2nd box at 6:28 needs to assume b is not 0. But then it runs into the problem that sa-tb could be 0. That's where I got stuck
It seems like you're switching between i*sqrt(19) and (1+i*sqrt(19))/2. For example, what you wrote at 25:19 seems like it works if you suppose w=i*sqrt(19), but if it's what you defined it as originally, there's no reason a and b need have opposite parities. If a=b=1, then (1+1/2(1+i*sqrt(19)))/c=1/4(3+i*sqrt(19)) isn't in the integer ring.
And I'm looking at Dummit and Foote now, and they write alpha/beta=(a+bsqrt(-19))/c, so I think that's probably what you meant to write, and we can replace that there from 17:16 to 31:52. Every other symbol on the board is correct, I think.
@@noahtaul thank you for clarifying what the confusion is...and thank for writing the name of book that this proof is contained in it.
I'm wondering why we construct the norm as at 10:25? This seems to be of no use in the following proof.
We're proving that an integral domain is a principal ideal domain if and only if it has some Dedekind-Hasse norm (in general a PID could have many different Dedekind-Hasse norms), which we later use to show that the domain Z[(1 + i sqrt(19))/2] is a PID, by exhibiting one (it doesn't matter that the norm exhibited for Z[(1 + i sqrt(19))/2] doesn't look like the norm in the abstract proof at 10:25; we just need to exhibit _some_ norm).
I don't even know what is going on
I just love to watch 👨💻
In proving the first lemma, can't we just pick the universal side divisor to be always u = 1?
The point is that 1 is not a useful side divisor, so we exclude it. We showed that Euclidean domains have non-unit side divisors, and that’s a more restrictive thing, so we just restrict to them. Basically we make things harder on ourselves because we know the theory can take it.
To elaborate on what noahtaul said, I think Michael meant to say u came from D-(D tilde), rather than D tilde. Obviously any unit divides everything, but it's the nonunits we are interested in. For example, in Z[i], a universal side-divisor is 1+i. This is a nonunit, but dividing by it always gives remainder 0 or a unit.
sir please suggest some analytical books of Abstract algebra
This is the ring of integers in Q(i sqrt(19)), yes?
I'm silly. Of course it's the ring of integers. It's a PID integral over the integers in the field.
@@CallMeIshmael999 Yeah it's a PID so in particular is integrally closed, and therefore the ring of integers.
Fantastic video :-)
Great video and content. One question: you mention that when a1, a2, a3,..., an in Z are comprime in then Bezout’s lemma applies I.e there are x1, x2,..., xn in Z such that the sum of ai.xi is equal to the gcd so 1 in case the ai s are coprime and you mention... elementary arithméticiens proof. Could you kindly show how this would be derived as this is not so obvious to me? Thank you
You do it via induction.
👍
I am early for this one.
Some of your content is off-topic such that they do not follow one another. For instance, polynomial rings are generally discussed if you have exhausted PID, ED, UFD and their properties.
I don't understand anything of it, no idea why it is still interesting
2nd