Abstract Algebra | A PID that is not a Euclidean Domain

[@shelipemaktadir483]

congratulation on reaching 20k! Amazing content on your channel

[@inf0phreak]

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 $k$.

[@JM-us3fr]

I've been meaning to look up this argument. Thanks for going over it.

[@peterusmc20]

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 ?

[@divisix024]

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

[@stenzenneznets]

It is indeed confusing

[@余淼-e8b]

Since r = ay - 19bx - cq is an integer, given c = 5, |r|

[@HolyAjax]

I'm wondering why we construct the norm as at 10:25? This seems to be of no use in the following proof.

[@schweinmachtbree1013]

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).

[@JM-us3fr]

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

[@noahtaul]

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.

[@noahtaul]

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.

[@mahmoudalbahar1641]

@@noahtaul thank you for clarifying what the confusion is...and thank for writing the name of book that this proof is contained in it.

[@何海安]

This is just an exercise problem in "Algebra" (GTM73) by Thomas W. Hungerford.

[@rahulkumarmishra4303]

sir please suggest some analytical books of Abstract algebra

[@f5673-t1h]

In proving the first lemma, can't we just pick the universal side divisor to be always u = 1?

[@noahtaul]

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.

[@JM-us3fr]

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.

[@CallMeIshmael999]

This is the ring of integers in Q(i sqrt(19)), yes?

[@CallMeIshmael999]

I'm silly. Of course it's the ring of integers. It's a PID integral over the integers in the field.

[@JM-us3fr]

@@CallMeIshmael999 Yeah it's a PID so in particular is integrally closed, and therefore the ring of integers.

[@marcfreydefont7520]

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

[@JM-us3fr]

You do it via induction.

[@maxamedaxmedn6380]

I don't even know what is going on I just love to watch 👨‍💻

[@paticdarko4584]

Fantastic video :-)

[@ojas3464]

👍

[@ryan-cole]

I am early for this one.

[@asht750]

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.

[@SANI-sp5gq]

2nd

[@lagduck2209]

I don't understand anything of it, no idea why it is still interesting