Abstract Algebra | A PID that is not a Euclidean Domain

We present an example of a principal ideal domain that is not a Euclidean domain. We follow the outline described in Dummit and Foote. In particular, we show that an integral domain D is a PID if and only if it has a Dedekind-Hasse Norm and that every Euclidean domain has a universal side divisor. Then we show that our example has a Dedekind-Hasse norm but no universal side divisor.

[infphreak]

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 $k$.

[divisix]

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<1

[peterusmc]

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 ?

[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 isn't in the integer ring.

[f-th]

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

[JM-usfr]

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

[shelipemaktadir]

congratulation on reaching 20k! Amazing content on your channel

[asht]

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.

[CallMeIshmael]

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

[何海安]

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

[JM-usfr]

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

[rahulkumarmishra]

sir please suggest some analytical books of Abstract algebra

[maxamedaxmedn]

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

[marcfreydefont]

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

[HolyAjax]

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

[lagduck]

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