Abstract Algebra | Introduction to Principal Ideal Domains (PIDs)

After introducing the notion of a principal ideal domain (pid), we give some examples, and prove some simple results.

[eamon_concannon]

6:50 It appears that b must be non-zero here to ensure that b(1- xy)= 0 implies 1 - xy = 0 by cancellation rule in an integral domain.

[eamon_concannon]

17:23 Suppose ab∈M and a ∉ M where M is a maximal ideal of the integral domain D.
The ideal <a, M> = <a> + M = D since <a, M> is larger than M. This means that 1 ∈ <a, M> so 1=ax+m for some x ∈ D and m ∈ M.
Therefore b= bax + bm ∈ M as ab ∈ M, so a maximal ideal of an integral domain is a prime ideal.

I have learnt a lot of new and interesting concepts from this abstract algebra series. Great work!

[peelysl]

Thanks this help me in my 8th grade math studies

[gaetanocascio]

After proving point 1 of the Lemma, the other two points actually follow by observing that saying that a and b are associates is equivalent to saying that a divides b and b divides a (hence (a) is included in (b) and (b) is included in (a)), and that the units of D are precisely the elements associated to the identity 1 (which of course generates the whole D as an ideal).

I wanted to thank you for your excellent videos, and ask you what is your favourite book (or books) to learn these topics. I'm currently studying them again on my own on the book Basic Algebra I by N. Jacobson, and although I find it an amazing book (I like its very terse style and its didactic approach, which leaves many important results as exercises), very often I found myself frustrated by some of its most difficult exercises, for which the author provides no hints.

[vikramsundara]

In the proof for 3 you can just use the fact that as a is a unit the ideal generated contains 1

[adenpower]

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[ImaginaryMdA]

Slight issue, you claim that in an arbitrary ring "prime < irreducible" but this is only true for domains.
You need the cancellation property to prove that every prime is irreducible.
p=ab => p|a => a=pr => p=prb *=>* 1=rb.