Abstract Algebra | Every PID is a UFD.

We prove the classical result in commutative algebra that every principal ideal domain is in fact a unique factorization domain. Along the way, we introduce the ascending chain condition and the notion of a Noetherian ring.
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[@DavesMathVideos]

A very well explained video, Professor. Thanks for taking the time to make these.

[@MichaelPennMath]

Thanks. Although most of my subscribers and views seem to come from contest problems and fun example videos the main goal of this channel is to create a very thorough resource for as many topics in "advanced" mathematics as possible.

[@DavesMathVideos]

@@MichaelPennMath I just started a channel recently. As my most immediate audience is going to be students in their final level of secondary and first year of university, I'm thinking of making some resources on intermediate level maths (calculus, multivariable, linear algebra, differential equations) as well as doing some contest problems in the future. We'll see how this goes.

[@pijames7021]

This was an excellent video. Your explanations are clear and your logic is very consistent. It has helped me tremendously learning about PIDs, UFDs, and how to make steps in higher mathematical proofs. Thank you!

[@noahtaul]

I’m concerned by your conclusion at 15:09. If you’re just really bad at choosing your a_n, you could pick a=a_1=a_2=...=4, and you can’t conclude that the stabilizing ideal is generated by an irreducible number. I would go by contrapositive: instead of choosing a_n arbitrarily, we suppose that there’s no irreducible that divides a and we can just keep nontrivially factoring the a_n’s which contradicts the ascending chain condition.

[@carl3260]

Yep, need to assume each a_n is either irreducible, so a_n, a_{n+1} are associates; or it’s not, so there is a ‘proper factorisation’, which you must take so that the a_I get “smaller” (otherwise could stay at a forever, which satisfies ACC). ACC implies the latter case occurs a finite number of times N, so a_N is irreducible, rename it p and rest follows.

[@akrishna1729]

thanks so much for this video - such a clear and detailed exposition!

[@MrTanorus]

Thank you sir, This whole video is like a highway.

[@divyenduroychoudhuri7267]

Thank you very much sir for such wonderful explanation. 😊

[@amin_moayedi]

Thanks a lot it was a brief proof and well explained

[@eamon_concannon]

23:15 If p_2 | u_1 then u_1 = k p_2 for some k in D and (u_1)^-1 = (p_2)^-1 (k^-1). Since p_2 is a non-unit by construction in first part of proof we get a contradiction, so p_2 ∤ u_1 and hence p_2 must divide one of the q terms.

[@inf0phreak]

I think from a presentation perspective it would work better to prove that a ring is a UFD iff it's atomic (non-zero non-units factor into irr.s, but not necessarily uniquely) and irr. implies prime. Then you wouldn't have to prove uniqueness of factorisation for PIDs directly.

[@MichaelPennMath]

Nice suggestion. I think I may make a video proving that classification of UFDs.

[@eamon_concannon]

Thanks

[@morelelfrancel6603]

I did advanced mathematics in french preparatory classes for two years before becoming an engineer a year ago. I still don't understand anything in this video :) This is amazing.

[@ilyass1260bj]

always tooop

[@user-jc2lz6jb2e]

I got this question in an exam 2 years ago. Didn't know how to do it.

[@MichaelPennMath]

This is a classic result. A must know for anyone preparing for a PhD preliminary exam in algebra.

[@papanujian7758]

Wobderful

[@OunegNebty]

J'ai rien compris...

[@eamon_concannon]

14:50 I found the next few minutes confusing. Here is another take. If a_i is reducible then we can write it as a_i = ef where e and f are non-units. We know that ⊆ (or ). Next show that ⫋ . Suppose that = , then since e(f+1) ∈ we have e(f+1) ∈ so e(f+1) = efg for some g ∈ D. e is non-zero so we get e(fg - (f+1))=0 implies fg-f-1 = 0 by cancellation property of D. Hence f(g-1) = 1 and so f is a unit. However f is not a unit so this contradiction means that we must have that is a proper subset of . So if a and all of the a_i's are reducible then we can have a chain only consisting of infinitely many proper subsets, which is a contradiction since all chains in a PID eventually stabilize. This means that we must have an irreducible term a_N where a ∈ and therefore a = a_N x for some x ∈ D, so we can factorize an irreducible term out of every non-unit element of a PID.

[@maxdickens9280]

You can add a new rule: if there exist x, y ∈ D, s.t. a_i = xy, then a_{i+1} and b_{i+1} must be both non-unit. (And, even this rule being added, we can still find an N, s.t. = for all n >= N. Thus, a_N is irreducible by definition)