Rings and modules 4 Unique factorization

[@kaushaltimilsina7727]

I think a series on chain complexes and spectral sequences and other objects with chains/sequences of modules and algebras over a ring would be really interesting. I have noticed lately that such objects are very useful in Algebraic Topology and Algebraic Geometry, in understanding and computing things and are used greatly alongside categorical techniques.

[@irrazionalex226]

A series on spectral sequences would be a dream 🤩

[@theflaggeddragon9472]

I third this, I'm trying to learn spectral sequences on and off for a couple of weeks now and it's very painful without guidance. Some worked examples with Borcherds wonderful style would be a life saver on this front

[@andreagrigo5923]

Totally agree, that would be awesome for anyone interested in the topic!

[@doyunnam3124]

Thank you, professor. note : in PID, an element that is irreducible is prime; and using this, we can show that PID is UFD.

[@Tehom1]

If we defined 1 as prime, the problem would be that we'd have to say "the primes except 1" for almost everything interesting. It's the same with the ring of 1 not being defined as a field. Extending the definition without breaking anything else is the easy part.

[@constantijndekker8343]

Great video! For anyone interested, at 34:40, I think the word Richard was looking for is "cuboid"

[@danielprovder]

Parallelapiped

[@constantijndekker8343]

@@danielprovder I think he almost accidentally used that word, but realized that it is more general than a cuboid.

[@Israel2.3.2]

Fun proof via Book VII of Euclid. [I think this works. Weil got me excited about Books VII through X and the effort has been well rewarded thus far.] Suppose that N is the smallest number with distinct factorizations, then N = AD = BC, where A and B are the largest powers of primes P and Q measuring N for distinct factorizations AD and BC of N. (See Comment III.) Either P and Q are distinct primes or they are not. If P and Q are distinct primes then the powers A and B are coprime [VII. 27], but A:B::C:D so that A measures C and B measures D by some number M [VII. 20]. From N = A(BM) = B(AM) we see that some M < N inherits distinct factorizations from N. If P and Q are not distinct then N/P < N inherits distinct factorizations from N. Comment I. The requirement that A and B be highest powers can be relaxed, it is sufficient to consider single prime divisors. Comment II. The argument can be dramatically shortened. Given Euclid's lemma [contrapositive of VII.24] we may exclude the case where P and Q are distinct. Very lovely. Surely this is what Borcherds means he says that Euclid proved this result. Comment III. I should say more about the factorizations AD and BC in the proof. Note that N is not prime, for by Euclid's definition a prime has but a single factorization. Also note that every composite number N has a decomposition EF where E and F are not units, by induction it follows that every composite number N has a prime decomposition.

[@caspermadlener4191]

This is also how I like to show unique factorisation, for groups for example.

[@Israel2.3.2]

@@caspermadlener4191 I still haven't learned group theory proper, this summer/fall I hope to finally learn the basics

[@caspermadlener4191]

@@Israel2.3.2 Well, assuming you know the axioms, here are my two favourite questions: When ab=ba, for every a and b in the group, that group is called abelian. 1. Is every group of 21 elements abelian? *My teacher offered €10 if anyone in the class could answer this within 20 seconds, and I could! 2. Suppose that for every g in a group, g³ is the identity. Is this group abelian? *This was one of the first questions in my syllabus, but still relatively hard.

[@Israel2.3.2]

@@caspermadlener4191 Not sure about the first one, although I'm curious about the solution. For the second one we see that ab = a(baa)a = a(aab)a = ba from the bijection E→E given by (a)→(aa), although I'm not sure how to show existence [ab = ba for all a,b iff aabb = bbaa for all a,b iff aac = caa for all a, c = bb (using the bijection)]. Interesting problems.

[@Israel2.3.2]

@@caspermadlener4191 Oh wait maybe I'm just assuming commutativity on the second problem. Nothing in the argument prevents a non-abelian case from arising.

[@lucianonotarfrancesco4443]

At 16:30 you mention a more general definition of Euclidean ring where the Euclidean function can go to a well-ordered set instead of only Z. I try to use always this more general definition because it can be useful, especially in computational algebra where you can now use the Euclidean division. For example with this definition the product of Euclidean rings is again an Euclidean ring, with quotient and remainder defined component-wise. Also many people say "Euclidean ring" without actually defining it properly, and they actually mean "Euclidean domain", this is kind of annoying... is better to avoid this confusion and allow zero divisors in Euclidean rings, and use "Euclidean domain" when you don't want zero divisors... for example Z/mZ has zero divisors but it is an Euclidean ring. It took me a while to find a satisfactory definition of Euclidean ring, the reference that clarified everything to me was Pierre Samuel, "About Euclidean Rings" (1970)

[@migarsormrapophis2755]

yeee

[@diribigal]

-1 is a product of the the unit -1 and the product of the empty set of primes, so it's a product of primes "up to units". I don't think a modification is needed for it.

[@eamonnsiocain6454]

If 1 were not prime, it would have at least two distinct factors different to each other and both different to itself. Let’s not be silly, though.