Commutative algebra 5 (Noetherian rings)

[@mouchenqi2859]

Thank you for the wonderful lectures. I have been enjoying a lot all the explicit examples in your lectures.

[@wreynolds1995]

17:40 is hilarious! Great lecture as always.

[@atomiccompiler9495]

Wonderful lecture as usual.

[@hausdorffm]

21:16 ideal of R/I is equivalent to ideal of R containing I. In fact, any ideal J of R/I corresponds to an ideal of R defined by the kernel of the map R -> R/I ->(R/I)/J.

[@pronaybiswas7524]

Thank you

[@maxwibert]

@7:00 aren't we just assuming the axiom of countable choice here? My understanding is that countable choice is significantly less controversial than arbitrary choice

[@annaclarafenyo8185]

The axiom used here is "dependent choice", which allows you to choose a sequence from a collection of sets, where each choice is dependent on the previous choice. Dependent choice is a kind of countable choice, but it's different from 'countable choice' or 'choose one element from a countable collection', because you can't quite reconstruct a dependent choice from uncountable sets using just a regular countable choice function with no dependence. Regardless, this 'dependent choice' is what people mean when they say 'countable choice' nearly all the time--- being able to choose a sequence step by step. This axiom is not controversial in the usual AC way, because, unlike the usual full uncountable AC, it can't be used to prove any false statements. That's because it's compatible with Lebesgue measurability of subsets of R. It's only 'controversial' in the less crucial intuitionistic way, in that you don't get constructive proofs using it, but that's not a big deal, you just have to qualify the use of this axiom when you are writing a theorem checker program.

[@ΕιρηνηΘεοδωρου-θ9υ]

Hello friends! Could anyone help me with an exercise? Let R is a field and S is a subring of R Is it always true that S is Netherian? Excuse me for my bad english Take care friends!

[@manousosp.koutsoukos395]

είσαι στο μαθηματικό; έχουμε μία ομαδική στο messenger για βοηθεια στην μεταθετικη στο ΕΚΠΑ αν θες να σε προσθεσω!

[@ΕιρηνηΘεοδωρου-θ9υ]

@@manousosp.koutsoukos395 Γεια σου φιλε.Ναι ναι στο μαθηματικο ειμαι! Αν μπορεις θα ηθελα να με προσθεσεις. Σε ευχαριστω πολυ!

[@soumyajitghosh9637]

u know Ring of polynomials in infinite variables S = K [ x1, x2, x3, ...... ] is not Noetherian as it is not finitely generated but its quotient field F is always Noetherian ring so u get a field such that its subring is not Noetherian .

[@The2378AlpacaMan]

At first, I thought "well, what if you have a function that oscillates infinitely and approaches zero at the endpoints?", but then I realized this is not analytic at the endpoints oops.