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I have problem with the proof of theorem 4 in this link

It says if $R$ is an Noetherian ring, we construct $\mathbb U$, the set of ideals generated by each element of $R$ that cannot be written as a product of irreducible elements of $R$. then $\mathbb U$ have an maximal element $(r)$.

But I'm not sure what is mean by the "set of ideals generated by each element of $R$"

I think it means $\mathbb U = \{(a) : a \text{ is not a product of irreducibles} \}$ But in that case, how can I make an ascending chain that gives maximal element of $\mathbb U$?

MrTanorus
  • 637
  • You don't need to "make an ascending chain". The property of being Noetherian guarantees that any nonempty set of ideals contains a maximal element. You don't need to prove Zorn's Lemma applies to the set $\mathbb{U}$: noetherianness tells you it does (because any ascending chain is finite). – Arturo Magidin Nov 15 '23 at 01:18
  • @ArturoMagidin Thank you. Then I'll search for that property. – MrTanorus Nov 15 '23 at 01:31
  • Here's a proof that (i) all submodules are finitely generated; (ii) all ascending chains of submodules stabilizer; and (iii) any nonempty collection of submodules has maximal elements; are equivalent. If $R$ is noetherian, then any finitely genreated $R$-module is noetherian; and $R$ is a finitely generated module over itself. Alternatively, just do the proofs with "ideals" substituting "submodules". – Arturo Magidin Nov 15 '23 at 01:55

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