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It is a consequence of a theorem known as Krull’s Hauptidealsatz that every non-unit element in a Noetherian domain is contained in a prime ideal of height 1. Assuming this, prove that a Noetherian domain R is a UFD if every prime ideal of height 1 in R is principal.

I want to use the following characterization of UFDs:

  1. ACCP holds
  2. every irreducible element is prime

The first point is obvious since R is noetherian, but I couldn't prove the second. I do not know whether the way I try is correct or not. Please help me, thank you.

user26857
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Mugenen
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1 Answers1

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This is based off the solution sketched in the comment but I elaborated here since there was a comment asking for elaboration.

Let $q$ be an irreducible element. By Krull’s Hauptidealsatz (given in the question), $q$ is contained in prime ideal of height 1, which is principal by hypothesis. Write $q \in (p)$, where $p$ is a prime element. Equivalently, $q=pa$ for some $a\in R$. Since $q$ is irreducible, we must have $p$ or $a$ being unit, but $p$ is non-unit. So $a$ is unit.

This makes $q$ a prime element, because $q|xy \Rightarrow p|xy \Rightarrow (p|x \text{ or } p|y) \Rightarrow (q|x \text{ or } q|y)$.

suncup224
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