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In an integral domain $R$ the prime ideals are principal ideals. (given)

So my attempt was using the Zorn lemma. We assume that there exists a chain of non principal ideals so by set inclusion the chain will be partially ordered by set inclusion.

Now the chain will have a maximal element. My reason is like this that let $I_1$ be the ideal generated by $(a, b) $ , then this ideal will be contained in the ideal $I_2=(a, b, c) $ so I think that it will have a upper bound which will be the whole ring itself. Now by Zorn lemma there will be a maximal element. Is this OK?

Guria Sona
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    Your first sentence is not true in general. – Dietrich Burde Mar 24 '20 at 11:50
  • I am confused. The ring $\mathbb Z[X]$ is an integral domain and the ideal $(X,2)$ is prime, however, cannot be generated by one element. Also, if you take $R$ to be non-Noetherian domain (e.g. a polynomial ring over countably many variables), one can find counterexamples to the statement you are trying to prove. – Marktmeister Mar 24 '20 at 11:51
  • The first statement is something that is mentioned in the question @Marktmeister – Guria Sona Mar 24 '20 at 11:53

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