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I want to check if my solutions for this problem are right.

Let $R$ be a factorial ring in which every ideal generated by two elements is a principal ideal.

Show that $R$ is a principal ideal ring.

What I thought is: every element of a factorial Ring $a$can be written as a product of irreducible elements $p_1,...,p_m$. We have $a=p_1\cdots p_m$ moreover $a=br$ for two $b,r \in R$. This means that $a$ is an ideal generated by two elements.

We know therefore that all $a \in R$ are principal ideal. For this exists a common divisor for all elements $a \in R$ so that each $a \in R$ can be written as $a=br$, where $b$ is the common divisor. All elements of $R$ are therefore generated from a common element $b$. This is why $R$ is a principal ideal.

I think that my solution is wrong. Can someone help me?

Marco Di Giacomo
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  • The following seem to be non-sequitur: "moreover $a=br$ for two $b,r\in R$. This means that $a$ is an ideal generated by two elements". "All elements of are therefore generated from a common element " – rschwieb Nov 06 '23 at 16:12
  • Then there's the broken writing "$a\in R$" and "$a$ is an ideal" "This is why is a principal ideal." I recognize a language barrier may be a contributing factor. Nevertheless, you should be able to find your way using he attached duplicates. – rschwieb Nov 06 '23 at 16:14
  • In a nutshell, a UFD doesn't have infinite strictly ascending chains of princpal ideals. The existence of a non-finitely-generated ideal, combined with the "2-generated ideals are principal" (this is called a Bezout ring) would imply the existence of an infinitely ascending chain of principal ideals. That contradiction would mean all ideals are finitely generated, and therefore principal. – rschwieb Nov 06 '23 at 16:34

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