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I am reading a proof which uses the fact that every PID has a prime element (it takes a prime element from a PID), and I am thinking whether every PID has a prime element.

If the PID is not a field, I believe there is an irreducible element which is a prime. If it is a field, $0$ is a prime element since $(0)$ is a prime ideal, but it's not an irreducible element. However, we claim that irreducible elements and prime elements in a PID are the same. Is $0$ in a field a prime element? How can we assume that every PID has a prime?

Coco
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  • I don't think fields have prime elements. They don't need any for the purposes of divisibility studies because every non-zero element is a unit. Yes, fields are PIDs. Exactly how is the text you are studying using the primeness? Would the result be trivial for fields? – Jyrki Lahtonen Sep 05 '23 at 18:36
  • @JyrkiLahtonen The text is Lang's $\textit{algebra}$ and when proving the fact that every free module over a PID has a dimension, he writes something like "take a prime element from the PID...." Lang doesn't exclude the possibility that the PID might be a field. – Coco Sep 05 '23 at 18:40
  • Ok. As I suspected, there is no need to cater for the case of a field. Vector spaces have a dimension. Actually, that particular claim is often proven by reducing it to the case of a field. – Jyrki Lahtonen Sep 05 '23 at 18:47
  • @JyrkiLahtonen Thank you. Do you mean that I can assume the PID is not a field? Is it that every PID which is not a field has an irreducible element, and thus a prime element? – Coco Sep 05 '23 at 23:00

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The correct statement is that every PID which is not a field has a prime element.

Hint: use the fact that every nonzero ring has at least one maximal ideal.

Alex Mathers
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  • And every maximal ideal is a prime ideal in a PID, which indicates that a PID has a prime element. Thank you for your answer! I think my confusion is more of: $0$ should be a prime element by the definition but then irreducible elements are actually not the same as prime elements? – Coco Sep 05 '23 at 18:34
  • @Coco the definition of prime element should exclude $0$. i.e. a prime element should be "a nonzero element $p$ such that $p|ab$ implies $p|a$ or $p|b$". Then primes and irreducibles are the same in a PID (but not in general) – Alex Mathers Sep 05 '23 at 20:49
  • Thank you. But this way, not every PID (such as a field) has a prime element, but the author writes "take a prime element from a PID.." @AlexMathers – Coco Sep 05 '23 at 22:57
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    @Coco It is likely that if the PID is a field then the result is trivial, and the author forgot to write this. You ought to separate into two cases for yourself in a more careful way than the author is doing: case 1) the PID is a field, and case 2) the PID is not a field. In the latter case you can apply the argument you see in the book, and in case 1) the result is likely simple – Alex Mathers Sep 06 '23 at 01:25
  • @Coco Alex forgot the "nonunit" hypothesis in the definition of a prime element. Primes are irreducible (atoms) but the converse (AP := atoms are prime) is not generally true (it is equivalent to uniqueness of factorizations into atoms, which does hold true in gcd domains (e.g. UFDs and PIDs). See here for many other closely related divisibility properties. – Bill Dubuque Sep 06 '23 at 02:00
  • @BillDubuque Thanks for catching that, my mistake – Alex Mathers Sep 06 '23 at 03:21
  • @AlexMathers Gosh you remind me that if the ring is a field, then a module (whether free or not) is a vector space, then the dimension is well-defined. Thank you! – Coco Sep 06 '23 at 04:11