PIDs are not Artinian?

Question

In my notes there is the following statement:

Let $A$ be a PID, then $A$ as an $A$-module is trivially Noetherian but not Artinian. In fact, take a prime element $p$ in $A$, then we have the chain $$(p)\supset (p^2) \supset (p^3) \supset \dots$$

There are a few things that I don't understand:

• The chain constructed uses a prime element, but wouldn't suffice to use an element $p$ which is neither a unit nor nilpotent?
• If we are assuming the existence of an element which is neither a unit nor nilpotent, aren't we implicitly assuming that $A$ is not local? What can we say about the general case? In other words, what can we say about local PIDs?

Thanks

Answer 1

1. Yes, any nonzero nonunit will work. The convenience of saying ``prime element'' is probably just that they are neither zero nor a unit. Also, you do not need to worry about nilpotents (besides zero) because PIDs are domains.

2. No, you are not assuming it is local. For instance, $\mathbb Z_{(p)}$, the integers localized at a prime $p$, is a local ring and a PID and not artin.

I think you are confusing being local with the dimension, where you'd run into problems if the ring has dimension 0 (i.e. is a field). Under that interpretation of your question you are correct: all fields are PIDs and also artinian. The argument only works for PIDs which are not fields.