Is the property that "a commutative domain $A$ is a PID" absolute between models of ZFC? The property that "a commutative domain $A$ is a PID" is downward absolute. In fact, it can be written by
$$(\forall I \subseteq A \text{ ideal})\ (\exists x \in A)\ (\forall y \in A)\ [y \in I \leftrightarrow (\exists z \in A)\ y = xz].$$
Is the property absolute?
That is, For every transitive models $V \subseteq W$ of ZFC and every domain $A \in V$, does it hold that $A$ is a PID in $V$ is equivalent to $A$ is a PID in $W$?
If the property is not absolute, what is an example of a domain $A$ which is a PID in $V$ and not a PID in some forcing extension $V[G]$?
How about other ring theoretic properties such as "a ring $A$ is Noetherian" (that is a downward absolute) or "a module $M$ over a ring $A$ is free" (that is upward absolute)? Are they absolute?
 A: Noetherianness is absolute.  Note that a ring $R$ is Noetherian iff there is no strictly increasing sequence $(I_n)_{n<\omega}$ of finitely generated ideals of $R$.  That is, $R$ is Noetherian iff the poset $P$ of finitely generated ideals ordered by reverse inclusion is well-founded.  This is upward absolute, since it is witnessed by the existence of a strictly order-preserving map $P\to Ord$.
It follows that being a PID is absolute, since a commutative domain is a PID iff it is Noetherian and every finitely generated ideal is principal.
Freeness of modules is not absolute.  For instance, the $\mathbb{Z}$-module $M=\mathbb{Z}^\mathbb{N}$ is not free.  However, if you take a forcing extension where $M$ becomes countable, it becomes free.  This follows from the fact that a countable torsion-free $\mathbb{Z}$-module $M$ is free iff for any finitely generated submodule $N\subseteq M$, the submodule $N'=\{x\in M:nx\in N\text{ for some nonzero }n\in\mathbb{Z}\}$ is finitely generated.  (Sketch of a proof that $M$ satisfies this condition: given such an $N$, you can find a finite subset $F\subset\mathbb{N}$ such that the projection $M\to\mathbb{Z}^F$ is injective on $N$, and then it will also be injective on $N'$ and thus $N'$ is finitely generated since it is isomorphic to a submodule of $\mathbb{Z}^F$.)
