I'm not very well acquainted with forcing, just with the basic ideas; but I thought of the following proof, and since I'm definitely not comfortable with forcing I don't know if it's right, so my question is :
Is the following proof valid ? If not, can a bit of work make it valid, or is there an essential mistake ?
The question is the following : suppose you have a property $P$ that can be expressed in the first order language of rings; and assume that ZFC proves "if a ring $R$ satisfies $P$ and is countable, then $R$ is noetherian".
I want to prove that then ZFC proves "if a ring $R$ satisfies $P$, then $R$ is noetherian".
Let $V$ be a model of ZFC, by Löwenheim-Skolem one may assume that $V$ is countable (we may assume $V$ exists because we may assume ZFC is consistent). Let $R$ be a $V$-ring (I can clarify what this means but I think it's clear) that satisfies $P$ and $S$ a set of ideals of $R$.
Now consider a forcing extension $V[G]$ that sees $R$ as countable (this can be done because $V$ is countable, and we can use the classical collapsing forcing).
In this extension $R$ still satisfies $P$ (this is where I'm no comfortable : though if I recall correctly, first order assertions about $R$ are preserved under extensions), and is countable. Hence, it is noetherian, and hence $S$ (which is still a set of ideals of $R$) has a maximal element $I$. But $I\in S$ so $I\in V$ because $V$ is a transitive submodel of $V[G]$. Moreover, "$I$ is maximal in $S$" is absolute (that seems clear but I'd also like to be sure about this) so $V$ sees that $S$ has a maximal element; To conclude, $V$ sees $R$ as noetherian.
This yields that "$R$ has $P$ $\implies$ $R$ is noetherian" is provable as well.
Is that correct ?