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Let $T$ be a countable first-order theory, and assume that $T$ has exactly one atomic model up to isomorphism in every uncountable cardinality. (By "atomic" I mean a model which omits the non-principal types).

Now let $\mathfrak{M}$ be a countable transitive model of set theory, and assume that $T$ is also (countable) in $\mathfrak{M}$.

Is the above property preserved in $\mathfrak{M}$? i.e. does it hold in $\mathfrak{M}$ that there exists only one atomic model of $T$ in every uncountable cardinality in $\mathfrak{M}$?

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This is really a comment, but I need a bit more space.

If $\phi$ is a sentence of ${\cal L}_{\omega_1,\omega}$ that is $\aleph_0$-categorial, there is a complete first order theory $T$ in an expanded vocabulary such that the models of $\phi$ are exactly the reducts atomic models of $\phi$. The expansion is done in such a way that two structures will be isomorphic in the original language if and only if they are isomorphic in the expanded language.

So your question is really the same as: For $\phi$ a sentence of ${\cal}L_{\omega_1,\omega}$ is ``$\phi$ is $\kappa$-categorical for all infinite $\kappa$" absolute?

This is, as far as I know, still an open question. It is also open if ``$\phi$ is $\aleph_1$-categorical" is absolute.

John Baldwin in his paper ``Amalgamation, Absoluteness, and Categoricity" addresses some issues around this. Here is a link http://homepages.math.uic.edu/~jbaldwin/pub/singsep2010rev.pdf

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    $\begingroup$ I think this is absolutely valuable as an answer, not a mere comment. $\endgroup$ – Todd Trimble♦ Dec 30 '13 at 19:00
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    $\begingroup$ Shelah showed that for an atomic class $K$ under the assumption $\forall n\in\omega,\; 2^{\aleph_n}<2^{\aleph_{n+1}}$ (called very weak GCH (VWGCH)), "categoricity in all uncountable powers" is equivalent to "excellence plus categoricity in $\aleph_1$". Excellence in an amalgamation property of countable models. Therefore, it is absolute by Shoefield absoluteness. If $\aleph_1$-categoricity is absolute (which is still open), then your question has a positive answer. $\endgroup$ – Ioannis Souldatos Oct 17 '16 at 17:15

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