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N. A larger model may collapse a singular cardinal number, destroying cardinalhood, so the assertion is not described by an upward absolute formula. It may instead add a short cofinal function to a regular cardinal in the smaller model, so it is not described by a downward absolute formula. This is the nonabsoluteness of singular cardinalhood.

Suppose that the countable ordinal $\alpha$ satisfied $V_{\alpha }\models \mathsf{ZFC}$. The axioms force $\alpha$ to be a limit ordinal above $\omega$, so choose an externally countable cofinal function $n\mapsto {\alpha }_n$ into $\alpha$, with ${\alpha }_n+1<\alpha$. The internal Axiom of choice gives a bijection in $V_{\alpha }$ between each $V_{{\alpha }_n}$ and some ordinal below $\alpha$. Every such ordinal is externally a countable set, hence every $V_{{\alpha }_n}$ is externally countable. The countable union of countable sets is countable, sowould be countable. But $V_{\alpha }$ contains the full power set $\mathcal{P}(\omega )$, which is uncountable by Cantor theorem. This contradiction is the result Countable rank-initial segment cannot model ZFC, and therefore

Suppose first that $\kappa$ is regular and uncountable in $M$. Part (b) and cardinal preservation by chain-condition forcing show that $\kappa$ remains a cardinal number in $M[G]$, while part (a) makes every infinite ground-model cardinal below $\kappa$ countable. Every ordinal below $\kappa$ has ground-model cardinality below $\kappa$ and is therefore countable in the extension. Thus $\kappa$ is the least uncountable ordinal there:

Conversely, suppose $M[G]\models \kappa ={\aleph }_1$ and $\kappa$ were singular in $M$. Let $\mu ={\mathrm{cf}}^M(\kappa )<\kappa$ and take in $M$ a cofinal function $f:\mu \to \kappa$. Part (a) makes $\mu$ countable in $M[G]$, while the same $f$ remains cofinal there. This would give $\kappa$ countable cofinality, contradicting first uncountable ordinal is regular. Hence $\kappa$ was regular in $M$, and for the intended uncountable $\kappa$,