# Are Sharps in Countable Models Really Sharps?

Suppose $M$ is a countable transitive model of $\mathsf{ZFC}$ (maybe more). Suppose $x, y \in {}^\omega\omega \cap M$ and $M \models y = x^\sharp$. Is it true (possible with additional assumptions), that $V \models y = x^\sharp$?

From Kanomori's book, $y = x^\sharp$ is $\Pi_2^1$. However, I can not apply Shoenfield absoluteness since $M$ is countable and does not have the countable ordinals of $V$.

An even stronger general question may be: are there conditions on $M$ or $V$ such that $\Pi_2^1$ absoluteness can hold between the universe and the countable model $M$.

Thanks for any insight on these questions.

$$\hbox{There is a countable transitive model of \text{ZFC}+x^\sharp exists}$$
is itself a $\Sigma^1_2(x)$ assertion, and thus it is absolute to $L[x]$, where the real $x^\sharp$ does not exist. So there will be a countable transitive model $M\in L[x]$ that thinks $x^\sharp$ exists, but it must have a fake $x^\sharp$.
In particular, if there is a transitive model of ZFC+$0^\sharp$ exists at all, then there is such a model inside $L$. And any such model inside $L$ cannot have the real $0^\sharp$.