TITLE: A limit to Shoenfield Absoluteness
QUESTION [20 upvotes]: Shoenfield's Absoluteness Theorem states that if $\phi$ is any $\Sigma^1_2$ sentence of second-order arithmetic, then $\phi$ is absolute between any two models of $ZF$ which share the same ordinals. This means that such $\phi$ are unaffected by forcing, or by Axiom of Choice-related considerations (since for any $V\models ZF$, the corresponding subclass $L$ is a model of $ZFC$).
What I'm looking for is a simple example of either a $\Sigma^1_3$ or $\Pi^1_3$ sentence $\psi$ which is not absolute in this way - ideally, such a $\psi$ which is true in some $V\models ZFC$ and false in a generic extension $V[G]$. This is a purely pedagogical question for me - I'm trying to internalize Shoenfield Absoluteness, and I feel that a nice example of why it can't be made much stronger would help.
Thank you all in advance.

REPLY [4 votes]: Jensen and Solovay dealt with the question. For example $0^\#$ is a $\Delta^1_3$ real number, which is obviously not absolute because it cannot go deeper than $L[0^\#]$, and clearly it can't be absolute enough to be in $L$.
Jensen and Solovay point out that assuming a measurable cardinal, then there is such example. But what if one doesn't want to assume large cardinals?
Solovay constructed the following example. $M$ is a transitive countable model of $V=L$. Then there is $N=M[a]$ where $a$ is a real number and there is a $\Pi^1_2$ predicate $S(x)$ such that $N\models(\exists!x\subseteq\omega)S(x)$, and $N\models S(a)$, and moreover $N=M[a]$.
Jensen extended this result and showed that we can have that every $y\in\mathcal P(\omega)^L$ is recursive in $a$.
Both these results appear in the paper:

Jensen, R. B.; Solovay, R. M. "Some applications of almost disjoint sets." Mathematical Logic and Foundations of Set Theory (Proc. Internat. Colloq., Jerusalem, 1968) (1970) pp. 84–104.