# Can Sacks forcing add a Cohen generic real over $L$?

Motivated by this question Forcing the negation of CH without adding Cohen reals over L and Todd Eisworth's comment, the question is the following:

1) Suppose $V$ has no Cohen generic reals over $L$. (i.e. every real is contained in a constructibly coded Borel meager set.) Let $\mathbb{S}$ be Sacks forcing. Let $G \subseteq \mathbb{S}$ be $\mathbb{S}$-generic over $V$. Can $V[G]$ have a Cohen generic real over $L$?

Of course, by the usual fusion argument, there are no Cohen generic reals over $V$ in $V[G]$. So in the particular case of $V = L$, the answer is no.

Also if $V$ has stronger properties like $V$ has no unbounded reals over $L$, then Sacks forcing over $V$ can not add an unbounded real over $L$ and in particular no Cohen real over $L$.

Also, I seem to recall it is an open question whether in countable support iterations $\langle P_\alpha : \alpha \leq \omega \rangle$ of $\langle \dot Q_n : n \in \omega \rangle$ such that $P_n \Vdash$ "$\dot Q_n$ adds no Cohen reals", $P_\omega$ adds a Cohen real over $V$.

This seems to suggest this question for finite length iterations is known. So a second question is

2) If $P$ does not add Cohen reals and $1_{P} \Vdash \dot Q$ does not add Cohen reals, does the two step iteration $P * Q$ add Cohen reals?

If the answer in 2 is no, then I think Question 1) has a negative answer when $V$ is a set-generic extension of $L$.

Thanks for any information on this question.

Zapletal's solution to the "Half-Cohen problem" gives an example of a 2-stage iteration $P*\dot Q$ such that $P$ does not add Cohen reals, and $P$ forces that $\dot Q$ does not add Cohen reals over $V^P$, yet the iteration $P*\dot Q$ does add a Cohen real over the ground model.