# Random reals preserving Cohen reals

Suppose we have a model (of $$\mathsf{ZFC}$$) $$M$$, and that $$x\in 2^\omega$$ is random over $$M$$, and that $$y\in 2^{\omega}$$ is Cohen over $$M$$. My question is whether $$y$$ is also Cohen over $$M[x]$$. In other words, if I have a real that's Cohen over a model, is it still Cohen over the model that results from performing Random forcing?

That depends on the particular random real $$x$$ and Cohen real $$y$$. On the one hand, I could first choose $$x$$ random over $$M$$ and then choose $$y$$ Cohen over $$M[x]$$. Then $$y$$ is also Cohen over the submodel $$M$$, so it's an example where the answer to your question is yes.
On the other hand, I could first choose $$y$$ Cohen over $$M$$ and then choose $$x$$ random over $$M[y]$$. Then $$x$$ is also random over the submodel $$M$$. I'll show that $$y$$ is not Cohen over $$M[x]$$.
Partition $$\omega$$ into intervals $$I_n$$ of rapidly increasing length. (In fact, it suffices to take $$I_n$$ of length $$n$$, but "rapidly" avoids the need for arithmetic.) Define a symmetric binary relation $$R$$ on $$2^\omega$$ by putting $$aRb$$ iff $$a$$ and $$b$$ agree on infinitely many of these intervals. Note that, for any $$a$$, the set of $$b$$'s $$R$$-related to $$a$$ is comeager but has Lebesgue measure 0. Note also that $$R$$ is a low-level Borel set with code in $$M$$ (provided you chose the sequence of $$I_n$$'s in $$M$$). Since $$x$$ is random over $$M[y]$$, we have that $$x$$ and $$y$$ are not $$R$$-related. But then $$y$$ cannot be Cohen over $$M[x]$$.