Let me supplement Mohammad's answer by supplying a proof, since the result is not difficult, and it is extremely useful.

**Theorem. (Solovay)** If $G\subset\mathbb{P}$ and $H\subset\mathbb{Q}$ are mutually $V$-generic, that is, $G\times H\subset \mathbb{P}\times\mathbb{Q}$ is $V$-generic, then $V[G]\cap V[H]=V$.

Proof. Clearly $V[G]\cap V[H]\supset V$, so suppose that $x\in V[G]\cap V[H]$. Thus, $x=\sigma_G=\tau_H$ for some $\mathbb{P}$-name $\sigma$ and $\mathbb{Q}$-name $\tau$. By $\in$-induction, we may assume $x\subset V$. Since $\sigma_G=\tau_H$, there must be some condition $(p,q)$ forcing that $\sigma$ and $\tau$, viewed as $\mathbb{P}\times\mathbb{Q}$-names in the canonical way, are naming the same set in that way. But now every condition $p'$ stronger than $p$ must force $\check y\in \sigma$ just in case $y\in x$, because otherwise we could find a different generic filter $G'$, mutually generic with $H$, so that $G'\times H$ contains $(p,q)$, but such that $\sigma_{G'}\neq \tau_H$. Thus, the condition $p$ already knows exactly which $y$ are elements of $x$, and so $x\in V$, as desired. QED

See further information on this MO question about mutual genericity.