Is it consistent that there is no $\omega_2$-saturated ideal on $\omega_1$, but one is introduced by an $\omega_2$-closed forcing?

**Some motivation:**

If $\delta$ is a Woodin cardinal, then it remains so after any $\delta$-closed forcing. It is a theorem of Woodin and Shelah that if $\delta$ is Woodin and $G \subseteq Col(\omega_1,<\delta)$ is generic, then in $V[G]$ there is a saturated ideal on $\omega_1$.

Jech and Prikry showed that if CH holds and there is a saturated ideal on $\omega_1$, then $2^{\omega_1} = \omega_2$. Thus if $\delta$ is inaccessible and $G \times H \subseteq Col(\omega_1,<\delta) \times Add(\omega_1,\delta^+)$ is generic, then $V[G][H]$ has no saturated ideals on $\omega_1$.

But if $K \subseteq Col(\delta,\delta^+)$ is generic, then $V[K] \models Col(\omega_1,<\delta) \times Add(\omega_1,(\delta^+)^V) \cong Col(\omega_1,<\delta)$. Thus if $G,H,K$ are mutually generic, then $V[G][H][K]$ has a saturated ideal on $\omega_1$.

By the $\delta$-c.c., $Col(\delta,\delta^+)^V$ remains $\delta$-distributive in $V[G][H]$. So we can force over $V[G][H]$ to add a saturated ideal on $\omega_1$ without adding subsets of $\omega_1$. But the question was whether we can do this with a $\delta$-closed forcing.