A countably complete ideal $I$ on a set $Z$ ideal is c.c.c. when there is no uncountable family of pairwise disjoint $I$-positive subsets of $Z$. If such an ideal exists, then there exists a weakly Mahlo cardinal $\kappa$ and a $\kappa$-complete, c.c.c. ideal $J$ on $\kappa$. An example of such an ideal is the collection of measure zero subsets of a real-valued measurable cardinal.
The existence of such an ideal, which is nowhere maximal ("ultra"), requires a large continuum. The standard proof of this is as follows. Build a tree of subsets of $\kappa$, with root $\kappa$. Suppose by induction that each level of the tree up to level $\alpha$ is a partition of $\kappa$. For every $J$-positive member of the partition, partition this into two disjoint $J$-positive sets. For members in $J$, don't split them. This gives you level $\alpha+1$ and a finer partition. At limit stages, take intersections. By the c.c.c., we must have all members of the partition in $J$ at level $\omega_1$, and no cofinal branches. There are $2^{<\omega_1} = 2^\omega$ many measure zero sets at that stage, so we have $\kappa$ equals a union of continuum many sets from $J$, so $2^\omega \geq \kappa$.
My questions is: Does this process of building the tree terminate before $\omega_1$? If it does not, then by the c.c.c., some subtree of what we've constructed is a normal Suslin tree. So in some cases, we know the process does halt at a countable stage:
(a) If we start with a measurable $\kappa$ and force MA$_\kappa$ with the Solovay-Tennenbaum forcing, we get no Suslin trees, and a c.c.c. ideal on $\kappa$.
(b) By a result of Laver, we can start with a model of MA$_{\omega_2}$ and a measurable above, and then force a real-valued measurable cardinal by random real forcing, keeping SH.
So the obvious question is, is there any way to get a c.c.c., $\kappa$-complete ideal $J$ on $\kappa$, where we can find a Suslin tree embedded in the algebra of $J$-positive sets? Can this be done by adding measurably many Cohen reals?
