We have learned from Joel David Hamkins and Monroe Eskew that:

**Answers:**
Having a measurable cardinal $\delta$ we can force a $\kappa$-Suslin tree for many $\kappa$'s above $\delta$.

But is the opposite true (I am interested in the case when $\delta$ is the least measurable):

**Question 1:**
If we have a set theory with (the least) measurable cardinal $\delta$ can we have a set theory with $\delta$ (still the least measurable) but without any $\kappa$-Suslin trees for $\kappa>\delta$?

Actually I hadn't expected Souslin trees above $\delta$... Since I seem to be unable to remove another assumption from my constructions let me ask:

**Question 2:**
Does the validity of statements in Answer and Question 1 above change if we assume that $\delta$ (the least measurable) is strongly compact?

**Edit:**
I edited the text to indicate that, to me, the most interesting case is when $\delta$ is the least measurable cardinal. Inserted pieces are in parentheses.

*Why am I interested in this?* Let $\mathbb{Z}^\kappa=A\oplus B$ be a direct sum decomposition of the product of the group of integers. By old results, if no measurable cardinal exists below $\kappa$ then both $A$ and $B$ are themselves isomorphic to some products of integers. For larger $\kappa$'s little is known except that $A$ or $B$ has to be of this form. I think that the problem of whether both $A$ and $B$ have to be products depends on the existence of Suslin trees above $\delta$.