First inaccessible Suslin trees in L, an interesting detail It's known (but quite nontrivial) that $V=L$ implies that if $\kappa$ is the 1st inaccessible cardinal then there are $\kappa$-Suslin trees $T$.
Such a tree $T$ can be considered as a forcing notion in $L$.
Being Suslin, $T$ does not collapse $\kappa$.
QUESTION. Can such $T$ be defined so that the $T$-forcing over $L$ preserves all smaller cardinals?
 A: This will always be true. That is, I claim that forcing with a $\kappa$-Suslin tree always preserves all smaller cardinals, and indeed, preserves all cardinals and cofinalities.
Theorem. Forcing with a $\kappa$-Suslin tree preserves all cardinals and cofinalities.
Proof. Suppose $G$ is $V$-generic for a $\kappa$-Suslin tree $T$, where $\kappa$ is regular. Note that $\kappa$ is preserved, and all cardinals and cofinalities above $\kappa$, since the forcing is $\kappa$-c.c. Meanwhile, the filter $G$ is determined by a branch through $T$, and this branch is $\kappa$-closed in $V[G]$. This is enough to see that the forcing will be ${<}\kappa$-distributive, as follows. If $\sigma$ is a name for a $\beta$-sequence over $V$, for some $\beta<\kappa$, then the value of every entry $\sigma(\check\alpha)$ is determined by some condition on the generic branch. Since $\kappa$ remains regular in the extension, these conditions will be bounded on the branch, and so there will be some piece of the branch, a condition in the forcing, that decides all the values $\sigma(\check\alpha)$ for $\alpha<\beta$. So this condition forces that $\sigma$ is equal to a $\beta$-sequence from the ground model. So the forcing adds no new ${<}\kappa$-sequences over the ground model, and hence preserves all cardinals and cofinalities below $\kappa$. $\Box$
The argument works more generally to show that whenever one is forcing with a normal tree of height $\kappa$, and $\kappa$ remains regular in the extension, then the forcing will add no new ${<}\kappa$-sequences.
