Let $R$ be a countable, principal ideal domain which is not a field and $A$ be a countable $R$-algebra which is free as an $R$-module. Then we will construct an $\aleph_1$-free $R$-module $G$ of rank $\aleph_1$ with endomorphism algebra End$_RG = A$. Clearly the result does not hold for fields. Recall that an $R$-module is $\aleph_1$-free if all its countable submodules are free, a condition closely related to Pontryagin's theorem. This result has many consequences, depending on the algebra $A$ in use. For instance, if we choose $A = R$, then clearly $G$ is an indecomposable `almost free' module. The existence of such modules was unknown for rings with only finitely many primes like $R = \hbox{\Bbbvii Z}_{(p)}$, the integers localized at some prime $p$. The result complements a classical realization theorem of Corner's showing that any such algebra is an endomorphism algebra of some torsion-free, reduced $R$-module $G$ of countable rank. Its proof is based on new combinatorial-algebraic techniques related with what we call {\it rigid tree-elements\/} coming from a module generated over a forest of trees.