Let k be a global field and let k_v be the completion of k with respect to v,
a non-archimedean place of k. Let \mathbf{G} be a connected, simply-connected
algebraic group over k, which is absolutely almost simple of k_v-rank 1. Let
G=\mathbf{G}(k_v). Let \Gamma be an arithmetic lattice in G and let C=C(\Gamma)
be its congruence kernel. Lubotzky has shown that C is infinite, confirming an
earlier conjecture of Serre. Here we provide complete solution of the
congruence subgroup problem for \Gamm$ by determining the structure of C. It is
shown that C is a free profinite product, one of whose factors is
\hat{F}_{\omega}, the free profinite group on countably many generators. The
most surprising conclusion from our results is that the structure of C depends
only on the characteristic of k. The structure of C is already known for a
number of special cases. Perhaps the most important of these is the
(non-uniform) example \Gamma=SL_2(\mathcal{O}(S)), where \mathcal{O}(S) is the
ring of S-integers in k, with S=\{v\}, which plays a central role in the theory
of Drinfeld modules. The proof makes use of a decomposition theorem of
Lubotzky, arising from the action of \Gamma on the Bruhat-Tits tree associated
with G.Comment: 27 pages, 5 figures, to appear in J. Reine Angew. Mat