The free group on two generators $F\_2=\langle x,y|\rangle$ is the fundamental group of $\mathbb P^1(\mathbb C)\setminus\{0,1,\infty\}$. Now, there are plenty of galois covers of this space whose galois group is not solvable. Thus the "maximal solvable cover" (i.e. the limit over all galois covers with solvable galois group) is not the universal cover, but rather a quotient thereof. In other words, the natural map:
$$F\_2\to\lim\_{\begin{smallmatrix}\longleftarrow\cr H\unlhd F\_2\cr F\_2/H\text{ finite solvable}\end{smallmatrix}}F\_2/H$$
is not injective.


Can someone exhibit an explicit element of the kernel? What about the shortest element (by word length in $F\_2$) in the kernel? In other words, the question is: what universal word in $x,y$ always vanishes when $x,y$ are specialized to elements of some solvable group $G$? (note that since $G$ is solvable, so is the subgroup generated by $x$ and $y$).


Such an element now has the following seemingly impossible property. Consider it as a closed path $\gamma$ in $\mathbb P^1(\mathbb C)\setminus\{0,1,\infty\}$. Now try to lift $\gamma$ to the cover of $\mathbb P^1(\mathbb C)\setminus\{0,1,\infty\}$ corresponding to the linking number with $0$ (i.e. we take the universal cover of $\mathbb P^1(\mathbb C)\setminus\{0,\infty\}$ and take the inverse image of $\mathbb P^1(\mathbb C)\setminus\{0,1,\infty\}$). Of course $\gamma$ lifts to a *closed* curve, since otherwise we just exhibited an abelian quotient of $F\_2$ in which $\gamma$ is not sent to zero. The cover we just considered is $\mathbb P^1(\mathbb C)\setminus\mathbb Z$, and we can try to lift $\gamma$ to some abelian cover of this, etc. Of course, $\gamma$ always lifts to a *closed* curve, since all these covers are solvable! I'm having a hard time visualizing what such a curve $\gamma$ would look like geometrically in $\mathbb P^1(\mathbb C)\setminus\{0,1,\infty\}$ (I once thought all elements of $F\_2$ could be "broken" by a sequence of such covers!)



Explicit element in free group which is killed by every solvable quotient
There are no such elements -- the intersection of the derived series of a free group is trivial. In fact, even more is true -- the intersection of the lower central series of a free group is trivial. This is a theorem of Magnus, and by now there are many proofs. The classical one is in the final chapter of Magnus-Karass-Solitar's book on combinatorial group theory.


By the way, a topological proof of this fact (lifting curves to covers to resolve self-intersections, etc) is contained in my paper "On the self-intersections of curves deep in the lower central series of a surface group" with Justin Malestein.


EDIT : I see that you really want finite solvable quotients, not general solvable quotients. It is still true. Fixing a prime $p$, there is a ``mod $p$ lower central series'' of a group whose quotients are $p$-groups (so finite nilpotent if the group is finitely generated). For a free group, Zassenhaus proved in his paper 


H. Zassenhaus, Ein Verfahren, jeder endlichen p-Gruppe eine Lie-Ring mit der
Charakteristik p zuzuordnen, Abh. Math. Sem. Hamburg Univ. 13 (1939), 200-207.


that the intersection of the mod $p$ lower central series of a free group is trivial. This can also be deduced from the paper I mentioned with Justin Malestein, at least for the prime $2$ (one of the proofs we give actually yields regular covers whose order is a power of $2$).


