Let me reiterate this: There can in principle be no purely algebraic proof of the FToA because the reals have no purely algebraic definition. (Unless you define them as a real closed field of transcendence degree c, but that leaves the FToA as a trivial consequence and cannot be what is wanted.) The proof I outlined, which someone showed me 45 years ago, uses only the fact that R is a complete ordered field. Given that that is the analytic definition of R, it is impossible to avoid. That fact is, of course, at the heart of the fact that the circle is not contractible in a punctured plane. Incidentally, even constructivists (well even Errett Bishop, anyway) agree that odd order real polynomials have a real root and that positive numbers have square roots, since there are obvious constructions for these things. Their real line is not complete (it is countable, but the missing numbers are not constructible), but these roots are there anyway. The argument I outlined is elementary, even if not especially easy. First you have to construct the reals, the least elementary part of the argument. Then comes the theorem on symmetric functions. It is not a deep result; it needs a careful proof, but a student can follow it without knowing anything sophisticated. The construction of a splitting field (without getting into UFDs) is a bit tricky. To adjoin a root to an irreducible polynomial p of degree n, you start with a vector space whose basis is called 1, u, u^2,..., u^{n-1} and define a multiplication, by having p(u) = 0. This is analogous to how you get from R to C. Of course, you use the division algorithm to show you get a field. Michael