Does the Abel-Ruffini Theorem contradict the Fundamental Theorem of Algebra?

Question

It is my understanding that the Abel-Ruffini Theorem implies that certain polynomial equations $(x^5-x+1=0$, for instance) have transcendental roots. However, the Fundamental Theorem of Algebra states that we can factorise any polynomial into quadratics, and we can then solve these with the quadratic equation (implying that the solution would be algebraic).

On second thought, the only way I can see this to be resolved is if the quadratic factors of our polynomials have transcendental coefficients. Is this the case?

Answer 1

Abel-Ruffini says that your polynomial has roots not expressible in terms of radicals. This is a much stronger condition than algebraicity: the field of algebraic numbers is a very large extension of the field of algebraic numbers expressible in radicals. This is in fact the very point of Abel's theorem.

Answer 2

The roots of $x^5-x+1=0$ are by definition algebraic. It is the product of a degree $1$ real polynomial, and two real quadratics. If we choose these factors to be monic (lead coefficient $1$), as we always can, then the coefficients of these three polynomials are algebraic.

This generalizes. Let $P(x)$ be a polynomial with real algebraic coefficients. Then in any factorization of $P(x)$ as a product of monic polynomials, all the coefficients of these polynomials are algebraic.

Answer 3

There isn't a contradiction here, but many people do get confused on this point. It is true that the Fundamental Theorem of Algebra allows us to factorize any polynomial into a product of quadratics; but here is the catch: the Fundamental Theorem of Algebra doesn't guarantee that the coefficients of the quadratics can be expressed in terms of radicals (remember that the Abel-Rufini Theorem states that there exist polynomial equations whose solutions cannot be expressed exactly with radicals.) All this assumes that we are working over $\mathbb{R}$.