Apparent contradiction to Abel-Rouffini Theorem, using implicit function theorem.

Question

Let $f:\mathbb{R}\times\mathbb{R}^{n+1}\rightarrow \mathbb{R}$, as $f(x,a)=\sum_{0\leq i\leq n} a_i x^i$.

If we assume that the polynomial that function $f$ represents has a simple zero $c_0$, i.e. $f(x,a)=p(x)=(x-c_0)q(x)$ with $q(c_0)\neq 0$, then after some calculations, we can see that the implicit function theorem allows us to find some neighbourhoods of $c_0$ and $a_0$ (where $f(c_0,a_0)=0$), and a $C^{\infty}$ function $\psi$ where $\psi(a)=c$, for all $c \in B_{c_0}$ and $a \in B_{a_0}$. Hence, $c_0=\psi(a_0)$.

In what way does this not contradict the Abel-Ruffini Theorem which states no general formula exists for pol. of degree equal or higher than 5?

Any help would be appreciated.

You have expressed the zero as a smooth function of the coefficients. The theorem is about a formula involving roots. You have not proved that your smooth function can be expressed in terms of simple roots. That would be a contradiction of the theorem. — ziggurism, Nov 11 '16 at 13:01

score 0 · Accepted Answer · edited Apr 13 '17 at 12:21

The Abel-Ruffini theorem does not state that there do not exist any closed-form formula, it only says that no such formula can be expressible in radicals. That is: there is no finite formula that can be written down using only the coefficents and rational numbers under compositions of the simple algebraic functions $+,-,*,/$ and $\sqrt[n]{}$ when the degree of the polynomial is larger than $4$ (and which holds for all polynomials of given degree. One can write down formulas for special cases e.g. the roots of $x^5 - a$ is expressible in radicals. For more info see Galois theory).

General formulas can be found using special functions which typically can only be expressed using infinite series or integrals. See e.g. the related MSE questions [1] and [2].

As for your argument: the implicit function theorem only tells us that $\psi$ is a locally smooth function, it does not say anything about it being expressible in radicals so there is no contradiction.

Apparent contradiction to Abel-Rouffini Theorem, using implicit function theorem.

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