Is a factorable polynomial invertible?

Question

The reason there exists no quintic formula that finds the roots of a quintic polynomial is simply because some quintic polynomials are irreducible.

But reducible quintic polynomials may be invertible in a case by case scenario:

$$f(x)=x^5-1$$

$$f^{-1}(x)=\sqrt[5]{x+1}$$

But are all reducible quintic polynomials invertible? Proof please?

Answer 1

$$-x^5-x$$ is clearly reducible, as its equal to $-x(x^4+1)$. However, $f^{-1}$ is the Bring radical, which is not expressible in terms of regular radicals.

If you want a quintic without any inverses at all, not just a quintic whose inverse is not expressible by radicals, any quintic that's not injective (such as $x^5-x$) will do.

Answer 2

I take it your "invertible" means that there is an expression in radicals $g(y)$ such that for all $y$ in some interval, $f(g(y)) = y$. But if $f$ is any polynomial of degree $d > 1$, there is a set of $y$ dense in some interval such that $f(x) - y$ is reducible: namely take $y = f(r)$ for rational $r$, in which case $f(x) - f(r)$ is divisible by $x-r$. Thus for any non-invertible $f$, $f(x) - f(r)$ for rational $r$ is a counterexample.