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I understand there is a result that is proved using abstract algebra like Galois theory that there isn't a formula for degree $5$ or higher polynomials.

Does this just apply to polynomials whose coefficients are integers, real numbers, complex numbers or any field?

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  • $\begingroup$ I have read the proof that it follows for integers, whence it follows for all supersets. I do not know about the question for finite fields, but then in finite fields you have to seek roots in the algebraic closure of the field right? So I am not aware of the answer. $\endgroup$ – астон вілла олоф мэллбэрг Mar 17 '17 at 10:59
  • $\begingroup$ Thanks for the comment. I don't like Villa though :( $\endgroup$ – Ben B Mar 17 '17 at 11:02
  • $\begingroup$ You are welcome (to ask further of me, and to not like Villa). $\endgroup$ – астон вілла олоф мэллбэрг Mar 17 '17 at 11:03
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This result (Abel's theorem) is valid for any field of characteristic $0$, and relies on the following result, proved by Galois:

A polynomial equation over a field of characteristic $0$ is solvable by radicals if and only if its Galois group is solvable.

Now with literal coefficients, the Galois group of a polynomial of degree $n$ is the symmetric group $S_n$, which is not solvable for $n\ge 5$. For details on what is a solvable group, you can see Wikipedia.

The situation is more complex in characteristic $p>0$, because if a root has order divisible to the characteristic, it gives rise to a so-called non-separable extension, i.e. it is not a simple root of its minimal polynomial, contrary to the case of characteristic $0$.

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  • $\begingroup$ Re: characteristic $p>0$. Adding the point that over a finite field any finite extension is cyclic, hence solvable. $\endgroup$ – Jyrki Lahtonen Mar 18 '17 at 7:38

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