Abel's theorem in problems and solutions - School of Mathematics

The complex numbers 57
214. Let us choose, amongst all expressions of type (2.6), the expression
of minimal degree that is vanishing in M: the corresponding
equation is
Prove that the polynomial
is not reducible over the field of real numbers.
In the sequel we will prove (cf., 272) that every polynomial with
real coefficients of degree higher than 2 is reducible over the field of real
numbers. Hence in Problem 214 cannot be higher than 2. But since
(otherwise we should have and should be equal to the
real number we obtain that
Consequently in the case (a) there exist two real numbers and in
M which satisfy
and for which the polynomial is irreducible over the field of
real numbers.
215. Prove that in the case (a) the field M contains an element
such that
From the results of Problems 215 and 213 it follows that in the case
(a) the field M contains a field isomorphic to the field of complex
numbers. Therefore if the field M is a minimal extension of the field of
real numbers then the field M must coincide with As a consequence,
in the case (a) any minimal field which represents a minimal extension
of the field of real numbers coincides (i.e., it is isomorphic) with the field
of complex numbers. So in the case (a) there is only one field (up to
isomorphism) which is a minimal extension of the field of real numbers,
namely the field of complex numbers.
216. Find all fields that are minimal extensions of the field of real
numbers in the case (b).