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

154 Problems of Chapter 2
and thus (cf., 197) at least one of the two polynomials above within
brackets is equal to zero. By dividing this polynomial by its leading
coefficient or we obtain a vanishing expression of the type of (2.6),
having a degree lower than We thus obtain a contradiction of being
the lowest degree for which an expression of the type of (2.6) vanishes.
Therefore the claim that the polynomial considered is reducible is not
true.
215. As we had proved, in the case (a) the element of the field M
satisfies the equation:
where and are some real numbers, and is not reducible
over the field of real numbers. We have
If then for some real number Thus
i.e., the polynomial should be reducible over the field of the
real numbers. It follows that i.e., for a certain
non-zero real real number Since in the field M in M
we have
Consequently
Hence the element
belonging to M, is the element sought.
216. Answer. The sole field satisfying the required properties is (up to
isomorphism) the field whose elements are fractions, having polynomials