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

Solutions 153
Let C be the field of complex numbers. Consider the mapping of
the field C into the field M such that
Comparing formulae (3.2) and (3.3) with formulae (2.4) and (2.5) of §2.2
we obtain that is a homomorphism of C into M with respect to the
addition and with respect to the multiplication. Since is
a subgroup (cf., 150) of M with respect to the addition and with respect
to the multiplication. As the operations of addition and multiplication
possess the properties of commutativity, associativity, and distributivity
in M, this obviously also holds in Hence is a field.
If then Moreover, if then
is a real number, but this cannot be true because the
square of any number is never equal to –1. Consequently and
thus also Hence and Consequently the elements
of the form are different for different pairs It follows that
the mapping defined above is a bijective mapping of the field C onto
the field Moreover, since is a homomorphism, is an isomorphism
of the field C in the field i.e., the field is isomorphic to the field
of complex numbers.
214. Let the polynomial considered be reducible, i.e.:
where all the and are real numbers and
Put in place of in the first member of this equation. Since
M is a field one can eliminate the brackets, carrying out the product as
usual. We obtain in this way the initial polynomial, in which is replaced
by i.e.:
By hypothesis