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

Solutions 149
is a field.
195. We have Since a field is a group
under addition, one can add to both members of the equations. One
obtains Since the multiplication in the field is commutative one
also has
196. 1) It follows
that In the same way one proves that
(cf., the point
197. If and then there exists an element the
inverse of the element Thus But
Hence
198. Answer. See Table 14
199. Let be a non-prime number‚ i.e.‚ where and
Thus modulo we have but and Since
this is not possible in a field (cf.‚ 197)‚ for non-prime the remainders
with the operations modulo do not form a field.
Observe now the following property. Let and be two integers
and and the remainders of their division by i.e.‚
and Thus and
We obtain that the numbers and
as well as and divided by give the same remainder. In other
words‚ we obtain the same result either if we first take the remainders of
the division of and by and afterwards their sum (or their product)
modulo or if we first take the sum (or the product) of the integers and
as usual‚ and later on we take the remainder of the division by of this
sum (or of this product). In this way‚ to calculate a certain expression
with the operations modulo one may take the remainders of the division
by not after each operation‚ but‚ after having made the calculations as
usual with integers‚ take only at the end the remainder of the division