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

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150 Problems <strong>of</strong> Chapter 2<br />

by <strong>of</strong> the number obta<strong>in</strong>ed. Let us use this property. Under addition<br />

modulo the rema<strong>in</strong>ders form a commutative group (see 40). S<strong>in</strong>ce for all<br />

<strong>in</strong>tegers the numbers <strong>and</strong> are equal‚ <strong>and</strong> consequently<br />

the rema<strong>in</strong>ders <strong>of</strong> their division by are also equal‚ the equality<br />

also holds modulo i.e.‚ distributivity is satisfied. In<br />

the same way one verifies the associativity <strong>and</strong> the commutativity <strong>of</strong><br />

the multiplication modulo <strong>and</strong> The unit<br />

element for the multiplication is 1. It rema<strong>in</strong>s to prove that for prime<br />

every rema<strong>in</strong>der different from 0 has an <strong>in</strong>verse‚ i.e.‚ that there exists a<br />

rema<strong>in</strong>der such that modulo<br />

Let Consider the numbers (under<br />

the usual multiplication). The difference between two <strong>of</strong> these numbers<br />

is not divisible by because is prime <strong>and</strong><br />

<strong>and</strong> As a consequence all these numbers divided by<br />

give dist<strong>in</strong>ct rema<strong>in</strong>ders‚ <strong>and</strong> thus all the rema<strong>in</strong>ders possible. This<br />

means that one <strong>of</strong> these numbers divided by gives as rema<strong>in</strong>der 1‚ i.e.‚<br />

for a certa<strong>in</strong> rema<strong>in</strong>der one has modulo Hence for prime<br />

all the properties <strong>of</strong> fields are satisfied.<br />

200. S<strong>in</strong>ce it follows that<br />

201. Subtract from<br />

the equality<br />

One obta<strong>in</strong>s<br />

so<br />

If the degree <strong>of</strong> the polynomial <strong>in</strong> the left member <strong>of</strong> the<br />

equality (3.1) is not lower than the degree <strong>of</strong> the polynomial But<br />

the degree <strong>of</strong> the polynomial <strong>in</strong> the right member is strictly lower than<br />

the degree <strong>of</strong> the polynomial From this contradiction it follows

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