A Computational Approach to Number Theory and Algebra - Solution

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Chapter 1

1.2 Ideals and greatest common divisors

Exercise 1.8

As is already closed under addition, it's sufficient to show that implies is closed under multiplication with any and verse versa.

On one hand, the closing under addition property already implies that by repeatedly adding for times. And this can be easily extended to negative number by repeatedly adding for times since . Simultaneously, by choosing any , , we can deduce that .

On the other hand, is an ideal instantly implies by multiplying with .

So is an ideal if and only if all conditions in the question holds.

Exercise 1.9

Observation 1:

Observation 2: : proof:

Suppose and . By definition of we have and , thus

implies , thus . When , we have . By observation 1 we have

  • The first equation: Firstly, . Secondly, by observation 1 and 2 its easy to show that is the maximum integer that divides . So .
  • The second equation: and is the only divisor of , thus

Suppose , by theorem 1.8, there exists integers so that , thus . We can easily show that is the minimal value for for integers by proof of contradiction, thus . So .

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