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 .