Question

Exercise 1.49. The norm of a is the product of a and its complex conjugate: N(a) = aa. Ifa = x +yi, then N(a) is the square of the distance from (0,0) to (x.y) in the complex plane. If a and b are complex numbers, then N(ab)- N(a)N(b). If a is in G, the norm of a is a nonnegative member of Z. If a and b are in G and a divides b in G, then N(a) divides N(b) in Z. Exercise 1.50. The ring G is an integral domain that is, a commutative ring with unity in which the product of any two nonzero members of G is nonzero. The same is true of Z. (One usually thinks of Z as the model for an integral domain.)

From last part of Q49

Answer 1

$a|b\text{ in G}\Rightarrow N(a)|N(b)\text{ in }\mathbb{Z}$

This is because $a|b\Rightarrow b=pa$ for some $p\in G$

And so $N(b)=b\bar{b}=(pa)(\overline{pa})=(pa)\left(\bar{p}\bar{a} \right )$

Which is $N(b)=\left (p\bar{p} \right )\left ( a\bar{a} \right )$ after rearrangement which we can do as the product of complex numbers is associative and commutative

So $N(b)=N(p)N(a)$

As where $p\in G$ must be a non-negative integer, we have

And so we have shown that

$a|b\text{ in G}\Rightarrow N(a)|N(b)\text{ in }\mathbb{Z}$ as required

$\blacksquare$

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