Let C={a+bi|a,b in Z}. Show that C is an integral domain.
Please show details of this proof. I'm very confused and this is a practice problem for my final exam.
Thank you!
More directly.
$\displaystyle (a+bi)(c+di)=ac-bd + (ad+bc)i=0$
1)
$\displaystyle ac=bd$
2)
$\displaystyle ad=-bc$
multiply 1) by d on both sides
$\displaystyle acd=bd^2$
Apply 2
$\displaystyle -bc^2=bd^2$
$\displaystyle b(c^2+d^2)=0$
Case1) b=0 then your first one a+bi is just an integer which is definitely not a zero divisor, I think you can check that easily enough
Case2) c^2+d^2=0 but this means both c and d must be 0 because integers squared are nonnegative. But that is exactly the condition that the second gaussian integer was 0.
Here is another way. You need to know that $\displaystyle |z_1z_2| = |z_1||z_2|$ where $\displaystyle z_1,z_2\in \mathbb{C}$.
If $\displaystyle (a+bi)(c+di) = 0 \implies |(a+bi)(c+di)| = 0 \implies |a+bi|\cdot |b+di| = 0 \implies$$\displaystyle (a^2+b^2)(c^2+d^2)=0$.
Thus, $\displaystyle a^2+b^2=c^2+d^2 = 0 \implies a=b=c=d=0$.
EDIT: It should be "a=b=0 OR c=d=0" not "a=b=0 AND c=d=0".