Prove that the Gaussian integers Z [i] form an integral domain

Question

This is all I have done. Is this correct? please explain.

Proof. 0 = 0 + 0i is in Z [i] : Also note that 1 = 1 + 0i is in z[i].

Then (a + bi)(c + di) = (ac) + (bd)i is in Z [i] and (a + bi)(c + di) = (ac-bd) + (ad + bc)i is in Z [i]. Therefore Z [i] is a subring of C.

Hence Z [i] is a commutative ring with unity. Furthermore, if (a + bi) (c + di) = 0; then (as elements of the integral domain C) either a + bi = 0 or c + di = 0. Therefore Z [i] is an integral domain.

Answer 1

Your explaination of $\mathbb Z[i]$ being a subring of $\mathbb C$ is confusing. Are you using $xy$ to both denote the additive and multiplication operations on $x$ and $y$ in the ring structure?

You don't need to show that $0+0i\in\mathbb Z[i]$. To show that $(\mathbb Z[i],+,\times)$ is a subring of $(\mathbb C, +,\times)$, you only need to show the following (called the Subring Test):

1. $\mathbb Z[i]\neq\emptyset$
2. $x+(-y)\in\mathbb Z[i]$ for all $x, y\in\mathbb Z[i]$
3. $xy\in\mathbb Z[i]$ for all $x, y\in\mathbb Z[i]$

Showing this:

1. Non-emptyness is clear.
2. $(a+bi)-(c+di)=(a-c)+(b-d)i\in\mathbb Z[i]$
3. You showed this.

No need to point out commutivity. Your explanation of why it is a domain is good. In fact, this same argument is used to show that every subring with unity of an integral domain is an integral domain.