91Ó°ÊÓ

To do this, we'll need to show that the addition and multiplication of two elements in \(\mathbb{Z}[i]\) are closed, associative, commutative, and have their respective identities and inverses. 1. Let \(z_1 = a_1 + b_1i\) and \(z_2 = a_2 + b_2i\) be two elements in \(\mathbb{Z}[i]\). Then, \((z_1 + z_2) = (a_1 + b_1i) + (a_2 + b_2i) = (a_1 + a_2) + (b_1 + b_2)i\), which is in the form of \(a + bi\), meaning \(\mathbb{Z}[i]\) is closed under addition. 2. Addition is associative and commutative in \(\mathbb{Z}[i]\) because it is associative and commutative in \(\mathbb{Z}\). 3. The additive identity is \(0 = 0 + 0i\), and the additive inverse of \(a + bi\) is \(-a - bi\). Since \(0, a, b \in \mathbb{Z}\), these elements also belong to \(\mathbb{Z}[i]\). 4. Let \(z_1 = a_1 + b_1i\) and \(z_2 = a_2 + b_2i\) be two elements in \(\mathbb{Z}[i]\). Their product, \((z_1 * z_2) = (a_1 + b_1i)(a_2 + b_2i) = (a_1a_2 - b_1b_2) + (a_1b_2 + a_2b_1)i\), which is also in the form of \(a + bi\) and thus shows closure under multiplication. 5. Multiplication is associative and commutative in \(\mathbb{Z}[i]\) because it is associative and commutative in \(\mathbb{Z}\). 6. The multiplicative identity is \(1 = 1 + 0i\), and each element \(z = a + bi \neq 0\) in \(\mathbb{Z}[i]\) has a multiplicative inverse \(\frac{1}{z} \notin \mathbb{Z}[i]\); however, this is not a requirement to be a commutative ring. Since \(\mathbb{Z}[i]\) is a commutative ring with unity, we can proceed to the next step.

Suppose we have non-zero elements \(x = a + bi\) and \(y = c + di\) in \(\mathbb{Z}[i]\) such that \(x * y = 0\). Let's find the norm \(N\), which is a multiplicative function, meaning \(N(x * y) = N(x)N(y)\). 1. For \(x = a + bi\), the norm is \(N(x) = a^2 + b^2\). 2. For \(y = c + di\), the norm is \(N(y) = c^2 + d^2\). Now, multiplying these elements, we have: \(xy = (a + bi)(c + di) = (ac - bd) + (ad + bc)i\). The norm of this product is \(N(xy) = (ac - bd)^2 + (ad + bc)^2\). Using the properties of the norm function, we can say that \(N(xy) = N(x)N(y) = (a^2 + b^2)(c^2 + d^2)\). Since \(x\) and \(y\) are non-zero, we have \(N(x) > 0\) and \(N(y) > 0\). Thus, \(N(xy) > 0\). However, we have assumed that \(x * y = 0\), which means \(N(xy) = N(0) = 0\). This is a contradiction since \(N(xy) > 0\) and \(N(xy) = 0\) cannot be simultaneously true. Therefore, there are no non-zero divisors in \(\mathbb{Z}[i]\). Since \(\mathbb{Z}[i]\) is a commutative ring with unity and has no non-zero divisors, we can conclude that the Gaussian integers, \(\mathbb{Z}[i]\), are an integral domain.