> [!NOTE] **Theorem** ($\mathbb{Z}/n\mathbb{Z}$ is a commutative ring) > Given an integer $m\geq 2$, the set of [[Integers modulo n|congruence classes mod m]] $\mathbb{Z}/m\mathbb{Z}$ forms a [[Commutative Ring]] form a ring under [[Modular arithmetic|modular addition and multiplication]] $+_{n}$ and $\times_{n}$. **Proof.** We've shown that [[Integers Modulo n is a Group|Z mod n Z is a group under addition]]. Clearly $\mathbb{Z} \text{/} m\mathbb{Z}$ is closed under $\times_{m}.$ Now given $a,b,c \in \mathbb{Z}$ we have: Associativity of multiplication as follows: $\begin{align}[a]_{m} \times_{m} ([b]_{m} \times_{m} [c]_{m}) &= [a_{m}] \times_{m} [bc]_{m} \\ &= [a(bc)]_{m} \\ &= [(ab)c]_{m} \\ &= [ab]_{m} \times_{m} [c]_{m} \\ &= ([a]_{m} \times_{m} [b]_{m}) \times_{m} [c]_{m} \end{align}$ Existence of multiplicative identity: $[1]_{m} \times_{m} [a]_{m} = [1 \cdot a]_{m} = [a]_{m} = [a \cdot 1]_{m} = [a]_{m} \times_{m} [1]_{m}$ Distributivity: $\begin{align}[a]_{m} \times_{m} ([b]_{m} +_{m} [c]_{m}) &= [a]_{m} \cdot [b+ c]_{m} \\ &= [a(b+c)]_{m} = [ab+ac]_{m} \\&= [ab]_{m} +[ac]_{m} \\ &=[a]_{m} \times_{m} [b]_{m} +_{m} [a]_{m} \times_{m} [c]_{m} \end{align}$