[proofplan] We first define the residue-class operations and prove that they do not depend on the chosen integer representatives. The only non-formal point is well-definedness, which follows from [compatibility of congruence with addition and multiplication](/theorems/9334). Once the operations are well-defined, every ring identity in $\mathbb{Z}/n\mathbb{Z}$ is obtained by choosing representatives, using the corresponding integer identity in $\mathbb{Z}$, and passing back to congruence classes. [/proofplan]

[step:Define the operations on residue classes and prove they are well-defined]For each $a \in \mathbb{Z}$, let $\bar a \in \mathbb{Z}/n\mathbb{Z}$ denote the congruence class of $a$ modulo $n$. Define binary operations \begin{align*} +: (\mathbb{Z}/n\mathbb{Z}) \times (\mathbb{Z}/n\mathbb{Z}) \to \mathbb{Z}/n\mathbb{Z} \end{align*} and \begin{align*} \cdot: (\mathbb{Z}/n\mathbb{Z}) \times (\mathbb{Z}/n\mathbb{Z}) \to \mathbb{Z}/n\mathbb{Z} \end{align*} by \begin{align*} (\bar a,\bar b) \mapsto \overline{a+b} \end{align*} and \begin{align*} (\bar a,\bar b) \mapsto \overline{ab}. \end{align*} We prove that these definitions are independent of representatives. Suppose $\bar a=\bar a'$ and $\bar b=\bar b'$, where $a,a',b,b' \in \mathbb{Z}$. Then $a \equiv a' \pmod n$ and $b \equiv b' \pmod n$. By [citetheorem:9334], \begin{align*} a+b \equiv a'+b' \pmod n \end{align*} and \begin{align*} ab \equiv a'b' \pmod n. \end{align*} Therefore \begin{align*} \overline{a+b}=\overline{a'+b'} \end{align*} and \begin{align*} \overline{ab}=\overline{a'b'}. \end{align*} Thus both operations are well-defined on $\mathbb{Z}/n\mathbb{Z}$.[/step]

[guided]The point that must be checked before any ring axiom is meaningful is that the formulas for addition and multiplication do not depend on the chosen representatives. A residue class such as $\bar a$ can also be written as $\bar a'$ whenever $a \equiv a' \pmod n$, so the formula $\bar a+\bar b=\overline{a+b}$ would be invalid unless replacing $a$ and $b$ by congruent integers gives the same output class. Define the maps \begin{align*} +: (\mathbb{Z}/n\mathbb{Z}) \times (\mathbb{Z}/n\mathbb{Z}) \to \mathbb{Z}/n\mathbb{Z} \end{align*} and \begin{align*} \cdot: (\mathbb{Z}/n\mathbb{Z}) \times (\mathbb{Z}/n\mathbb{Z}) \to \mathbb{Z}/n\mathbb{Z} \end{align*} by sending $(\bar a,\bar b)$ to $\overline{a+b}$ and $\overline{ab}$, respectively. To prove these are well-defined maps, assume that the same input classes are represented in another way: \begin{align*} \bar a=\bar a' \end{align*} and \begin{align*} \bar b=\bar b'. \end{align*} By the definition of congruence classes, this means \begin{align*} a \equiv a' \pmod n \end{align*} and \begin{align*} b \equiv b' \pmod n. \end{align*} The compatibility theorem for congruence, [citetheorem:9334], applies exactly to these two congruences. It gives \begin{align*} a+b \equiv a'+b' \pmod n \end{align*} and \begin{align*} ab \equiv a'b' \pmod n. \end{align*} Passing from congruence to equality of residue classes, we obtain \begin{align*} \overline{a+b}=\overline{a'+b'} \end{align*} and \begin{align*} \overline{ab}=\overline{a'b'}. \end{align*} Hence the output of each operation depends only on the classes $\bar a$ and $\bar b$, not on the particular integers $a$ and $b$ used to represent them. This proves that addition and multiplication modulo $n$ are well-defined binary operations on $\mathbb{Z}/n\mathbb{Z}$.[/guided]

[step:Verify that addition makes $\mathbb{Z}/n\mathbb{Z}$ an abelian group] Let $\bar a,\bar b,\bar c \in \mathbb{Z}/n\mathbb{Z}$ with $a,b,c \in \mathbb{Z}$. Using associativity of addition in $\mathbb{Z}$, \begin{align*} (\bar a+\bar b)+\bar c=\overline{(a+b)+c}=\overline{a+(b+c)}=\bar a+(\bar b+\bar c). \end{align*} Using commutativity of addition in $\mathbb{Z}$, \begin{align*} \bar a+\bar b=\overline{a+b}=\overline{b+a}=\bar b+\bar a. \end{align*} The class $\bar 0$ is an additive identity because \begin{align*} \bar a+\bar 0=\overline{a+0}=\bar a \end{align*} and \begin{align*} \bar 0+\bar a=\overline{0+a}=\bar a. \end{align*} Finally, $\overline{-a}$ is an additive inverse of $\bar a$ because \begin{align*} \bar a+\overline{-a}=\overline{a+(-a)}=\bar 0 \end{align*} and \begin{align*} \overline{-a}+\bar a=\overline{(-a)+a}=\bar 0. \end{align*} Thus $(\mathbb{Z}/n\mathbb{Z},+)$ is an abelian group. [/step]

[step:Verify that multiplication is associative, commutative, and has identity $\bar 1$] Let $\bar a,\bar b,\bar c \in \mathbb{Z}/n\mathbb{Z}$ with $a,b,c \in \mathbb{Z}$. Using associativity of multiplication in $\mathbb{Z}$, \begin{align*} (\bar a\bar b)\bar c=\overline{(ab)c}=\overline{a(bc)}=\bar a(\bar b\bar c). \end{align*} Using commutativity of multiplication in $\mathbb{Z}$, \begin{align*} \bar a\bar b=\overline{ab}=\overline{ba}=\bar b\bar a. \end{align*} The class $\bar 1$ is a multiplicative identity because \begin{align*} \bar a\bar 1=\overline{a \cdot 1}=\bar a \end{align*} and \begin{align*} \bar 1\bar a=\overline{1 \cdot a}=\bar a. \end{align*} Therefore multiplication on $\mathbb{Z}/n\mathbb{Z}$ is associative and commutative, and has identity element $\bar 1$. [/step]

[step:Verify distributivity and conclude the ring axioms] Let $\bar a,\bar b,\bar c \in \mathbb{Z}/n\mathbb{Z}$ with $a,b,c \in \mathbb{Z}$. Using the distributive law in $\mathbb{Z}$, \begin{align*} \bar a(\bar b+\bar c)=\overline{a(b+c)}=\overline{ab+ac}=\bar a\bar b+\bar a\bar c. \end{align*} Similarly, \begin{align*} (\bar a+\bar b)\bar c=\overline{(a+b)c}=\overline{ac+bc}=\bar a\bar c+\bar b\bar c. \end{align*} The preceding steps show that addition is an abelian group operation, multiplication is associative with identity $\bar 1$, multiplication distributes over addition on both sides, and multiplication is commutative. Hence $\mathbb{Z}/n\mathbb{Z}$ is a commutative ring with identity $\bar 1$. [/step]