[proofplan]
We first prove that the proposed addition and multiplication do not depend on the chosen representatives of the cosets. Addition uses that $I$ is an additive subgroup of $R$, while multiplication uses both left and right absorption by the two-sided ideal $I$. Once the operations are well-defined, every ring axiom follows by applying the corresponding axiom in $R$ to representatives and then passing to cosets. Finally, the quotient map preserves the ring operations by definition, is surjective because every coset has a representative, and has kernel exactly $I$.
[/proofplan]
[step:Show that addition on cosets is independent of representatives]
Let $a,a',b,b' \in R$ satisfy $a+I=a'+I$ and $b+I=b'+I$. By equality of additive cosets, this means $a-a' \in I$ and $b-b' \in I$. Since $I$ is an additive subgroup of the additive group of $R$, it is closed under addition. Hence
\begin{align*}
(a+b)-(a'+b')=(a-a')+(b-b') \in I.
\end{align*}
Therefore $(a+b)+I=(a'+b')+I$. Thus the rule
\begin{align*}
(a+I)+(b+I)=(a+b)+I
\end{align*}
is well-defined.
[/step]
[step:Show that multiplication on cosets is independent of representatives]
Let $a,a',b,b' \in R$ satisfy $a+I=a'+I$ and $b+I=b'+I$. Then $a-a' \in I$ and $b-b' \in I$. Since $I$ is a two-sided ideal, $(a-a')b \in I$ and $a'(b-b') \in I$. Using distributivity in $R$, we compute
\begin{align*}
ab-a'b'=(a-a')b+a'(b-b').
\end{align*}
The right-hand side lies in $I$ because $I$ is closed under addition. Hence $ab-a'b' \in I$, so $ab+I=a'b'+I$. Thus the rule
\begin{align*}
(a+I)(b+I)=ab+I
\end{align*}
is well-defined.
[guided]
We must prove that multiplication of cosets gives the same answer no matter which representative of each coset is chosen. Suppose $a,a',b,b' \in R$ are representatives with
\begin{align*}
a+I=a'+I
\end{align*}
and
\begin{align*}
b+I=b'+I.
\end{align*}
By the definition of equality of additive cosets, these two equalities mean
\begin{align*}
a-a' \in I
\end{align*}
and
\begin{align*}
b-b' \in I.
\end{align*}
The goal is to prove that the products represent the same coset, namely that $ab+I=a'b'+I$. Again using the coset criterion, it is enough to prove
\begin{align*}
ab-a'b' \in I.
\end{align*}
We rewrite the difference in a form where the ideal property can be applied:
\begin{align*}
ab-a'b'=(a-a')b+a'(b-b').
\end{align*}
This identity follows by expanding the right-hand side:
\begin{align*}
(a-a')b+a'(b-b')=ab-a'b+a'b-a'b'=ab-a'b'.
\end{align*}
Now the hypothesis that $I$ is two-sided is used exactly here. Since $a-a' \in I$ and $I$ absorbs multiplication on the right by arbitrary elements of $R$, we have $(a-a')b \in I$. Since $b-b' \in I$ and $I$ absorbs multiplication on the left by arbitrary elements of $R$, we have $a'(b-b') \in I$. Because $I$ is closed under addition, their sum lies in $I$:
\begin{align*}
(a-a')b+a'(b-b') \in I.
\end{align*}
Therefore $ab-a'b' \in I$, and hence $ab+I=a'b'+I$. This proves that multiplication on $R/I$ is well-defined.
[/guided]
[/step]
[step:Verify the ring axioms by passing the identities from $R$ to cosets]
The operations are now well-defined binary operations on $R/I$. Let $a,b,c \in R$. Additive associativity follows from associativity of addition in $R$:
\begin{align*}
((a+I)+(b+I))+(c+I)=((a+b)+c)+I=(a+(b+c))+I=(a+I)+((b+I)+(c+I)).
\end{align*}
Additive commutativity follows from commutativity of addition in $R$:
\begin{align*}
(a+I)+(b+I)=(a+b)+I=(b+a)+I=(b+I)+(a+I).
\end{align*}
The additive identity is $0_R+I$, since
\begin{align*}
(a+I)+(0_R+I)=(a+0_R)+I=a+I.
\end{align*}
The additive inverse of $a+I$ is $(-a)+I$, since
\begin{align*}
(a+I)+((-a)+I)=(a-a)+I=0_R+I.
\end{align*}
Multiplicative associativity follows from associativity of multiplication in $R$:
\begin{align*}
((a+I)(b+I))(c+I)=((ab)c)+I=(a(bc))+I=(a+I)((b+I)(c+I)).
\end{align*}
The multiplicative identity is $1_R+I$, since
\begin{align*}
(1_R+I)(a+I)=(1_Ra)+I=a+I
\end{align*}
and
\begin{align*}
(a+I)(1_R+I)=(a1_R)+I=a+I.
\end{align*}
Left distributivity follows from left distributivity in $R$:
\begin{align*}
(a+I)((b+I)+(c+I))=(a(b+c))+I=(ab+ac)+I=(a+I)(b+I)+(a+I)(c+I).
\end{align*}
Right distributivity follows from right distributivity in $R$:
\begin{align*}
((a+I)+(b+I))(c+I)=((a+b)c)+I=(ac+bc)+I=(a+I)(c+I)+(b+I)(c+I).
\end{align*}
Thus $R/I$ is a unital ring.
[/step]
[step:Prove that the quotient map is a surjective unital ring homomorphism]
Define
\begin{align*}
\pi:R&\to R/I
\end{align*}
by $\pi(a)=a+I$ for each $a \in R$. For all $a,b \in R$, the definitions of the quotient operations give
\begin{align*}
\pi(a+b)=(a+b)+I=(a+I)+(b+I)=\pi(a)+\pi(b)
\end{align*}
and
\begin{align*}
\pi(ab)=ab+I=(a+I)(b+I)=\pi(a)\pi(b).
\end{align*}
Also,
\begin{align*}
\pi(1_R)=1_R+I,
\end{align*}
which is the multiplicative identity of $R/I$. Hence $\pi$ is a unital ring homomorphism.
To prove surjectivity, let $x \in R/I$. By definition of $R/I$, there exists $a \in R$ such that $x=a+I$. Therefore $x=\pi(a)$, so $\pi$ is surjective.
[/step]
[step:Identify the kernel of the quotient map]
The kernel of $\pi$ is
\begin{align*}
\ker \pi=\{a \in R:\pi(a)=0_R+I\}.
\end{align*}
For $a \in R$, we have
\begin{align*}
a \in \ker \pi \iff a+I=0_R+I \iff a-0_R \in I \iff a \in I.
\end{align*}
Therefore $\ker \pi=I$. This completes the construction of the quotient ring and the verification of the stated properties of the quotient map.
[/step]
