[proofplan]
Use comaximality to choose elements $a \in I$ and $b \in J$ with $a+b=1$. These elements let us construct, for any pair of residue classes $(x+I,y+J)$, a single element $r \in R$ with the prescribed residues modulo $I$ and modulo $J$. We then identify the kernel as $I \cap J$, prove directly that comaximality forces $I \cap J=IJ$, and finally define the induced quotient map from $R/(IJ)$ to the product and prove it is an isomorphism.
[/proofplan]
[step:Check that the canonical map is a ring homomorphism]
The quotient rings $R/I$ and $R/J$ are well-defined because $I$ and $J$ are ideals of the commutative ring $R$. The product $R/I \times R/J$ is equipped with componentwise addition and multiplication.
For $r,s \in R$, the map $\Phi: R \to R/I \times R/J$ satisfies
\begin{align*}
\Phi(r+s)=((r+s)+I,(r+s)+J)=(r+I,r+J)+(s+I,s+J)=\Phi(r)+\Phi(s).
\end{align*}
It also satisfies
\begin{align*}
\Phi(rs)=(rs+I,rs+J)=(r+I,r+J)(s+I,s+J)=\Phi(r)\Phi(s).
\end{align*}
Finally,
\begin{align*}
\Phi(1_R)=(1_R+I,1_R+J),
\end{align*}
which is the multiplicative identity of $R/I \times R/J$. Hence $\Phi$ is a ring homomorphism.
[/step]
[step:Construct a preimage for every pair of residue classes]
Since $I+J=R$, there exist $a \in I$ and $b \in J$ such that
\begin{align*}
a+b=1_R.
\end{align*}
Let $(x+I,y+J) \in R/I \times R/J$ be arbitrary, where $x,y \in R$. Define
\begin{align*}
r:=bx+ay \in R.
\end{align*}
Then
\begin{align*}
r-x=bx+ay-x=(b-1_R)x+ay=-ax+ay \in I,
\end{align*}
because $a \in I$ and $I$ is an ideal. Therefore $r+I=x+I$. Similarly,
\begin{align*}
r-y=bx+ay-y=bx+(a-1_R)y=bx-by \in J,
\end{align*}
because $b \in J$ and $J$ is an ideal. Therefore $r+J=y+J$.
Thus
\begin{align*}
\Phi(r)=(x+I,y+J).
\end{align*}
Since the target element was arbitrary, $\Phi$ is surjective.
[guided]
The point of comaximality is that it gives a partition of the identity inside the two ideals. Since $I+J=R$, the element $1_R \in R$ belongs to $I+J$, so there are elements $a \in I$ and $b \in J$ with
\begin{align*}
a+b=1_R.
\end{align*}
Now take an arbitrary element of the product quotient ring. Such an element has the form $(x+I,y+J)$ for some $x,y \in R$. We need one element $r \in R$ that has residue $x$ modulo $I$ and residue $y$ modulo $J$. The element
\begin{align*}
r:=bx+ay
\end{align*}
does exactly this, because $b$ behaves like $1_R$ modulo $I$ and $a$ behaves like $1_R$ modulo $J$.
Indeed, since $a+b=1_R$, we have $b-1_R=-a$. Hence
\begin{align*}
r-x=bx+ay-x=(b-1_R)x+ay=-ax+ay.
\end{align*}
Both $-ax$ and $ay$ lie in $I$, because $a \in I$ and $I$ is an ideal. Therefore $r-x \in I$, so $r+I=x+I$.
Likewise $a-1_R=-b$, so
\begin{align*}
r-y=bx+ay-y=bx+(a-1_R)y=bx-by.
\end{align*}
Both $bx$ and $-by$ lie in $J$, because $b \in J$ and $J$ is an ideal. Therefore $r-y \in J$, so $r+J=y+J$.
Thus
\begin{align*}
\Phi(r)=(r+I,r+J)=(x+I,y+J).
\end{align*}
Since every pair of residue classes has such a preimage, $\Phi$ is surjective.
[/guided]
[/step]
[step:Identify the kernel as the intersection of the two ideals]
For $r \in R$, the element $r$ lies in $\ker \Phi$ precisely when
\begin{align*}
\Phi(r)=(I,J).
\end{align*}
This means exactly that $r+I=I$ and $r+J=J$, which is equivalent to $r \in I$ and $r \in J$. Hence
\begin{align*}
\ker \Phi=I \cap J.
\end{align*}
[/step]
[step:Use comaximality to prove that the intersection equals the product]
First, $IJ \subset I \cap J$. Indeed, each generator $uv$ with $u \in I$ and $v \in J$ lies in $I$ because $I$ is an ideal, and lies in $J$ because $J$ is an ideal. Since $I \cap J$ is an ideal, every finite sum of such products lies in $I \cap J$.
Conversely, let $z \in I \cap J$. Using $a+b=1_R$, we compute
\begin{align*}
z=z1_R=z(a+b)=za+zb.
\end{align*}
Since $z \in J$ and $a \in I$, commutativity gives $za=az \in IJ$. Since $z \in I$ and $b \in J$, we also have $zb \in IJ$. Therefore $z=za+zb \in IJ$. Thus $I \cap J \subset IJ$.
Combining both inclusions,
\begin{align*}
I \cap J=IJ.
\end{align*}
[/step]
[step:Define the quotient isomorphism induced by the canonical map]
Define
\begin{align*}
\Psi: R/(IJ) \to R/I \times R/J
\end{align*}
by
\begin{align*}
\Psi(r+IJ):=(r+I,r+J).
\end{align*}
This is well-defined: if $r+IJ=s+IJ$, then $r-s \in IJ$. Since $IJ=I \cap J$, we have $r-s \in I$ and $r-s \in J$, so $r+I=s+I$ and $r+J=s+J$.
The map $\Psi$ is a ring homomorphism because addition, multiplication, and the multiplicative identity are inherited from the already verified componentwise identities for $\Phi$. It is surjective because $\Phi$ is surjective. Its kernel consists of all classes $r+IJ$ such that
\begin{align*}
(r+I,r+J)=(I,J).
\end{align*}
By the kernel computation above, this is equivalent to $r \in I \cap J=IJ$, which means $r+IJ$ is the zero element of $R/(IJ)$. Hence $\ker \Psi=\{IJ\}$, so $\Psi$ is injective.
Therefore $\Psi$ is a bijective ring homomorphism, hence a ring isomorphism. Consequently
\begin{align*}
R/(IJ) \cong R/I \times R/J.
\end{align*}
This proves the theorem.
[/step]
