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Let $C_K = \{ (r,s) \in R \times R \mid r-s \in K \}$ be the congruence defined by an ideal $K$.

Then, we have three maps defined on $S = C_I \otimes_R C_J$ :

• $\pi_0 : S \to C_I$ induced by the first projection $C_J \to R$ and the identity on $C_I$
• $\pi_1 : S \to C_J$ induced by the first projection $C_I \to R$ and the identity on $C_J$
• $\mu : S \to R \times R$ induced by the inclusions $C_I \to R\times R$ and $C_J \to R \times R$.

Letting $\Delta \subseteq R \times R$ be the image of the diagonal, define $T = \{ x \in S \mid \pi_0(x) \in \Delta \wedge \pi_1(x) \in \Delta \}$. I claim that $\mu(T) = C_{IJ}$.

On the level of $R$ modules, we have isomorphisms $C_K \cong R \oplus K$, such as $(r,s) \mapsto (r, s-r)$, and so we have

$$ S \cong R \oplus I \oplus J \oplus (I \otimes_R J) $$

In this form, the maps $\pi_i$ become projections onto the relevant summands, so $T$ is precisely the submodule $R \oplus (I \otimes_R J)$, so we've eliminated the $I$ and $J$ summands you were having trouble with.

By splitting the $R$-module maps, $T$ is genenerated as an $R$-module by elements of the form $$ (r, r+i) \otimes (s, s+j) - (0,i) \otimes (s,s) - (r,r) \otimes (j,0) $$ and applying $\mu$ to such a thing gives the element $(rs, rs + ij)$, and now it's easy to see that $\mu(T) = C_{IJ}$ as claimed.

The unitalization approach can be made to work.

Let $C_K = \{ (r,s) \in R \times R \mid r-s \in K \}$ be the congruence defined by an ideal $K$.

Then, we have three maps defined on $S = C_I \otimes_R C_J$ :

• $\pi_0 : S \to C_I$ induced by the first projection $C_J \to R$ and the identity on $C_I$
• $\pi_1 : S \to C_J$ induced by the first projection $C_I \to R$ and the identity on $C_J$
• $\mu : S \to R \times R$ induced by the inclusions $C_I \to R\times R$ and $C_J \to R \times R$.

Letting $\Delta \subseteq R \times R$ be the image of the diagonal, define $T = \{ x \in S \mid \pi_0(x) \in \Delta \wedge \pi_1(x) \in \Delta \}$. I claim that $\mu(T) = C_{IJ}$.

On the level of $R$ modules, we have isomorphisms $C_K \cong R \oplus K$, such as $(r,s) \mapsto (r, s-r)$, and so we have

$$ S \cong R \oplus I \oplus J \oplus (I \otimes_R J) $$

In this form, the maps $\pi_i$ become projections onto the relevant summands, so $T$ is precisely the submodule $R \oplus (I \otimes_R J)$, so we've eliminated the $I$ and $J$ summands you were having trouble with.

By splitting the $R$-module maps, $T$ is genenerated as an $R$-module by elements of the form $$ (r, r+i) \otimes (s, s+j) - (0,i) \otimes (s,s) - (r,r) \otimes (j,0) $$ and applying $\mu$ to such a thing gives the element $(rs, rs + ij)$, and now it's easy to see that $\mu(T) = C_{IJ}$ as claimed.

Let $C_K$ be the congruence defined by an ideal $K$.

Then, we have three maps defined on $S = C_I \otimes_R C_J$:

• $\pi_0 : S \to C_I$ induced by the projection $C_J \to R$
• $\pi_1 : S \to C_J$ induced by the projection $C_I \to R$
• $\mu : S \to R \times R$ induced by the inclusions $C_I \to R\times R$ and $C_J \to R \times R$.

Let $T = \{ x \in S \mid \pi_0(x) \in R \wedge \pi_1(x) \in R \}$. I claim that $\mu(T) = C_{IJ}$.

On the level of $R$ modules, we have the evident isomorphism $C_K \cong R \oplus K$, and we have

$$ S \cong R \oplus I \oplus J \oplus (I \otimes_R J) $$

The maps $\pi_i$ are projections onto the relevant summands, so $T$ is precisely the submodule $R \oplus (I \otimes_R J)$, so we've eliminated the $I$ and $J$ summands you were having trouble with.

By splitting the $R$-module maps, $T$ is genenerated as an $R$-module by elements of the form $$ (r, r+i) \otimes (s, s+j) - (0,i) \otimes (s,s) - (r,r) \otimes (j,0) $$ and applying $\mu$ to such a thing gives the element $(rs, rs + ij)$, and now it's easy to see that $\mu(T) = C_{IJ}$ as claimed.

Let $C_K = \{ (r,s) \in R \times R \mid r-s \in K \}$ be the congruence defined by an ideal $K$.

Then, we have three maps defined on $S = C_I \otimes_R C_J$  :

• $\pi_0 : S \to C_I$ induced by the first projection $C_J \to R$ and the identity on $C_I$
• $\pi_1 : S \to C_J$ induced by the first projection $C_I \to R$ and the identity on $C_J$
• $\mu : S \to R \times R$ induced by the inclusions $C_I \to R\times R$ and $C_J \to R \times R$.

Letting $\Delta \subseteq R \times R$ be the image of the diagonal, define $T = \{ x \in S \mid \pi_0(x) \in \Delta \wedge \pi_1(x) \in \Delta \}$. I claim that $\mu(T) = C_{IJ}$.

On the level of $R$ modules, we have isomorphisms $C_K \cong R \oplus K$, such as $(r,s) \mapsto (r, s-r)$, and so we have

$$ S \cong R \oplus I \oplus J \oplus (I \otimes_R J) $$

In this form, the maps $\pi_i$ become projections onto the relevant summands, so $T$ is precisely the submodule $R \oplus (I \otimes_R J)$, so we've eliminated the $I$ and $J$ summands you were having trouble with.

By splitting the $R$-module maps, $T$ is genenerated as an $R$-module by elements of the form $$ (r, r+i) \otimes (s, s+j) - (0,i) \otimes (s,s) - (r,r) \otimes (j,0) $$ and applying $\mu$ to such a thing gives the element $(rs, rs + ij)$, and now it's easy to see that $\mu(T) = C_{IJ}$ as claimed.

If you can construct augmentations, this argument should work.

The maps $\text{id}: R \oplus I \to R \oplus I$ annd $R \oplus J \to R \oplus I : (r,j) \mapsto (r,0)$ induce

$$ R \oplus I \oplus J \oplus (I \otimes_R J) \to R \oplus I : (r,i,j,k) \mapsto (r,i,0,0) $$

and similarly, for the maps to $R \oplus J$. The $I \otimes_R J$ summand is characterized as precisely the subgroup of things that vanish under both maps. The subring $R \oplus (I \otimes_R J)$ is precisely the elements whose image under both maps lies in $R$.

Let $C_K$ be the congruence defined by an ideal $K$.

Then, we have three maps defined on $S = C_I \otimes_R C_J$:

• $\pi_0 : S \to C_I$ induced by the projection $C_J \to R$
• $\pi_1 : S \to C_J$ induced by the projection $C_I \to R$
• $\mu : S \to R \times R$ induced by the inclusions $C_I \to R\times R$ and $C_J \to R \times R$.

Let $T = \{ x \in S \mid \pi_0(x) \in R \wedge \pi_1(x) \in R \}$. I claim that $\mu(T) = C_{IJ}$.

On the level of $R$ modules, we have the evident isomorphism $C_K \cong R \oplus K$, and we have

$$ S \cong R \oplus I \oplus J \oplus (I \otimes_R J) $$

The maps $\pi_i$ are projections onto the relevant summands, so $T$ is precisely the submodule $R \oplus (I \otimes_R J)$, so we've eliminated the $I$ and $J$ summands you were having trouble with.

By splitting the $R$-module maps, $T$ is genenerated as an $R$-module by elements of the form $$ (r, r+i) \otimes (s, s+j) - (0,i) \otimes (s,s) - (r,r) \otimes (j,0) $$ and applying $\mu$ to such a thing gives the element $(rs, rs + ij)$, and now it's easy to see that $\mu(T) = C_{IJ}$ as claimed.