A proof using Yoneda lemma

Martin Brandenburg pointed out elsewhere in the comments that he could give a one line proof, using the Yoneda lemma, of
$$\frac{\mathbf{C}[x_1,\ldots,x_{n+m}]}{I(X)^e+I(Y)^e} \cong \frac{\mathbf{C}[x_1,\ldots,x_n]}{I(X)} \otimes_\mathbf{C} \frac{\mathbf{C}[x_{n+1},\ldots,x_{n+m}]}{I(Y)}$$ (for $X,Y$ affine algebraic varieties), but apparently the proof was too long for his margin. How can this be done?

Fundamental question aside: why is there no abstract nonsense tag?

Solutions Collecting From Web of "A proof using Yoneda lemma"

The claim, I suppose, is that for commutative $k$-algebras $R$ and $S$ and ideals $I \trianglelefteq R$ and $J \trianglelefteq S$, we have
$$(R \otimes_k S) / (I^e + J^e) \cong (R / I) \otimes_k (S / J)$$
where $I^e$ is the extension of $I$ along $R \to R \otimes_k S$ and similarly for $J^e$.

Well,
$$\mathbf{CAlg}_k ((R / I) \otimes_k (S / J), T) \cong \mathbf{CAlg}_k (R / I, T) \times \mathbf{CAlg}_k (S / J, T)$$
and
$$\mathbf{CAlg}_k (R / I, T) \cong \{ f \in \mathbf{CAlg}_k (R, T) : f (I) = 0 \}$$
$$\mathbf{CAlg}_k (S / J, T) \cong \{ g \in \mathbf{CAlg}_k (S, T) : g (J) = 0 \}$$
but
$$\mathbf{CAlg}_k (R, T) \times \mathbf{CAlg}_k (S, T) \cong \mathbf{CAlg}_k (R \otimes_k S, T)$$
so
$$\mathbf{CAlg}_k ((R / I) \otimes_k (S / J), T) \cong \{ h \in \mathbf{CAlg}_k (R \otimes_k S, T) : h (I S) = 0, h (R J) = 0 \}$$
but $I S = I^e$ and $R J = J^e$, and we have $h (I^e) = 0$ and $h (J^e) = 0$ if and only if $h (I^e + J^e) = 0$. Hence,
$$\mathbf{CAlg}_k ((R / I) \otimes_k (S / J), T) \cong \mathbf{CAlg}_k ((R \otimes_k S) / (I^e + J^e), T)$$
as required.

$$\hom(R/I \otimes S,-) = \hom(R/I,-) \times \hom(S,-) = \{f \in \hom(R,-):f(I)=0\} \times \hom(S,-)$$
$$ = \{h \in \hom(R \otimes S,-) : h(I \otimes 1)=0\}=\hom((R \otimes S)/I^e,-).$$
Hence, $R/I \otimes S = (R \otimes S)/I^e$. Then also $R/I \otimes S/J = (R \otimes S/J)/I^e = (R \otimes S)/(J^e,I^e)$.