$\DeclareMathOperator{\quot}{Quot}\DeclareMathOperator{\tr}{tr}$ Let $I_1$ and $I_2$ be nontrivial ideals in $\mathbb C[x_1,\ldots,x_k]$ and $\mathbb C[y_1,\ldots,y_m]$, respectively. Define $$ R_1 = \mathbb C[x_1,\ldots,x_k] / I_1, \quad R_2 = \mathbb C[y_1,\ldots,y_m] / I_2, \\ R_3 = \mathbb C[x_1,\ldots,x_k,y_1,\ldots,y_m]/(I_1+I_2). $$ If we denote by $n_1$, $n_2$ and $n_3$ the transcendence degree of $\quot(R_1)$, $\quot(R_2)$, $\quot(R_3)$ over $\mathbb C$, respectively, then $n_1+n_2 = n_3$. How to show this equality?
I tried to select transcendence bases for $\quot(R_1)$ and $\quot(R_2)$ over $\mathbb C$ and then to show that their union will be a transcendence basis for $\quot(R_3)$. But unfortunately I don't see how to show the algebraic independence and the generating property for this union of bases. Maybe my approach is incorrect and there is some other way to show the desired equality?
P.S. The equality in the question implies that the dimension of the product of algebraic varieties is the sum of their dimensions since the dimension can be defined as the transcendence degree of the functions field over $\mathbb C$.
Update. As Georges Elencwajg wrote in his answer and comments below, $R_3 = R_1 \otimes_{\mathbb C} R_2$ and $$\quot(R_3) = \quot(\quot(R_1)\otimes_{\mathbb C} \quot(R_2)).$$ So it sufficient to show that for transcendental extensions $F_1$ and $F_2$ of $\mathbb C$ we have $$\tr\deg \quot(F_1 \otimes_{\mathbb C} F_2) = \tr\deg F_1 + \tr\deg F_2.$$ Is this equality well-known or at least does it hold in general?
$\DeclareMathOperator{\quot}{Quot}\DeclareMathOperator{\tr}{tr}$Since you mention $\quot(R_i)$, the $R_i$'s must be domains or equivalently the $I_i$'s must be prime, which I now assume. The keys to your question are then:
a) The transcendence degree of $K_i=\quot(R_i)$ over $\mathbb C$ is precisely the dimension of the variety $V_i=\text {Spec}(R_i)$ defined by $R_i$.
b) The $\mathbb C$-algebra $R_3$ is the tensor product $R_3=R_1\otimes_\mathbb CR_2$, so that $V_3=V_1\times V_2$.
The required equality $n_3=n_1+n_2$ is then just the translation of the equality
c) $\operatorname {dim}V_3=\operatorname {dim}V_1+ \operatorname {dim}V_2$.
The equality c) is immediate from Noether's normalization theorem, an incredibly useful result in algebraic geometry, which is also the main ingredient for the proof of a) .