Apologies for my limited algebra knowledge.
I previously asked why polynomial rings $k[x,y]/\langle y-x^2\rangle$ and $k[x]$ with $k$ a field are isomorphic. I ran into a related answer where the accepted answer used a suitable substitution to make the ideal zero and hence obtain a description of the quotient ring.
In the present case the suitable substitution would be $y=x^2$ as the ideal $\langle y-x^2\rangle$ we are taking the quotient against vanishes under that substitution. Does this sort of thing work in general? Consider a quotient
$$
k[x_1, \dots, x_n] / \langle f_1, \dots, f_m \rangle,
$$
with $f_1, \dots, f_m \in k[x_1, \dots, x_n]$. Assume that we are able to find substitutions $x_i = g_i(y_1, \dots, y_r)$ in terms of $1 \leq r < n$ free parameters $y_i$ such that each polynomial $f_i$ vanishes. Does that imply the isomorphism
$$
k[x_1, \dots, x_n] / \langle f_1, \dots, f_m \rangle \simeq k[x_1, \dots, x_r]?
$$
Best Answer
Did you mean $x_i=g_i(x_1,x_2,\ldots,x_r)$? Even so, the answer is definitely no. Even if you have such substitutions, you cannot conclude that $ k[x_1, \dots, x_n] / \langle f_1, \dots, f_m \rangle \cong k[x_1, \dots, x_r] $. For example, $k[x,y]/\langle x^2-y^2\rangle$ is not isomorphic as a ring to $k[x]$ even if you have a substitution $y=x$ that kills $x^2-y^2$.
However, if you have polynomials $f_i$ and $g_i$ for $i=1,2,\ldots,n-r$ such that $$f_i(x_1,x_2,\ldots,x_n)=x_{r+i}-g_i(x_1,x_2,\ldots,x_r)$$ or $$f_i(x_1,x_2,\ldots,x_n)=x_{r+i}-g_i(x_1,x_2,\ldots,x_{r+i-1})\,,$$ then the answer is yes, namely, $$k[x_1,x_2,\ldots,x_n]/\langle f_1,f_2,\ldots,f_{n-r}\rangle \cong k[x_1,x_2,\ldots,x_r]\,.$$ This is the case with your original problem.