Does this ring isomorphism hold for rings that aren't necessarily fields? The problem I'm working on is: Show that $R[x,y]/(x^2-y^3)\cong A$ where $A=\{\sum a_ix^i\in R[x]:a_1=0\}$. A similar question has been posted on the website multiple times, but the explanation is either confusing (to me at least) or requires $R$  to be a field (which is not assumed to necessarily be true in the problem).
I have constructed a homomorphism $\varphi:R[x,y]\to A; \varphi(x):=x^3;\varphi(y):=x^2;\varphi(r):=r,r\in R$. There is a unique homomorphism for which all of the previous equalities hold and it is surjective. To prove the claim it is sufficient to show that $ker\varphi\subset(x^2-y^3)$. The other inclusion is trivial. Once proven, the hypothesis easily follows from the  first isomorphism theorem. The inclusion above could be proven using division with rest if $R$ were a field, however that isn't necessarily the case.
I've gotten this far on my own:
We can split the terms of any $f\in R[x,y]$ in the following way: $f=g_0+g_1+...+g_n$ where $\varphi(g_k)=c_kx^k$ for some $c_k\in R$ and all $k$. This partition of terms is clearly unique. Let $f\in ker\varphi$. In this case $c_k=0$ for all $k$. The goal is to prove that $g_k(x,y)=(x^2-y^3)h_k(x,y)$ for some polynomial $h_k.$ It is easy to see that $g_0$ consists of only constant terms, therefor $g_0=0$. To find the others let us take a look at the following equation: $\varphi(x^iy^j)=x^k$. This equation clearly holds if and only if $3i+2j=k$. It's easy to find all of the solutions of this equation. If $k$ is even then $i=0,j=\frac{k}{2}$ is a solution. If, on the other hand $k$ is odd, then $i=1,j=\frac{k-3}{2}$ is a solution (we're obviously looking for solutions where $k\neq1$). All the other solutions can be found by subtracting $3$ from $j$ and adding $2$ to $i$. Using this method we can find that:
\begin{equation}
\begin{split}
&g_0=a_{0,0}\\
&g_1=0\\
&g_2=a_{2,0}y\\
&g_3=a_{3,0}x\\
&g_4=a_{4,0}y^2\\
&g_5=a_{5,0}xy\\
&g_6=a_{6,0}x^2+a_{6,1}y^3...
\end{split}
\end{equation}
Where $a_{i,j}\in R$. From here we can see why $g_0=g_1=...=g_5=0$. For example $0=\varphi(g_2)=a_{2,0}x^2\Rightarrow a_{2,0}=0\Rightarrow g_2=0$. Similarily, we can conclude that $g_k=0$ whenever the equation $3i+2j=k$ has a unique solution in the natural numbers (holds also for $k=7$ and $k=8$). If the equation has two solutions (like in case $k=6$) we can do the following:
$$
0=\varphi(g_6)=(a_{6,0}+a_{6,1})x^6
$$
This means that $g_6(x,y)=a_{6,0}x^2-a_{6,0}y^3=(x^2-y^3)a_{6,0}$. So the claim holds for $k=6$. Similar can be shown for all other $k$ for which the equation above has only two solutions (like for $k=9,10,11$) using the equality $cx^iy^j-cx^{i-2}y^{j+3}=(x^2-y^3)\cdot cx^{i-2}y^j$. Furthermore it can also be easily factored in the case where the equation has three solutions ($k=12$ for example). That's because if
$$
g_k(x,y)=c_0x^iy^j+c_1x^{i-2}y^{j+3}+c_2x^{i-4}y^{j+6}
$$
then $\varphi(g_k)=0\Rightarrow c_0+c_1+c_2=0$ therefor
\begin{equation}
g_k(x,y) = c_0x^iy^j - c_2x^{i-2}y^{j+3} - c_0x^{i-2}y^{j+3} + c_2x^{i-4}y^{j+6} = \\
(x^2-y^2)(c_0x^{i-2}y^j - c_2x^{i-4}y^{j+3})
\end{equation}
This is as far as I've come. I can't find a solution for when the equation with $k$ has four solutions. All of the attempts at factoring the general case have failed.

We want to show that the kernel of $\varphi$ is $(x^2-y^3)=: I$, with $I$ being a subset of said kernel.
Let $h \in R[x,y] \backslash I$ with $\varphi(h)=0$ (assume there exists one for the sake of contradiction). We can write $h \in g+I$, with $\varphi(g)=0$, and $g(x,y)=g_1(x)+yg_2(x)+y^2g_3(x)$, $g_1,g_2,g_3 \in R[x]$, and thus $g \notin I$.
But $0=\varphi(g)=g_1(x^3)+x^2g_2(x^3)+xh_3(x^3)$ where $h_3(x)=xf_3(x)$.
Each term of the RHS only contributes to disjoint classes of powers of $x$: from left to right, the $x^{3d}$, the $x^{3d+2}$, the $x^{3d+1}$. As the RHS is zero, it follows that all three terms are zero and thus $g=0 \in I$, a contradiction.
