Let $\mathbb Z_{\geq 0}$ denote the set of non-negative integers. Let $\mathbb Z_{\geq 0}^n$ denote the set of $n$-tuples of non-negative integers.
(Theorem 2.1.5, Herzog, 1969) Given a finitely generated commutative semigroup $S$ of rank $n$ and any surjective semigroup homomorphism $\varphi : \mathbb Z_{\geq 0}^n \to S,$ the congruence $\sim_\varphi \, : \mathbb Z_{\geq 0}^n \times \mathbb Z_{\geq 0}^n \to \mathbb Z_{\geq 0}^n$ defined by $\mathbf a \sim_\varphi \mathbf b$ if and only if $\varphi(\mathbf a) = \varphi(\mathbf b)$ induces an isomorphism of semigroup rings $$R[S] \cong R[\mathbb Z_{\geq 0}^n / \sim_\varphi] \cong R[x_1, \dots, x_n] / (x^{\mathbf a} - x^{\mathbf b} \,|\, \mathbf a \sim_\varphi \mathbf b).$$
One can establish by the First Isomorphism Theorem that there is an isomorphism $k[x, y] / (x^3 - y^2) \cong k[t^2, t^3]$; however, I am interested in using the above theorem of Herzog to prove the same result. For this purpose, we may consider the numerical semigroup $\mathbb Z_{\geq 0} \langle 2, 3 \rangle = \{2r + 3s \,|\, r, s \in \mathbb Z_{\geq 0}\}.$ Of course, there is a natural surjection $\varphi : \mathbb Z_{\geq 0}^2 \to \mathbb Z_{\geq 0} \langle 2, 3 \rangle$ defined by $\varphi(r, s) = 2r + 3s,$ hence we have a congruence $\sim_\varphi$ on $\mathbb Z_{\geq 0}^2 \times \mathbb Z_{\geq 0}^2$ defined by $\mathbf a = (r_1, s_1) \sim_\varphi (r_2, s_2) = \mathbf b$ if and only if $\varphi(r_1, s_1) = \varphi(r_2, s_2)$ if and only if $2r_1 + 3s_1 = 2r_2 + 3s_2$ if and only if $r_1 \equiv r_2 \text{ (mod } 3)$ and $s_1 \equiv s_2 \text{ (mod } 2).$
From here, I am not sure how to conclude that $(\mathbf x^{\mathbf a} - \mathbf x^{\mathbf b} \,|\, \mathbf a \sim_\varphi \mathbf b) = (x^3 - y^2).$ I would be very gratefully for any feedback, comments, or suggestions about how to prove this equality.
Let $2r_1+3s_1=2r_2+3s_2$ and assume that $r_1 > r_2$ and $s_1 < s_2$. It follows that $\frac{s_1-s_2}{2}=\frac{r_2-r_1}{3}$ and we denote this number by $N$. We have $$ x^{r_1}y^{s_1}-x^{r_2}y^{s_2}=x^{r_2}y^{s_1} (x^{r_1-r_2}-y^{s_2-s_1})=x^{r_2}y^{s_1} (x^{3N}-y^{2N})= \\ =x^{r_2}y^{s_1} (x^{3}-y^{2})(x^{3(N-1)}+x^{3(N-2)}y^2+x^{3(N-3)}y^4+ \dots + x^3 y^{2(N-2)}+y^{2(N-1)}) $$ and, then, $x^{r_1}y^{s_1}-x^{r_2}y^{s_2} \in (x^3-y^2)$. Now it is enough to note that (3,0) and (0,2) are in the same congruence class.