I want to show that the homomorphism $\mathbb C[x,y]/(x^m-y^n) \rightarrow \mathbb C [t]$, given by $x \mapsto t^n, y \mapsto t^m$ is injective if $n$ and $m$ are coprime.
I know that I must show that $Ker(f)={0}$, but I don't know how to choose an element in $\mathbb C[x,y]/(x^m-y^n)$.
Since $x^m-y^n$ is monic in $x$, the $\mathbb{C}[y]$-module $\mathbb{C}[y][x]/(x^m-y^n)$ is free of rank $m$ with basis $\{x^i : 0 \leq i < m\}$. It follows that the $\mathbb{C}$-module $\mathbb{C}[y][x]/(x^m-y^n)$ is free with basis $\{x^i y^j : 0 \leq i < m, ~0 \leq j\}$. (Slight abuse of notation: $x$ denotes the variable in $\mathbb{C}[x,y]$ but also its image in the quotient $\mathbb{C}[y][x]/(x^m-y^n)$.)
So any element in $\mathbb{C}[y][x]/(x^m-y^n)$ has a unique expression as $\sum_{j=0}^{\infty} \sum_{i=0}^{m-1} \lambda_{ij} x^i y^j$, where $\lambda_{ij} \in \mathbb{C}$ (almost all zero). It lies in the kernel iff $\sum_j \sum_i \lambda_{ij} t^{ni+mj}=0$. Now show that the $ni+mj$ are pairwise distinct (this is where one uses that $n,m$ are coprime!) and conclude that $\lambda_{ij}=0$.