Looking the quotient ring $A:=\mathbb{C}[x,y]/(xy-1)$, we define a homomorphism $\phi:\mathbb{C}[t]\rightarrow A$ with $t \mapsto \bar{x}+\bar{y}$, where $R:=\operatorname{Im} (\phi)$.
We've said in the classes that $\phi$ is injective because when we pick up $f(t)\in \mathbb{C}[t]$ such that $\phi(f(t))=f(\bar x+\bar y)=0$ in $A$, then there is $g(x,y)\in\mathbb{C}[x,y]:f(x,y)=g(x,y)(xy-1)$.
Now we were said that when $f(x,y)$ has maximal $y$-degree $m$, then this would be contradiction because $g(x,y)$ is also of $y$-degree $m$ and $g(x,y)(xy-1)$ is of maximal $y$-degree $m+1$. So $g(x,y)=0$.
I don't understand this argumentation. Can't we elect a $g(x,y)$ with degree $m-1$, so that everything is fine? And why because of injectivity of $\phi$ it finally follows that $R$ is isomorphic to $\mathbb{C}[t]$? Any help is appreciated.
We can write $f(x,y)=a_0+a_1(x+y)+...+a_m(x+y)^m=g(x,y)(xy-1)=g(x,y)xy-g(x,y)$ it results that $g(x,y)$ has a monomial of the form $a_mx^m$, this implies that $-a_mx^mxy$ is in $g(x,y)(xy-1)$ impossible since the degree of $g(x,y)(xy-1)$ is $m$.