injectivity of concatenation of the canonical homomorphisms Given the quotient ring $A:=\mathbb{C}[x,y]/(xy-1)$, 
$\phi:\mathbb{C}[x]\rightarrow A $, with $x \mapsto \bar{x} $, where $\phi$ is the concatenation of the  canonical homomorphisms $\mathbb{C}[x]\hookrightarrow \mathbb{C}[x,y] \twoheadrightarrow A$.
I have to show that $\phi$ is injective by the following idea:
Suppose there exists $f(x)\in\mathbb{C}[x]$ such that $f(\bar x)=0$ in $A$. Then there is a $g(x,y) \in \mathbb{C}[x,y]: f(x)=g(x,y)(xy-1)$ in $\mathbb{C}
[x,y]$. That is equivalent to $0=g(x,y)xy-g(x,y)-f(x)$. Now I've got problems to show that all polynomials have to vanish. I've got a hint that I should consider it as a polynomial of $y$, that is in $\mathbb{C}[x][y]$. But I don't know how to go on. Can I put in random values for all $x$ and $y$ to show a contradiction
?
 A: The way to go from here is indeed to consider the polynomials as polynomials in $y$ with coefficients in $\Bbb{C}[x]$, i.e. as elements of $(\Bbb{C}[x])[y]$. From this perspective $f(x)$ is a constant, and hence so is
$$f(x)=g(x,y)(xy-1).$$
But because $\Bbb{C}[x]$ is an integral domain, the degree of the product of two nonzero polynomials over $\Bbb{C}[x]$ is the sum of the degrees of those two polynomials. That is, if $g(x,y)\neq0$ then
$$\deg_y f(x)=\deg_y(g(x,y)(xy-1))=\deg_yg(x,y)+\deg_y(xy-1).$$
But $\deg_yf(x)=0$ and $\deg_y(xy-1)=1$, a contradiction. Hence $g(x,y)=0$ and so also $f(x)=0$.
A: The quotient ring ${\Bbb C}[x,y]/\langle xy-1\rangle$ has the (monomial) standard basis $1,\bar x,\bar x^2,\ldots, \bar y,\bar y^2,\ldots$ (as vector space or algebra), since $xy=1$ in this ring.
Thus your mapping $\phi$ maps $f(x)$ in ${\Bbb C}[x]$ to $f(x)$ in ${\Bbb C}[x,y]$ and then to $f(\bar x)$ in ${\Bbb C}[x,y]/\langle xy-1\rangle$.
By the standard basis, this homomorphism is one-to-one.
