A question on algebraic geometry.

Question

In a course about algebraic geometry I am taking, a question appeared in class which claimed that $A=\mathbb{C}[x,y]/(xy-1)$ is a field. In order to prove this, one concluded that $xy-1=0 \Leftrightarrow y=1/x$ and so $A=\mathbb{C}[x,y]/(xy-1) = \mathbb{C}[x,\frac {1}{x}]= \text{Frac}A$, and so $A$ is a field since it equals its field of fractions.

However there is something I am missing here, I understand that one can obtain alot of information about $A$ by considering the curve $xy-1=0$ but why is it OK to substitute $y$ with $1/x$ in the actual ring? This seems fishy to me, is it really true that $\mathbb{C}[x,y]/(xy-1) = \mathbb{C}[x,\frac {1}{x}]$? (This is notes from class, I assume equality means isomorphic, is this true?)

Answer 1

Yes, $\mathbb{C}[x,y]/(xy-1) \cong \mathbb{C}[x,x^{-1}]$ (since both algebras satisfy the same universal property), but certainly this is not a field. $x+1$ is not invertible for instance.

Answer 2

$\newcommand{\C}{\mathbb{C}}$ In the ring $A=\C[x,y]/(xy-1)$ we have, as you observed, $xy=1$, so $y$ is an inverse to $x$. Since inverses are unique, it is safe to superficially rename $y$ to $1/x$.

For a more elaborate argument, we can construct an isomorphism between the localized ring $\C[x]_x$ and $A$ by sending $x^{-1}$ to $y$.