I want to show that hyperbola $\textrm Z(xy-1)$ is not isomorphic to an affine line $\mathbb A^1.$
My reasoning: let $f:\textrm Z(xy-1)\rightarrow\mathbb A^1\setminus0$ be a morphism defined by $f(x,y)=x.$ It is invertible: $f^{-1}(x)=(x,\frac{1}{x}).$ Hence, $\textrm Z(xy-1)\cong\mathbb A^1\setminus0$ and $k[\textrm Z(xy-1)]\cong k[\mathbb A^1\setminus0].$
But $\mathbb A^1\setminus0\subset\mathbb A^1$, so $k[\mathbb A^1]\subset k[\mathbb A^1\setminus0]$. That implies $\textrm Z(xy-1)\not\cong\mathbb A^1.$
Is my proof correct? I am really new to algebraic geometry and not sure if my solution has any sence.
It is enough to show that the affine ring of $\mathbb A^1$, which is $k[x]$, is not isomorphic to the affine ring of $Z(xy-1)\cong\mathbb A^1\setminus0$, which is $k[x,y]/(xy-1)\cong k[x,x^{-1}]$.
To show this, suppose there is an isomorphism $f\colon k[x,x^{-1}]\to k[x]$. Then since $f(x)\cdot f(x^{-1})=f(1)=1$, we must have $f(x)\in k[x]^\times=k^\times$. However, then, clearly $x\in k[x]$ is not in the image of $f$.