Show that there exists a bijection between prime ideals of $\mathbb{C}[x,y]$ which contains the ideal $(xy-1)$ and the prime ideals in $\mathbb{C}[x]$ which does not contain $x$.
I know: $$ \mathbb{C}[x,y]\Big/(xy-1)\cong \mathbb{C}\left[x,\frac{1}{x}\right] $$ and also, this.
I will post my answer for completeness.
First, for a commutative ring $R$, let $\operatorname{Spec}(R)$ be the set of prime ideals of $R$. By Krull's theorem (about the existence of maximal ideals), the above set isn't empty.
In order to get our desire bijection I will use three results, letting the proofs as exercises for the OP. The first two are given below, whereas the third one is the bijection given in my answer linked in the question above.
Theorem 1: If $\phi\colon R\rightarrow R'$ is an isomorphism of commutative rings, then there is a bijection $\overline{\phi}\colon \operatorname{Spec}(R)\rightarrow \operatorname{Spec}(R')$.
Informally, by the isomorphism we can think as $R$ and $R'$ being the same ring, so they will have the same prime ideals.
Theorem 2: Let $R$ be a commutative ring and $I$ be an ideal of $R$. If we denote by $\operatorname{Var}(I)=\{P\in \operatorname{Spec}(R): I\subseteq P\}$, then there is a bijection $\psi\colon \operatorname{Var}(I)\rightarrow \operatorname{Spec}(R/I)$.
This follows from the correspondence theorem for rings and the third isomorphism theorem for rings.
Finally, we're ready to find our bijection. We can work more generally in an arbitrary field $K$. As I indicated (following the OP), we have the following isomorphisms $$ K[x,y]\big/\langle xy-1\rangle\cong K[x,x^{-1}]\cong K[x]_x.$$
Now, set $A=\bigl\{\mathfrak{P}\in \operatorname{Spec}\bigl(K[x]\bigr): x\notin \mathfrak{P}\bigr\}$, i.e., the set of prime ideals of $K[x]$ which don't contain $x$, and $B=\operatorname{Var}\bigl(\langle xy-1\rangle\bigr)$, i.e., the set of prime ideals of $K[x,y]$ which contain the ideal $\langle xy-1\rangle$.
By the theorem given in my answer linked in the question we have the bijection: $$\alpha\colon A\longrightarrow \operatorname{Spec}(K[x]_x).$$
On the other hand, by theorem 1 we have the bijection: $$\beta\colon \operatorname{Spec}\bigl(K[x]_x\bigr)\longrightarrow \operatorname{Spec}\Bigl(K[x,y]\big/\langle xy-1\rangle\Bigr).$$
Analogously, by theorem 2 we have the bijection: $$\theta\colon \operatorname{Spec}\Bigl(K[x,y]\big/\langle xy-1\rangle\Bigr)\longrightarrow B.$$
Finally, our desired bijection is $\theta\circ \beta\circ \alpha\colon A\longrightarrow B$ and we're done.
