This was an extra tutorial question in my algebraic geometry course,
We have the following subsets of $\mathbb{C^2}$
$$X_1 = \{( (x,y) \ | \ y = 0 \}$$
$$X_2 = \{( (x,y) \ | \ y -x^2 = 0 \}$$
$$X_3 = \{( (x,y) \ | \ xy = 1 \}$$
so $X_1$ is the $x$-axis, $X_2$ is the parabola and $X_3$ is the hyperbola in $\mathbb{C^2}$. We're asked to calculate the function algebras $\mathcal{O}(X_1),\mathcal{O}(X_2), \mathcal{O}(X_3)$, and show that $X_1$ and $X_2$ have isomorphic function algebras, but that $\mathcal{O}(X_3)$ is not isomorphic to either.
For the function algebras I first get the ideals
$$ I_1 = \langle y \rangle, \quad I_2 = \langle y-x^2 \rangle, \quad I_3 = \langle xy-1 \rangle $$ immediate from the definition of the zero sets. Then the function algebras are,
$$\mathcal{O}(X_1) = \mathbb{C}[x,y]/\langle y \rangle \cong \mathbb{C}[x] $$
$$\mathcal{O}(X_2) = \mathbb{C}[x,y]/\langle y-x^2 \rangle \cong \mathbb{C}[x] $$
but for the third,
$$\mathcal{O}(X_3) = \mathbb{C}[x,y]/\langle xy-1 \rangle \cong \ ? $$
I think I understand $\mathcal{O}(X_3)$ is the quotient of $\mathbb[x,y]$ with no mixed monomial terms, but I don't know how to actually write that as a quotient. I'm hesitant to claim it's isomorphic to $\mathbb{C}[x,\frac{1}{x}]$? (Is this then the same as the fraction field?)
Finally for the isomorphisms, $\mathcal{O}(X_1) \cong \mathcal{O}(X_2)$ is clear assuming I got those quotients right, but how do I show $\mathcal{O}(X_3)$ is not isomorphic to the other two, precisely?
(Also one other more broad question. This course has emphasized the equivalence of algebra and geometry (the geometry should be reflected in the algebra, and vice versa) so what does it mean geometrically for the function algebras of the parabola and the $x$-axis to be isomorphic?)