Birational map

Remarks

Equivalently, there are nonempty open subsets $U \subseteq X$ and $V \subseteq Y$ and an isomorphism of varieties $U \to V$ (a morphism with inverse morphism). Also equivalently, there is an isomorphism of $k$-algebras between the function fields $k(Y)$ and $k(X)$.

Example: $\mathbb{A}^2$ and $\mathbb{P}^2$ are birational, since $\mathbb{A}^2$ is isomorphic to the open subset $\set{z \ne 0}$ of $\mathbb{P}^2$. They are not isomorphic, since every regular function on $\mathbb{P}^2$ is constant and $\mathbb{A}^2$ has the nonconstant regular function $x$.

Example: the conic $C = V(x^2 + y^2 - 1) \subseteq k^2$, with $\operatorname{char} k \ne 2$, is birational to $\mathbb{A}^1$ through $t \mapsto \left(\frac{1 - t^2}{1 + t^2}, \frac{2t}{1 + t^2}\right)$, whose inverse is $(x, y) \mapsto y/(1 + x)$.

Non-example: the rational map $\mathbb{P}^2 \dashrightarrow \mathbb{P}^1$, $[x : y : z] \mapsto [x : y]$, is dominant but not birational, since $\dim \mathbb{P}^2 = 2 \ne 1 = \dim \mathbb{P}^1$ and birational varieties have the same dimension.