Irreducible affine varieties are birational exactly when their function fields are isomorphic

Suppose $\eta :X⇢Y$ is birational, with inverse $\theta :Y⇢X$. By [L2], both maps induce injective pullbacks ${\eta }^{\ast }:k(Y)\hookrightarrow k(X)$ and ${\theta }^{\ast }:k(X)\hookrightarrow k(Y)$. Because pullback is functorial, $(\theta \circ \eta {)}^{\ast }={\eta }^{\ast }\circ {\theta }^{\ast }$ and $(\eta \circ \theta {)}^{\ast }={\theta }^{\ast }\circ {\eta }^{\ast }$. Since the two compositions are identity rational maps by [L1], the two field homomorphisms are inverse isomorphisms.

Conversely, let $\sigma :k(Y)\stackrel{\sim }{\to }k(X)$ be a field isomorphism. By [L2], the map $\sigma$ comes from a dominant rational map $\eta :X⇢Y$, and ${\sigma }^{-1}$ comes from a dominant rational map $\theta :Y⇢X$. Applying [L2] again to the compositions shows that $\theta \circ \eta$ and $\eta \circ \theta$ induce the identity on function fields, so they are the identity rational maps. Hence $\eta$ is birational.

Steps 1.1 and 1.2 prove that $X$ and $Y$ are birationally equivalent exactly when $k(X)\cong k(Y)$ as $k$-extensions.