Dominant rational maps to an affine variety correspond to injective homomorphisms of function fields

By [L1], every dominant rational map $\eta :X⇢Y$ determines an injective $k$-homomorphism ${\eta }^{\ast }:k(Y)\hookrightarrow k(X)$.

Conversely, let $\sigma :k(Y)\hookrightarrow k(X)$ be an injective $k$-homomorphism. For the coordinate classes ${\bar{y}}_1,\dots ,{\bar{y}}_m\in k[Y]$ from [L3], write $\sigma ({\bar{y}}_i)=\frac{a_i}{d_i}$ with $a_i,d_i\in k[X]$ and $d_i\mathrm{\ne }0$ in the domain $k[X]$. Put $d=d_1\cdots d_m$ and $U=D_X(d)$. By [L5], each $a_i\mathrm{/}{d}_i$ is regular on $U$, so each $\sigma ({\bar{y}}_i)$ lies in $k[U]$.

Define a $k$-algebra homomorphism $\alpha :k[Y]\to k[U]$ by sending ${\bar{y}}_i$ to $\sigma ({\bar{y}}_i){\mathrm{\mid }}_U$. This is well defined because [L4] identifies $I(Y)$ as the defining ideal of $Y$, and if $g\in I(Y)$, then its class is $0$ in $k[Y]$ by [L3], hence $g(\sigma ({\bar{y}}_1),\dots ,\sigma ({\bar{y}}_m))=\sigma (g)=0$ in $k(X)$. Therefore every relation of $I(Y)$ is respected.

By [L6], the homomorphism $\alpha$ corresponds to a morphism $\phi :U\to Y$. Its rational-map class gives a rational map ${\eta }_{\sigma }:X⇢Y$.

The rational map ${\eta }_{\sigma }$ is dominant. If its image were not dense in $Y$, then some nonzero $h\in k[Y]$ would vanish on $\phi (U)$. Then $\alpha (h)=0$ in $k[U]$, so $\sigma (h)=0$ in $k(X)$ by step 2.1, contradicting injectivity of $\sigma$.

The pullback $({\eta }_{\sigma }{)}^{\ast }$ agrees with $\sigma$ on the coordinate classes ${\bar{y}}_i$, hence on the whole coordinate ring $k[Y]$, and therefore on its fraction field $k(Y)$ by [L2]. Thus the construction of steps 1.2-4.1 is inverse to the construction in step 1.1.

Steps 1.1 and 5.1 give the stated bijection between dominant rational maps $X⇢Y$ and injective $k$-homomorphisms $k(Y)\hookrightarrow k(X)$.