Dominant rational maps to an affine variety correspond to field embeddings

Let $\sigma :k(Y)\hookrightarrow k(X)$ fix k. Put $A=k[X]$ and $B=k[Y]=k[y_1,\dots ,y_m]$. Write $\sigma (y_i)=a_i\mathrm{/}{b}_i$ with $a_i,b_i\in A$ and $b_i\mathrm{\ne }0$. The finite product $d={\prod }_i{b}_i$ is nonzero since A is a domain; for $m=0$ put $d=1$. Each $b_i$ is invertible in $A_d$ with inverse $({\prod }_{j\mathrm{\ne }i}{b}_j)\mathrm{/}d$, so $\sigma (B)\subseteq A_d$. F2 and F3 realize this algebra map as $\varphi :D(d)\to Y$.

If the image of $\varphi$ were contained in a proper closed $C⊊Y$, F5 would give a polynomial function $0\mathrm{\ne }b\in B$ vanishing on C: choose an element of $I(C)\setminus I(Y)$. Then the coordinate dictionary makes $\sigma (b)={\varphi }^{\ast }(b)=0$ in $A_d$, hence in k(X), contrary to injectivity of $\sigma$. Thus $\varphi$ is dominant. F4 identifies the source field with k(X); F1 and the equality on B show its field pullback equals $\sigma$ on every ratio.

If two dominant rational maps have the same field pullback, take representatives on U and V. Their pullbacks of each y_i agree as elements of k(X), so the faithful open-section embedding F7 makes their values agree on $U\cap V$. This intersection is nonempty by F8, hence the representatives determine the same rational map. Conversely equal rational maps have equal pullbacks by F1. Thus step 2.1 proves surjectivity and this argument proves injectivity.

Finally F1 gives identity preservation and reversal of composition, so the bijection is natural with respect to dominant rational composition. This construction asserts a dense image, and does not require an image-constructibility theorem.