Let $X$ be an irreducible variety over a field $k$. Let $k(X)$ be its field of rational functions. Given a subfield $E\subseteq k(X)$, I wish to construct a dominant rational map $X\to Y$ for some (irreducible) variety $Y$ such that $k(Y)\cong E$.
The reason for this is that I am trying to understand the proof of Lemma 2.7 of this paper. I am not really understanding how they use the equivalence of categories between dominant rational maps $X\to Y$ and field extensions, because in one direction they do not have a variety $Y$. Moreover, in the proof of this equivalence—and any such proof for that matter—they seem to already have the variety $Y$. Namely, they generally start with varieties $X$ and $Y$, then from a dominant rational map $\varphi:X\to Y$ create a monomorphism $\varphi^*:k(Y)\hookrightarrow k(X)$, and then create an inverse of the map $\varphi\mapsto\varphi^*$, but they seem to already assume that their field extension is of the form $k(Y)\hookrightarrow k(X)$ for varieties $X$ and $Y$.
In order to understand the proof of Lemma 2.7, I am trying to prove it myself using the things mentioned in the proof.
First some definitions and terminology from the paper. All field are assumed to contain a fixed field $k$.
Let L/K be a separable field extension of degree $n$. Let $F\subseteq K$ be a subfield. We say that $L/K$ is defined over $F$, if there exists a separable field extension $E/F$ of degree $n$ with $E\subseteq L$ such that $EF=L$.
The essential dimension of $L/K$ is the minimal transcendence degree of $F$ over $k$, where $F$ runs over all fields over which $L/K$ is defined. This is denoted by $\operatorname{ed}_k(L/K)$.
All varieties are assumed to be irreducible.
Let $G$ be a finite group. Let $X$ be a faithful $G$-variety. That is, the $G$-action $G\times X\to X$ is a morphism of varieties and $G$ acts faithfully on $X$. The essential dimension of $X$ is the minimal dimension of faithful $G$-varieties $Y$ for which there exists a dominant rational $G$-map $\varphi:X\to Y$. This is denoted by $\operatorname{ed}_k(X)$.
The $G$-map part means that $\varphi(\sigma x)=\sigma\varphi(x)$ for all $x\in X$ and $\sigma\in G$.
Lemma 2.7:
Let $G$ be a finite group and let $X$ be a faithful $G$-variety. Let $L:=k(X)$ and $K:=L^G$. Then $\operatorname{ed}_k(X)=\operatorname{ed}_k(L/K)$.
(I'm assuming that $L/K$ is separable, as the paper only works over characteristic $0$.) Here $G$ acts on $k(X)$ via $\sigma\cdot f(x):=f(\sigma^{-1}x)$. Onr can show that the action on $k(X)$ is faithful if and only if the action of $G$ on $X$ is faithful.
Let $X\to Y$ be a dominant rational $G$-map to a faithful $G$-variety $Y$, such that $\operatorname{ed}_k(X)=\dim(Y)$. Then we have an extension of fields $k(Y)\hookrightarrow k(X)$ as mentioned above. Let $E:=k(Y)$ and $F:=E^G$. Then $F\subseteq K$, as $F$ is fixed by $G$, and $E\subseteq L$. Also, since $G$ acts faithfully on $L$ and $E$, $[L:K]=\#G=[E:F]$.
Finally, one can show that $L/K$ is a Galois extension, and we see that no element of $G$ fixes $EK$, wherefore $L=EK$. Hence $L/K$ is defined over $F$. Since $\operatorname{trdeg}_k(F)=\operatorname{trdeg}_k(E)=\dim(Y)$, $\operatorname{ed}_k(X)\geq\operatorname{ed}_k(L/K)$.
For the other inequality I am stuck. Now let $E/F$ be an extension of fields for the essential dimension of $L/K$. That is, $E\subseteq L$, $F\subseteq K$, $EK=L$, $[E:F]=[L:K]$, $\operatorname{trdeg}_k(F)=\operatorname{ed}_k(L/K)$, and we may assume that $E/F$ is Galois with Galois group $G$ (Lemma 2.2 of the paper). Here is were my question comes in: I need to construct a faithful $G$-variety $Y$ such that $k(Y)\cong E$ to get a dominant rational $G$-map $X\to Y$. Then we get $\operatorname{ed}_k(X)\leq\dim(Y)=\operatorname{trdeg}_k(E)=\operatorname{ed}_k(L/K)$, which would prove the lemma.
Thank you in advance!
Edit: I have realized that in the proof of aforementioned equivalence they do mention how to obtain such a variety. My question therefore mainly boils down to: How do they obtain a $k$-algebra $A\subseteq E$ and why is $Y$ then irreducible and admits a faithful $G$-action?
I have tried the following:
We can replace $X$ by an affine dense open, so we may assume that $X$ is affine. Then the regular functions on $X$ are $\mathcal{O}(X)=k[X_1,\ldots,X_n]/I$ for some ideal $I\subseteq k[X_1,\ldots,X_n]$ of the polynomial ring over $k$ in $n$ variables. Since $L=\operatorname{Frac}(\mathcal{O}(X))$ is then finitely generated, so is $E$. Say, $E=k(y_1,\ldots,y_m)$. Then consider the $k$-subalgebra $A:=k[y_1,\ldots,y_m]\subseteq L$, which is clearly contained in $E$, so it is a domain. Also, $\operatorname{Frac}(A)=E$. Now $A\cong k[Y_1,\ldots,Y_m]/J$ for some ideal $J\subseteq k[Y_1,\ldots,Y_m]$ of the polynomial ring over $k$ in $m$ variables. Let $Y:=Z(J)$. Then $k(Y)=E$. Thus it remains to show that $Y$ is irreducible and that $G$ acts on $Y$ in such a way that the action induced from this on $k(Y)$ coincides with the already existing action on $E$, which will imply that $G$ acts faithfully on $Y$.