I am reading the Eisenbud, Commutative Algebra, Corollary 13.13 ( Emmy Noether ) and stuck at some statement.
First, I propose a relavant question.
Q. Let $R:=k[x_1 , \dots , x_d]$ with $\operatorname{char}(k)=p >0$ so that its fraction field is $K=K(R) = k(x_1, \dots ,x_d)$. Let $L'/K$ be a finite purely inseparable extension. Then there exists a power $q$ of $p$ such that $L'$ is obtained from $K$ by adjoining finitely many $q$-th roots of elements $f_i \in K$ ( $\because$ Since $L'/K$ is finite dimensional purely inseparable extension, $L'=K(u_1, \dots , u_m)$, where $u_1 , \dots , u_m$ are purely inseparable element over $K$ ; i.e., $u_1^{p^{n_1}}, \dots , u_m^{p^{n_m}} \in K=k(x_1, \dots ,x_d)$ for some $n_1, \dots , n_m \ge 0$. Let $n:=n_1 + \cdots + n_m$. Then $u_1^{p^n} , \dots , u_m^{p^n} \in K$ ).
Let $L'':=k'(x_1^{1/q} , \dots , x_d^{1/q})$ where $k'$ is the finite extension of $k$ obtained by adjoining all $q$-th roots of all coefficients of the $f_i$. Note that $L' \subseteq L''$ ( and finite extension? )
Then, the integral closure $\bar{R}^{L''}$ of $R$ in $L''$ is $T:=k'[x_1^{1/q},\dots,x_d^{1/q}]$ and $T$ is finite over $R$?
This question originates from following proof of Eisenbud's book ( corollary 13.13 ).
Corollary 13.13 (Emmy Noether). Let $R$ be an affine domain over a field $k$ ( ; i.e., finitely generated domain over $k$ ). Set $K=K(R)$ and let $L$ be a finite extension field of $K$. If $T$ is the integral closure of $R$ in $L$, then $T$ is a finitely generated $R$-module ; in particular, $T$ is again an affine domain.
Proof. By Noether normalization, $R$ is a finite module over a polynomial ring $k[x_1, \dots, x_d] \subset R$, and it suffices to prove the corollary after replacing $R$ by this subring ( $\because$ The integral closure $T$ of $R$ in $L$ is also the integral closure of $k[x_1, \dots,x_d]$ in $L$, and if $T$ is finite over $k[x_1, \dots x_d]$ it is also finite over $R$.). Further, since a submodule of a finitely generated module is finitely generated, it suffices to prove the corollary after making a finite extension of $L$, and thus we may replace $L$ by its normal closure and assume that $L/K$ is a normal extension in the sense of Galois theory.
Let $L'$ be the fixed field of the Galois group of $L$ over $K$, so that $L/L'$ is Galois and $L'/K$ is purely inseparable ( C.f. If $L/K$ normal and $H = \operatorname{Aut}(L/K)$, then $L/L^H$ is separable and $L^H/K$ is purely inseparable. ). We shall first show that the integral closure $R'$ of $R$ in $L'$ is a finitely generated $R$-module.
If $L'=K$ this is trivial, so we suppose that $L' \neq K$. Let $p$ be the characteristic of $L$, which is necessarily nonzero since $L'/K$ is purely inseparable.
For some power $q$ of $p$, the field $L'$ is generated by finitely many $q$-th roots of elements $f_i \in K = k(x_1 ,\dots x_d)$ ( $\because$ Since $L'/K$ is finite dimensional purely inseparable extension, $L'=K(u_1, \dots , u_m)$, where $u_1 , \dots , u_m$ are purely inseparable element over $K$ ; i.e., $u_1^{p^{n_1}}, \dots , u_m^{p^{n_m}} \in K=k(x_1, \dots ,x_d)$ for some $n_1, \dots , n_m \ge 0$. Let $n:=n_1 + \cdots + n_m$. Then $u_1^{p^n} , \dots , u_m^{p^n} \in K$ ). Extending $L'$ further by adjoining $q$-th roots of their coefficients, we may assume that
$$ L' =k'(x_1^{1/q},\dots ,x_d^{1/q})$$
where $k'$ is obtained from $k$ by adjoining the $q$-th roots of the coefficients ( $\because$ Note that $k'(x_1^{1/q},\dots ,x_d^{1/q})$ is really extension of $L'$ (?). Check this for simplest case that $R=k[x]$ (so that $K=k(x)$) and $L'= K(f^{1/q})$, where $f=\frac{a_mx^m + \cdots + a_0}{b_nx^n + \cdots + b_0} \in K$ ; using the Freshman's dream ( We have $\operatorname{char}L =p$) ). The integral closure of $R$ in $L'$ is $T=k'[x_1^{1/q} , \dots, x_d^{1/q}]$, since this ring is integrally closed, has quotient field $L'$, and is finite over $R$. Since $R' \subset T$, this shows that $R'$ is finite over $R$.
In Proposition 13.14 we shall see that the integral closure of the normal ring $R'$ in the Galois extension $L$ of $L'$ is finitely generated over $R'$. Since $R'$ is itself finitely generated over $R$, this completes the proof.
Proposition 13.14. Suppose that $R$ is a Noetherian normal domain with quotient field $K$. If $L$ is a finite separable extension of $K$, then the integral closure of $R$ in $L$ is a finitely generated $R$-module.
QED.
Why the bold statement is true?
Q. Why the integral closure of $R=k[x_1, \dots, x_d]$ in $L'$ is $T=k'[x_1^{1/q} , \dots, x_d^{1/q}]$ and why $T$ is finite over $R$ ?
Can anyone help?
Best Answer
Unless I am mistaken this should be rather simple.
Since $x_1,\dotsc,x_n$ are algebraically independent over $k$, they are also algebraically independent over $k'$ (this just a finite extension of $k$ and has nothing to do with the variables), and then also $x_1^{1/q},\dotsc,x_d^{1/q}$ are algebraically independent over $k'$. This means, actually, $T$ is a polynomial ring over $k'$ in $d$ variables, therefore factorial and hence integrally closed in $Q(T)$.
The elements of $k'$ are integral over $k$ and hence integral over $R$. The elements $x_1^{1/q},\dotsc,x_d^{1/q}$ are integral over $R$ as well (consider $T^q - x_i$). So we see that $T$ is a finitely generated $R$-algebra that is generated by integral elements. Hence, $T$ is finite and integral over $R$.
The integral closure $C$ of $R$ in $L''$ satisfies $T \subseteq C$, but since $T$ is integrally closed in $Q(T) = L''$, we must have $T=C$.