TITLE: Is $k[x_1, \ldots, x_n]$ always an integral extension of $k[f_1, \ldots, f_n]$ for a regular sequence $(f_1, \ldots, f_n)$?
QUESTION [9 upvotes]: The elements of a regular sequence in $k[x_1, \ldots, x_n]$ are algebraically independent over $k$ (see for example Matsumura ex. 16.6), and so for a length n regular sequence $(f_i)$ of homogeneous elements, $k(x_1, \ldots, x_n)$ will be algebraic over $k(f_1, \ldots, f_n)$.
My question is whether it's also true that $k[x_1, \ldots, x_n]$ is integral over $k[f_1, \ldots, f_n]$.  Aside from experimental evidence (dozens of randomly generated regular sequences in Macaulay with n=3 and low-ish degrees), I don't really have any reason to think that it should be true, and I haven't been able to prove or disprove it (or find a mention of it anywhere).
Thanks.
Edit: added homogeneous condition

REPLY [7 votes]: Stumbled by coincidence over this question. Therefore the late answer. 
In the graded case it is true that $k[x_1,...,x_n]$ is integral over $k[f_1,...,f_n]$ ($k$ a field). But even more is true: 
Let $k$ be a field and let $R = \oplus_{i\ge 0}R_i $ be a finitely generated graded commutative $k$-algebra with $R_0= k$. Assume moreover that $R$ is Cohen-Macaulay and that 
$y_1,...,y_n$ is a regular sequence in $R$. Set $I := \oplus_{i> 0}R_i$. Then 

$\sqrt{I} = (y_1,...,y_n)$ 
There is $k > 0$ with $I^k \subseteq (y_1,...,y_n)$. 
$R$ is a free $k[y_1,...,y_n]$-module. 

Proof: 2) It's well-known that if $y \in R$ is a homogeneous regular element of positive degree, then Krull-dim $R/(y) =$ Krull-dim$(R) - 1$. Applying this repeatedly to $R/(y_1,...,y_i)$ one obtains that $\bar{R} :=  R/(y_1,...,y_n)$ has dimension $0$. Hence $\bar{R}$ is Artinian and thus there is $k > 0$ such that $\bar{R}_i = 0$ for all $i \ge k$. So, if $x \in I$ then $\bar{x}^k = 0$ in $\bar{R}$, i.e. $x^k \in (y_1,...,y_n)$. 
1) Follows from 2) 
3) Induction on $n$. Case $n=0$: $R$ is Artinian and hence a finite dim. $k$ vector space. 
Assume the case $n-1$ is true. Since $\tilde{R} := R/(y_1)$ is again CM and $\tilde{y}_2,...,\tilde{y}_n$ a regular sequence, there are $\tilde{a}_i \in \tilde{R}$ with $\tilde{R} = \oplus_i k[\tilde{y}_2,...,\tilde{y}_n]\tilde{a}_i$. 
Let's show $R = \oplus_i k[y_1,...,y_n]a_i$: Let $f_i \in k[y_1,...,y_n]$ with $\sum_i f_i a_i = 0$. 
We proceed by induction on $m := \max_i\; \deg_{y_1}(f_i)$: Reduction modulo $y_1$ shows $f_i \in (y_1)$ for all $i$. If $m=0$, then $f_i = 0$ for all $i$. If $m> 0$ let $f'_i = f_i/x_1$. Thus $y_1\sum_i f'_i a_i = 0$ and since $y_1$ is regular, we conclude $\sum_i f'_i a_i = 0$ with $m' < m$. Then by induction hypothesis $f'_i = 0$ and hence $f_i = 0$. 
q.e.d.