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

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

Answer 1

No. For example, the element $y \in k[x,y]$ is not integral over $k[x-1, xy]$. (To see this, write $k[x-1,xy] = k[x,xy]$ and reduce modulo $x$).

Answer 2

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

1. $\sqrt{I} = (y_1,...,y_n)$

2. There is $k > 0$ with $I^k \subseteq (y_1,...,y_n)$.

3. $R$ 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.

Answer 3

In the homogeneous case that is true, I believe. To show that $A=k[x_1,\ldots,x_n]$ is a finitely generated $k[f_1,\ldots,f_n]$-module, let's notice that for a regular sequence $f_1,\ldots,f_n$, the corresponding Koszul complex $\Lambda_A^*(g_1,\ldots,g_n)$ (with the differential being the unique graded derivation taking $g_k\in\Lambda_A^1(g_1,\ldots,g_n)$ to $f_k\in A=\Lambda_A^0(g_1,\ldots,g_n)$) is acyclic in positive degrees and having $B=A/(f_1,\ldots,f_n)$ as its zero degree homology. Now, writing down the graded Euler characteristics, we observe that the generating series for dimensions of graded components of $B$ is equal to $$ \frac{(1-t^{\deg(f_1)})(1-t^{\deg(f_2)})\ldots (1-t^{\deg(f_n)})}{(1-t)^n}, $$ which evaluates to $\deg(f_1)\deg(f_2)\ldots\deg(f_n)$ for $t=1$. Therefore, $\dim(B)<\infty$, - but clearly a (lifting of any) basis of $B$ forms a system of generators of $A$ over $k[f_1,\ldots,f_n]$ in the graded case, and we are done.