TITLE: If a polynomial ring is finite free over a subring, is the subring polynomial?
QUESTION [21 upvotes]: Let $R = k[x_1, \ldots, x_n]$ for $k$ a field of characteristic zero and let $S \subset R$ be a graded sub-$k$-algebra (for the standard grading: $\deg x_i = 1$) such that $R$ is a free $S$-module of finite rank. Does this imply $S \cong k[y_1,\ldots,y_n]$?

REPLY [8 votes]: It turns out that this result appears in Bourbaki, Groupes et algebres de Lie IV-VI, Ch. 5, $\S$ 5, Lemme 1, with an elementary proof. (However, this proof uses characteristic zero, which Neil's answer does not.)  A similar proof of a different theorem occurs as Section 3.5 in Humphrey's Reflection Groups and Coxeter Groups. Warning for those who read the originals: their $R$ and $S$ are reversed from mine!
Notation: Let $R_+$ and $S_+$ be the maximal graded ideals of $R_+$ and $S_+$. We will use the following obvious property of free modules: If $M$ is a free $S$-module, and $x \in M$ does not lie in $S_+ M$, then there is a graded $S$-linear map $\lambda: M \to S$ such that $\lambda(x) \not\in S_+$. If $x$ is assumed homogenous, this means that $\lambda(x)$ is a nonzero scalar. 
Let $y_1$, ..., $y_p$ be homogenous elements of $S_+$ which are minimal generators for the $R$-ideal $S_+ R$. We first claim that 
the $y_i$ generate $S$ as a $k$-algebra. To this end, let $\lambda_0 : R \to S$ be a graded $S$-linear map with $\lambda_0(1) = 1$, which exists since $R$ is a free $S$-module. We prove by induction on $i$ that $S_i \subset k[y_1, \ldots, y_p]$. For $i=0$, this holds because $S_0=k$. Let us assume the result proved in all degrees $<i$. Let $z \in S_i$. Then $z = \sum f_j y_j$ for some $f_j \in R$ with $\deg f_j = i - \deg y_j$. But then $z = \sum \lambda_0(f_j) y_j$ and $\lambda_0(f_j) \in S_{i - \deg y_j}$. So by induction, the $\lambda_0(f_j)$ lie $k[y_1, \ldots, y_p]$, so $z$ lies in $k[y_1,\ldots, y_p]$.
Suppose for the sake of contradiction that there is an algebraic relation $H(y_1, \ldots, y_p)=0$ for some $H \in k[T_1,\ldots, T_p]$. Without loss of generality, we may assume $H$ is homogenous in the grading where $T_i$ is given degree $\deg y_i$; choose $H$ of minimal possible degree. Set
$ h_i = \left. \tfrac{\partial H}{\partial T_i} \right|_{T_j = y_j} . $
So $h_i$ is a homogenous element of $R$.
Since we are in characteristic zero, there is some $i$ for which $(\partial H)/(\partial T_i) \neq 0$ and then, by the minimality of $H$, we have $h_i \neq 0$.
Let $\vec{h}$, $\vec{x}$ and $\vec{y}$ be the vectors whose entries are the $h_i$, $x_i$ and $(\deg y_i) y_i$. Let $D$ be the matrix $[ \partial y_j/\partial x_i ]$. By the chain rule, $\vec{h}^T D=\vec{0}$. For any symmetric polynomial $y$, we have $\sum x_i \tfrac{\partial y}{\partial x_i} = (\deg y) y$, so $D \vec{x} = \vec{y}$. 
Let $\mathfrak{H}$ be the ideal generated by the $h_i$. Reorder so the $h_i$ that $\{ h_1, h_2, \cdots, h_m \}$ is a minimal set of generators for $\mathfrak{H}$. Since not all the $h_i$ are zero, $m \geq 1$.
Let $\vec{h}_{\mathrm{gen}} = [ h_1 \ h_2 \ \cdots h_m]^T$. So 
$$ \vec{h} = \begin{bmatrix} \mathrm{Id} \\ G \end{bmatrix}  \vec{h}_{\mathrm{gen}}  $$
for some $m \times (p-m)$ matrix $G$ with entries in $S$.
We thus have $\vec{h}_{\mathrm{gen}}^T [ \mathrm{Id} \ G^T ] D=\vec{h}^T D=\vec{0}$. 
We put $\tilde{D} = [ \mathrm{Id} \ G^T ] D$.
Case 1: Some entry of $\tilde{D}$ does not lie in the ideal $S_+ R$ of $R$. 
Then there is an $S$-linear map $\lambda : R \to S$ such that some $\lambda(\tilde{D}_{ij})$ is a nonzero scalar. Since the $h_i$ are in $S$, we get $\vec{h}_{\mathrm{gen}}^T \lambda(\tilde{D})=0$. Looking at the $j$-th entry in this product, we can write $h_i$ as an $S$-linear combination of the other $h_{i'}$, contradicting the minimality of $\{ h_1, h_2, \cdots, h_m \}$.
Case 2: Every entry of $\tilde{D}$ lies in $S_+ R$. Now, $\tilde{D} \vec{x} =  [ \mathrm{Id} \ G^T ] \vec{y}$, and all the $x_i$ lie in $R_+$. So the entries of $[ \mathrm{Id} \ G^T ] \vec{y}$ lie in $S_+ R_+$ and, in particular, are $0$ in $(S_+ R) \otimes_R k$. So we have relations $(\deg y_i) y_i + \sum_{j=m+1}^p G_{ij} (\deg y_j) y_j =0$ in $(S_+ R) \otimes_R k$. But the $y_i$'s were chosen minimal generators for $S_+ R$, so their images in $(S_+ R) \otimes_R k$ are a basis; contradiction. (We have used characteristic zero to divide out the scalars $\deg y_i$.) $\square$

I would love to understand this proof on a conceptual level. Large parts of it are clearly playing with Kahler's differentials and normal bundles and would work in any regular ring, but the use of $\sum x_i (\partial y/\partial x_i) = (\deg y) y$ seems very special.