If a polynomial ring is finite free over a subring, is the subring polynomial?

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 $S \cong k[y_1,\ldots,y_n]$?

• Are you assuming a particular grading group $G$ such that $S$ is $G$-graded? – Ali Taghavi Oct 5 '17 at 18:26
• Can we take $R=\mathbb{Q}(\sqrt{2})$ and $S=\mathbb{Q}$, with trivial grading? – Sam Hopkins Oct 5 '17 at 18:35
• @SamHopkins Good point, sub-$k$-algebra. – David E Speyer Oct 5 '17 at 19:03
• @AliTaghavi Standard grading on $R$. – David E Speyer Oct 5 '17 at 19:04

Note that $R$ and $S$ each have only one graded maximal ideal, so they are local in the graded sense, so most of the standard results for ungraded local rings are applicable. There is an obvious finite resolution of $k$ by modules that are finitely generated and free over $R$ and thus also over $S$. This implies that $S$ has finite global dimension, and so is a regular local ring by a theorem of Serre (Theorem 19.2 in Matsumura). Together with the grading assumptions this forces $S$ to be polynomial.

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.