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Let $R$ be an integral $\bar{k}$-algebra of finite type. Let $V(I) \subseteq \mathbb{A}_R^n$ be a reduced (closed) subgroup scheme such that $V(I)\backslash \{0\} \neq \emptyset$ and the $\mathbb{G}_m^R$-action on $\mathbb{A}_R^n \backslash \{0\}$ restricts to a free action on $V(I)\backslash \{0\}$. Is it true that $I$ can generically be generated by linear polynomials? (If $R=k$ is any field, and $V(I)$ as above, can $I$ be generated by linear polynomials?)

If $R=\bar{k}$ this is clear, because we can just check it on $\bar{k}$-points (should not even need the $\mathbb{G}_m^R$-action). In general, intuitively this seems good to me (I would just think of 'generalised' subvector spaces), but I know too few about group schemes to verify it. I guess if there was a counterexample, then it would be in the non-perfect world.

Edit: As Jason pointed out there are indeed counterexamples. What if we additionally assume that $V(I)$ has reduced (closed) fibers?

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    $\begingroup$ What is your definition of a "free action"? Since every $\mathbb{G}_m$-orbit contains the origin in its closure, every nonempty $\mathbb{G}_m$-invariant closed subscheme contains the origin. The stabilizer of the origin is not trivial, it is all of $\mathbb{G}_m$. $\endgroup$ – Jason Starr Jul 16 '16 at 17:03
  • $\begingroup$ You are right of course. $V(I) \backslash \{0\}$ is supposed to be non-empty and the action on $V(I) \backslash \{0\}$ is supposed to be free. $\endgroup$ – Louis Jul 16 '16 at 17:13
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    $\begingroup$ In that case, as you predicted, there are counterexamples arising from non-perfect function fields. Let $R$ be $\overline{k}[t]$. Let $\mathbb{A}^2_R$ be $\text{Spec} R[x,y]$. Let $I\subset R[x,y]$ be the ideal $\langle x^p-ty^p \rangle$. This is a radical ideal, and the associated closed subscheme is a subgroup scheme. $\endgroup$ – Jason Starr Jul 16 '16 at 17:16
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    $\begingroup$ Denote the ring of regular functions on $\mathbb{A}^n_R$ by $S=R[x_1,\dots,x_n]$. The ideal $I$ is homogeneous, by hypothesis. Thus $I$ is the direct sum of its graded components $I_d$. In particular, since direct sum decompositions are compatible with base change, $I_1$ is compatible with base change of $R$. Similarly, the quotient $S$-module $I/(I_1\cdot S)$ is compatible with base change of $R$. If this $S$-module is nonzero, then its support contains a $\overline{k}$-point in $\mathbb{A}^n_R$. This maps to a $\overline{k}$-point of $\text{Spec}(R)$. Thus, some fiber is nonreduced. $\endgroup$ – Jason Starr Jul 16 '16 at 18:29
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    $\begingroup$ That is neat. To recap: we used 'group action' for 'I is homogeneous'; '$V_x$ is reduced group scheme for all closed $x \in V$' and '$R$ is $\bar{k}$-algebra' for '$I \otimes k(x)$ is generated in degree $1$'. If you want to post this comment as a an answer, I will gladly accept it. Thanks a lot! (and suddenly the question seems stupid after all...) $\endgroup$ – Louis Jul 16 '16 at 18:41
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I am posting my comments above as an answer. If one does not impose a condition on fibers, then there are counterexamples such as when $R=\overline{k}[t]$ and the ideal $I$ in $R[x,y]=\Gamma(\mathbb{A}^2_R,\mathcal{O})$ equals $\langle x^p-ty^p\rangle$.

On the other hand, if the fiber over every $\overline{k}$-point of $\text{Spec}(R)$ is reduced, then the ideal $I\subset R[x_1,\dots,x_n]$ is generated by $I_1=I\cap R[x_1,\dots,x_n]_1$. The homogeneity condition on $V(I)$ implies that $I$ is homogeneous, i.e., $I$ equals the direct sum over every integer $d$ of $I_d=I\cap R[x_1,\dots,x_n]_d$. In particular, $I_1$ is compatible with arbitrary base change of $R$, so that also the cokernel $I/I_1 R[x_1,\dots,x_n]$ is compatible with arbitrary base change. If the support of this module is nonempty, then this closed subset of $\mathbb{A}^n_R$ contains a $\overline{k}$-point. This maps to a $\overline{k}$-point of $\text{Spec}(R)$. Thus, the base change of $I$ over this $\overline{k}$-point of $R$ is also not generated by linear polynomials. That implies that the fiber is nonreduced.

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