Subgroup schemes of $\mathbb{A}^n$ 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?
 A: 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. 
