Criterion for a group scheme to be smooth Let $G$ be an algebraic group over a field $k$ and fix a $k$-point $x \in G$ (let's just take the unit $ e \in G(k) $).
$G \to \mathrm{Spec}(k)$ is smooth iff it is smooth in $e$ (see proposition 1.22 in Algebraic Groups (Milne)). That is, take an affine open neighbourhood $ U \cong \mathrm{Spec} (k[x_1, \dots, x_n] / (f_1, \dots, f_r)) $ of $ e $. $ G $ is smooth iff for some choice of $ r $, $ x_1, \dots, x_n $ and $ f_1, \dots, f_r $, the determinant $ \left\vert \frac{\partial f_i}{\partial x_j}(e) \right\vert_{1 \leq i, j \leq r} $ is a unit in $ k $.
Does there exist an analogue of this result for group schemes over arbitrary rings $ R $? Like: take an affine open neighbourhood $ U \supseteq e $. If there exists a choice for $ r $, $ x_1, \dots, x_n $ and $ f_1, \dots, f_r $ such that $ U \cong \mathrm{Spec}(R[x_1, \dots, x_n] / (f_1, \dots, f_r)) $ and the determinant $ \left\vert \frac{\partial f_i}{\partial x_j}(e) \right\vert_{1 \leq i, j \leq r} $ is a unit in $ R $, then $ G \to \mathrm{Spec}(R) $ is smooth.
I found it difficult to extend Milnes proof to arbitrary rings, because for fields, a closed point of $ G $ can be identified with a $ k $-point (or $ \overline k $-point for $ k \not = \overline k $) of $ G $, but over arbitrary rings this does not work anymore.
I am currently working with an affine group scheme over $\mathbb Z_p$, if that helps.
 A: Question: "I found it difficult to extend Milnes proof to arbitrary rings, because for fields, a closed point of G can be identified with a k-point (or k¯¯¯-point for k≠k¯¯¯) of G", but over arbitrary rings this does not work anymore. I am currently working with an affine group scheme over Zp
, if that helps."
Answer: @TempestasLudi - If $G/k$ is an affine algebraic group over an algebraically closed field $k$ there is a canonical left action $\sigma: G\times_k G \rightarrow G$, and for any element $h\in G(k)$ there is an automorphism
$$f:=\sigma_h:G \rightarrow G$$
defined by $\sigma_h(u):=hu$. In particular $\sigma_h(e)=he=h$. The inverse map is $g:=\sigma_{h^{-1}}$. The maps $f,g$ induce isomorphisms of tangent spaces
$$T_e(G) \cong T_h(G)$$
hence $G$ is "regular at $h$" iff $G$ is "regular at $e$" and the element $h$ is arbitrary. If $x\in G$ is any point it follows $\mathcal{O}_{G,x}$ is the localization of $\mathcal{O}_{G,h}$ for $h$ a closed point, hence it follows $G$ is regular at any point $x$, since the localization of a regular local ring is regular:
Lemma: $G$ is regular at any point iff $G$ is regular at the identity $e$.
For group schemes over rings $R$ that are not fields you may try to use a similar approach. Notions such as "regularity", "smoothness" are usually not that "well behaved" for schemes over $R$.
Example: If $A:=\mathbb{Z}[x_{ij},t]/(tdet(x_{ij})-1)$ and $G:=SL(n,\mathbb{Z}):=Spec(A)$, it follows $\kappa(e)\cong \mathbb{Q}$ where $e$ is the identity.  Hence $e$ is not a closed point. To use the above approach, your "point" $h\in G$ must have $\mathbb{Q} \subseteq \kappa(h)$. Closed points $h\in G$ always have finite residue field. If you chose any point $h\in G(\mathbb{Q})$ the maps $f,g$ defined above will induce isomorphisms of vector spaces
$$T_e(G) \cong T_h(G)$$
hence $G$ is regular at $h$ iff $G$ is regular at $h$. The point $h\in G(\mathbb{Q})$ is arbitrary. If $k$ is a field, you get an action
$$\sigma_k:G(k) \times G(k) \rightarrow G(k)$$
and you may compare tangent spaces for points $x,y\in G(k)$: $G$ is regular at $x$ iff $G$ is regular at $y$.
