I'm reading the book Representations of Algebraic Groups of Jantzen, chapter 1, part II. When discussing big cells of a split, connected reductive group $G$ over an integral domain $k$, the author states that, because the big cell $U^+B$ is dense in $G$, so we can regard $k[G]$ as a subalgebra of $k[U^+B]$ (this is justifiable because the inclusion map is dominant in this case). Because $k[U^+B]$ is isomorphic to a product $G_m$ and $G_a$, it must be infinitesimally flat ($G$ is infinitesimally flat if $k[G]/I^n$ is finitely presented and projective for all $n \in \mathbb{N}$, here $I$ is the augmentation ideal of $G$).
But from then, the author states that $G$ must also be infinitesimally flat, which confuses me a lot. Is it obvious here? Do we have any lemma states that, if $A$ is a $k$-algebra, $B$ is a subalgebra of $A$, $I$ is an ideal of $A$, then $A/I^n \simeq B/(I \cap B)^n$, so if $A/I^n$ is finitely presented and projective, then $B/(I \cap B)^n$ must be so?
The second statement in the book is also confusing due to the same reason, which states that $Dist(G) \simeq Dist(U^+B)$. Here $Dist(G) = \bigcup Dist_n(G)$ and $Dist_n(G) = (k[G]/I^{n+1})^*$. So the second statement seems to be deduced from the fact that (assume the same hypothesis as above) $(A/I^n)^* \simeq (B/(I \cap B)^n)^*$. If any one has any idea about this, please help me. Thanks a lot.
Question: "The second statement in the book is also confusing due to the same reason, which states that $Dist(G)≃Dist(U^+B)$....If any one has any idea about this, please help me. Thanks a lot."
Answer: If $x\in X$ is a point and if $\mathcal{O}_{X,x}$ is the local ring of $\mathcal{O}_X$ at $x$ with maximal ideal $\mathfrak{m}$, you define
$$Dist_n(X,x):=\mathcal{O}_{X,x}/\mathfrak{m}^{n+1}$$
(Jantzen, Representations of algebraic groups, page 111). If $x\in U \subseteq X$ is an open subscheme you get
$$Dist_n(X,x) \cong Dist_n(U,x)\text{ and } Dist(X,x) \cong Dist(U,x) ,$$
since $\mathcal{O}_{X,x}\cong \mathcal{O}_{U,x}$. Hence if $U^+B \subseteq G$ contains an open neighborhood $U \subseteq U^+B$ of $x$, you get
$$Dist_n(X,x)=Dist_n(U,x)=Dist_n(U^+B,x)$$
and
$$Dist(X,x)=Dist(U,x)=Dist(U^+B,x).$$
By Jantzen, page 182 it follows the $wU^+B$ form an open cover of $G$ (see note $(7)$), hence it may be this proves the result.
In general if $G/S$ is a group scheme and $e\in G$ is the "multiplicative identity" and $U \subseteq G$ an open subgroup scheme, it follows $e\in U \subseteq G$. The algebra of distributions $Dist(G)$ is defined in terms of the local ring $\mathcal{O}_{G,e}$, and since $\mathcal{O}_{G,e} \cong \mathcal{O}_{U,e}$ (since $U$ is an open subgroup scheme) it follows $Dist(G) \cong Dist(U)$ for any such open $U$.
Note: You find an "elementary introduction" to affine group schemes, Hopf algebras and the algebra of distributions in the exercises in Waterhouse "An introduction to affine group schemes", Ex.12.10. This may be helpful. Since the algebra of distrubtions is defined locally in terms of $e$ you may restrict to affine group schemes.
Note also that from the above it follows the group scheme $G/S$ is not determined by its algebra $Dist(G)$ of distributions.