is finite) then a necessary and sufficient condition for $u$ to be faithfully flat is that $\dim{{\mathscr{O}}}(G)^H=1$.

We leave the proof of necessity in the statement to the reader and from now on assume that $\dim{{\mathscr{O}}}(G)^H=1$. Clearly, the equality $I=G$ is equivalent to faithful flatness of $u$. Since ${{\mathscr{O}}}(G)^H={{\mathscr{O}}}(G)^I$, the assumptions translate into $\dim{{\mathscr{O}}}(G)^I=1$. Now, we know that ${{\mathscr{O}}}(G)$ is a projective ${{\mathscr{O}}}(G)^I$-module whose rank equals $\dim{{\mathscr{O}}}(I)$ (see [@mumford70 §11, Theorem 1(B), p.111] or III.2.4 of [@demazure-gabriel70]). Then, since $\dim{{\mathscr{O}}}(G)^I=1$, we conclude that $\dim{{\mathscr{O}}}(I)=\dim{{\mathscr{O}}}(G)$, which means that $I= G$.

\[21.03.2019–1\] Let $u:H\to G$ be an arrow in $({\mathbf{FGSch}}/A)$ and assume that $G$ is pseudo-finite. Then the following conditions are equivalent.

(1) Both equalities $$A{\cdot}1= \left(A[G]_{\rm right}\right)^H \quad \text{and}\quad k{\cdot}1= \left(k[G]_{\rm right}\right)^H,$$ are true.

(2) The morphism $u$ is faithfully flat.

(3) Let $${\mathfrak{s}}(G)=
    \left\{ \begin{array}{c} \text{$V\subset A[G]_{\rm right}$ is $G$-invariant, is finitely generated }
    \\
    \text{ as an $A$-module and contains the constants}
    \end{array}
    \right\}$$ and $${\mathfrak{s}}_0(G)=
    \left\{ \begin{array}{c} \text{$M\subset k[G]_{\rm right}$ is $G$-invariant, is finitely generated }
    \\
    \text{ as a $k$-space and contains the constants}
    \end{array}
    \right\}.$$ Then, for each $V\in{\mathfrak{s}}(G)$ and $M\in{\mathfrak{s}}_0(G)$ we have $$A{\cdot}1= V^H \quad \text{and}\quad k{\cdot}1= M^H.$$

\(1) $\Rightarrow$ (2). The assumptions show that 