faithful flatness for the group scheme Let $K$ is an algebraic closed field, $$f:S{\longrightarrow}T$$ be a map of closed set in $K^n$. Moreover, assume that $f(S)$ is dense in $T$. Show $f(S)$ contains an open dense subset of $T$.
It is an exercise from the book of Waterhouse with a hint for the theorem 13.4 of generic faithful flatness, maybe I do not catch what is the meaning of faithful flatness for the group scheme, so I need some help.
 A: Assume that $T$ is irreducible.  Let $R$ be the coordinate ring of the affine variety $T$ and let $A$ be the coordinate ring of $S$.  Then $f$ gives a map $\phi:R\rightarrow A$.  Since $f$ is dominant and $R$ is reduced, it follows that $\phi$ is injective. Hence we can assume $R\subseteq A$.  Since $T$ is irreducible, $R$ is a domain.  Then the theorem you mentioned states that there exists $r\in R$ and $a\in A$ such that $R_r\rightarrow A_a$ is faithfully flat.  
But $R_r$ is the coordinate ring of the open subvariety $D(r)\subseteq T$ and $A_a$ is the coordinate ring of $D(a)\subseteq S$.  So we have shown that we get a faithfully flat morphism $D(a)\rightarrow D(r)$ by restricting the domain and codomain of $f$.  But faithfully flat morphisms of varieties are surjective, so $D(r)\subseteq T$ is the image of $D(a)$ under $f$, and hence is contained in $f(S)$.  Since $T$ is irreducible, the open set $D(r)$ is dense, so is is the desired dense open set.
For the case where $T$ may not be irreducible, let $T'$ be an irreducible component.  The map $f:f^{-1}(T')\rightarrow T'$ has dense image, so we may apply the case considered above.  We have a dense open set $U\subseteq T'$ which is contained in $f(f^{-1}(T'))\subseteq f(S)$.
Do this for each component $T_1,\ldots T_n$ to get dense open sets $U_i\subseteq T_i$ with $U_i\subseteq f(S)$.  Then $\bigcup U_i$ is dense in $T=\bigcup T_i$, and can be seen to be the desired dense open subset.
