Open orbits under the action of an algebraic group Let $k$ be a field, $X$ an algebraic variety, and $G$ a smooth algebraic group, acting on $X$ via $(g,x)\mapsto g\cdot x$.
Fixing $x$ in $X$ a $k$-point, there is a map $f_x:G\rightarrow X$ sending $g\mapsto g\cdot x$.
Now, assuming that $df_x:T_eG\rightarrow T_xX$ is surjective, how can one show that the orbit $G\cdot x$ is open in $X$?
NB: I expect that some restrictions on the field are necessary, but I'd like to keep them as minimal as possible!
 A: If $X$ is additionally assumed to be smooth, the orbit $G\cdot x$ is open in $X$ under the assumption that $T_eG \rightarrow T_xX$ is surjective.
Namely, the property "being an open subset" satisfies fpqc-descent, so we may (using the fpqc-cover $X \otimes_k k^\text{alg} \rightarrow X$) wlog assume that $k$ is algebraically closed. The condition imposed on tangent spaces implies that the map on tangent bundles $\mathcal{T}_G \rightarrow f_x^\ast\mathcal{T}_X$ induced by $f_x \colon G \rightarrow X$ is surjective in the fiber at $e$. This surjectivity extends to an open subset of $G$, i.e., there is some open $U\subseteq G$, such that $f_x|_U \colon U \rightarrow X$ induces surjections on tangent spaces at each point of $U$. By the last part of Alexander Bett's answer to Smooth morphism (algebraic geometry) vs. Submersion (differential geo) & Ehresman's Lemma (essentially, miracle flatness is used), $f_x|_U \colon U \rightarrow X$ is flat and as it is also of finite presentation (being a map between $k$-varieties), it follows that its image is open in $X$. Thus the $G$-orbit $G\cdot x$ contains an open subset of $X$. On the other side, $G\cdot x$ is a locally closed subset of $X$ (this holds for any action of a smooth algebraic group on a variety). Now, a locally closed subset of an irreducible variety (we may wlog assume that $X$ is irreducible) containing an open subset must be open itself.
I am not sure what happens for non-smooth $X$, but the only place where smoothness was used above is the application of miracle flatness. So, the smoothness hypothesis probably can be weakened.
A: $\DeclareMathOperator\Im{Im}$I don't think that smoothness of $X$ is necessary, just the fact that it is geometrically integral (I assume your varieties are geometrically integral). (As noted by Jason Starr, smoothness of $G$ is necessary.)
Here is a possible argument, please let me know if there are any mistakes.
As noted by AlexIvanov, it suffices to prove openness after passing to the algebraic closure of $k$, so we can assume that $k$ is algebraically closed.
Let $Z$ be the scheme theoretic closure of $G$ in $X$. The set theoretic image $\Im(G)$ of $G$ is open inside $Z$ by Chevalley's theorem and translating using the group action. After replacing $X$ with the $G$-stable open subvariety $X \setminus (Z \setminus \Im(G))$, we can assume that the natural map $G \to Z$ is surjective, in other words the orbit is closed. What we need to show at this point is that $Z = X$. We will argue by contradiction. Assume $Z\subset X$ is a proper closed subset.
We know that the tangent space of $Z$ at the base point $x$ agrees with that of $X$ (at $x$) by assumption. Using the fact that $k$ is algebraically closed, we can translate to see that the tangent space of $Z$ at any of its closed points coincides with that of $X$. But note that since $G$ is reduced (because $G$ is smooth), $Z$ is reduced. Since $k$ is algebraically closed, $Z$ is smooth over an open subset. In other words, the dimension of the tangent space of $Z$ for all points in an open dense agrees with the Krull dimension of $Z$. If $Z \subset X$ is a proper closed subset, the dimension of $Z$ is strictly smaller than that of $X$. This is a contradiction, because the dimension of the tangent space of $X$ at any closed point is at least the Krull dimension of $X$.
