Polynomial representation of $Gl_n(k)$ with finitely many orbits I am currently reading Etingofs notes on representation theory and I am struggling to solve the Problem 5.2. leading up to the proof of Gabriels theorem: 
Let $k$ be an algebraically closed field, and $G = GL_n(k)$. Let $V$ be a polynomial representation of $G$. Show that if $G$ has finitely many orbits on $V$ then $dim(V) \leq n^2$. Namely:
(a) Let $x_1, ..., x_N$ be linear coordinates on $V$ . Let us say that a subset $X$ of $V$ is Zariski dense if any polynomial $f(x_1, ..., x_N)$ which vanishes on $X$ is zero (coefficientwise). Show that if $G$ has
finitely many orbits on $V$ then $G$ has at least one Zariski dense orbit on $V$.
(b) Use (a) to construct a field embedding $k(x_1, ..., x_N) \to k(g_{pq})$, then use Problem 5.1.
Now if one shows (a) the rest of the problem is trivial, so the main problem is the first part.
 A: One natural way to think about this uses two basic facts: 
(1) If $V$ is an irreducible variety and $U \subseteq V$ is a closed subvariety of the same dimension, then $U=V$. 
(2) The dimension of the union of finitely many (locally closed) subvarieties of a variety $V$ is the maximum of their dimensions.
Evidently $V$ is the disjoint union of the orbits of $G$, fact (2) implies that one of them has dimension equal to the dimension of $V$. Thus fact (1) finishes the problem.
A: (a) Let the orbits be $O_1, \ldots, O_n$. Suppose by way of contradiction that none of the $O_i$ are Zariski dense. Then for each $i$ there exists a nonzero polynomial $p_i$ which vanishes on $O_i$. Then$$P = \prod_{i = 1}^m p_i$$is a nonzero polynomial that vanishes on all $V$.
Lemma. If $k$ is algebraically closed and $P$ is a nonzero polynomial in $k[x_1, \ldots, x_n]$, then there exist $t_1, \ldots, t_n$ such that $$P(t_1, \ldots, t_n) \neq 0.$$ Proof. Note that $k$ has infinitely many elements, since it is algebraically closed.
Induct on $n$. For $n = 1$, the polynomial can have at most $n$ roots so the assertion holds. Suppose it is proved for $n = 1$ and $P \in k[t_1, \ldots, t_n]$. Since $k[t_1, \ldots, t_{n - 1}]$ has infinitely many elements, thinking of $P$ as a polynomial of $t_n$ with coefficients in $k[t_1, \ldots, t_{n - 1}]$, some element in $k[t_1, \ldots, t_{n - 1}]$ is not a zero of $P$. Set $t_n$ to be this element to get a nonzero element of $k[t_1, \ldots, t_{n - 1}]$. By the induction hypothesis we can find values for $t_1, \ldots, t_{n - 1}$ so that the polynomial does not evaluate to $0$; substitute these values into the polynomial for $t_n$ to get $t_n$.$$\tag*{$\square$}$$By the lemma, $P$ does not vanish on $V$, contradiction. Hence some $O_i$ must be dense.
(b) From (a), $V$ has a dense orbit. Suppose the orbit of $v$ is dense. Since $V$ is a polynomial representation, $\rho(g)$ has entries that are polynomials in the entries of $g$, and$$\rho(g)v = \begin{pmatrix} q_1 \\ \vdots \\ q_N\end{pmatrix}$$for some polynomials $q_i$ in the entries of $g$ and $g^{-1}$. Since the entries in $g^{-1}$ are rational functions of the entries in $g$, each $q_i$ is a rational function in the $g_{pq}$, i.e. $q_i \in k(g_{pq} : 1 \le p,q \le n)$. Define $f(x_i) = q_i$ to get a map from $k(x_1, \ldots, x_N)$ to $k(g_{pq})$.
If$$P(q_1, \ldots, q_N) = 0,$$then the polynomial $P$ vanishes on the orbit of $v$. Since the orbit is dense, $P = 0$. This shows that $f$ is injective, hence a field embedding.
By Problem 5.1,$$\text{dim}(V) = N \le n^2,$$and so we are done.

For reference, I reproduce the text of Problem 5.1 here.

Problem 5.1. Field embeddings. Recall that $k(y_1, \ldots, y_m)$ denotes the field of rational functions of $y_1, \ldots, y_m$ over a field $k$. Let$$f: k[x_1, \ldots, x_n] \to k(y_1, \ldots, y_m)$$be an injective $k$-algebra homomorphism. Show that $m \ge n$. (Look at the growth of dimensions of the spaces $W_N$ of polynomials of degree $N$ in $x_i$ and their images under $f$ as $N \to \infty$.) Deduce that if$$f: k(x_1, \ldots, x_n) \to k(y_1, \ldots, y_m)$$is a $k$-linear field embedding, then $m \ge n$.

