How to show that $\mathbb{C}[SL_n]^N$ is generated by flag minors? Let $SL_n$ be the special linear group and $N$ the subgroup of $SL_n$ consisting of all unipotent upper triangular matrices. By definition, we have 
\begin{align}
\mathbb{C}[SL_n]^N = \{ f\in \mathbb{C}[x_{11}, x_{12}, \ldots, x_{nn}]: f(gn)=f(g), \forall \ g \in SL_n, n \in N \}.
\end{align}
It is easy to see the flag minors: $\Delta_I: x \to \det(x_{ij}: i \in I, j \in \{1, \ldots, |I|\})$, $I \subsetneq \{1, \ldots, n\}$, $I \neq \emptyset$, are in $\mathbb{C}[SL_n]^N$. 
How to show that $\mathbb{C}[SL_n]^N$ are generated by flag minors? Thank you very much.
 A: Let $k = \mathbb{C}$.
Let $V$ be the n-dimensional vector space over $k$ with the natural action of $SL_n(k)$. Let $e_1, e_2, \dots, e_n$ be the standard basis of the vector space $V$. Consider the following map $$\pi : SL_n \rightarrow \wedge^*V = \oplus_{1 \leq i \leq n} \wedge^iV$$ given by $g \mapsto (ge_1, ge_1 \wedge ge_2, \dots, ge_1\wedge ge_2 \wedge \dots\wedge ge_n)$ which is the natural induced representation of $SL_n$ on the alternating algebra. Note that this map does not have zero in the image and hence could be further composed with the quotient map to the projective space which is the map which gives an embedding of $G/B$ in a projective space.
It follows easily that the stabilizer of the point $(e_1, e_1\wedge e_2,\dots, e_1 \wedge e_2 \wedge \dots \wedge e_n)$ is exactly $U_n$. Computing $d\pi$ we see that $\ker(d\pi) = \mathrm{Lie}(U_n)$ and hence $\pi$ is a quotient map. The image of the map is a locally closed subvariety of $\wedge^*V$ which is an affine variety and hence the image is a quasi affine variety. 
Consider the induced co-morphism $\pi^* : k[\wedge^* V] \rightarrow k[SL_n]$. Now $\mathrm{Im}(\pi^*(k[\wedge^* V])) = k[SL_n/U_n] = k[SL_n]^{U_n}$. This last equality follows from the definition of the quotient (in the sense of Borel's Linear Algebraic Groups).
Now we have to find the image of the co-morphism. Note that the coordinate corresponding to $e_I = \wedge_{i \in I} e_i$, call it $x_{I}$ goes to $\Delta_I$ a flag minor. Since the coordinates $x_I$ generate $k[\wedge^*V]$, hence the image will be generated by $\Delta_I$ which is exactly $k[SL_n]^{U_n}$.
