What is the coordinate ring of $G/U$? Let $G$ be an algebraic group and $U$ its subgroup consisting all upper triangular matrices. For example, $G=GL_n(k)$ and $U$ the subgroup consisting of all upper triangular unipotent matrices in $GL_n(k)$, where $k$ is an algebraically closed field. 
It is said that the coordinate ring of $G/U$ is $k[G/U]= k[A]$, where $A$ is the set consisting of all minors of a matrix which contains continuous rows of the first column. This is because these minors are invariant under the action of multiply elements in $U$ from the right to a matrix in $G$. 
I am trying to understand this statement. Let $$G=GL_3(k)=\left\lbrace\left(\begin{matrix} x_{11} & x_{12} & x_{13} \\ x_{21} & x_{22} & x_{23} \\ x_{31} & x_{32} & x_{33}  \end{matrix}\right) : x_{ij} \in k, \det(x_{ij}) \neq 0 \right\rbrace.$$, 
$$
U=\left\lbrace\left(\begin{matrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1  \end{matrix}\right) : a, b, c \in k \right\rbrace.
$$ 
Now we want to verify that $$k[G/U] = k[x_{11}, x_{21}, x_{31}, x_{11}x_{22}-x_{12}x_{21}, x_{11}x_{23}-x_{21}x_{13}, x_{21}x_{32}-x_{31}x_{22},x_{21}x_{33}-x_{23}x_{31}, \det(x_{ij})] \quad (1)$$. Is this correct? 
In general, for an affine variety $X$, $k[X] = k[A^n]/I = k[x_{i}: i\in\{1, \ldots, n\}]/I$, where $I = \{f: f(x)=0, \forall x \in X\}$. I think that the coordinate ring of $GL_n(k)$ is $k[x_{ij}, y: i,j\in\{1, \ldots, n\}]/(y\det(x_{ij})-1)=k[x_{ij}, \det(x_{ij})^{-1}: i,j\in\{1, \ldots, n\}]$.
Thank you very much.
Edit: Maybe the statement should be as follows. 
The coordinate ring of $G/U$ is $k[G/U]= k[x_{ij}, A]$, where $A$ is the set consisting of the inverse of minors of a matrix which contains continuous rows of the first column. This is because these minors are invariant under the action of multiply elements in $U$ from the right to a matrix in $G$. 
Let $$G=GL_3(k)=\left\lbrace\left(\begin{matrix} x_{11} & x_{12} & x_{13} \\ x_{21} & x_{22} & x_{23} \\ x_{31} & x_{32} & x_{33}  \end{matrix}\right) : x_{ij} \in k, \det(x_{ij}) \neq 0 \right\rbrace.$$, 
$$
U=\left\lbrace\left(\begin{matrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1  \end{matrix}\right) : a, b, c \in k \right\rbrace.
$$ 
Now we want to verify that $$k[G/U] = k[x_{ij}, i, j \in \{1, \ldots, n\}, x_{11}^{-1}, x_{21}^{-1}, x_{31}^{-1}, (x_{11}x_{22}-x_{12}x_{21})^{-1}, (x_{11}x_{23}-x_{21}x_{13})^{-1}, (x_{21}x_{32}-x_{31}x_{22})^{-1},(x_{21}x_{33}-x_{23}x_{31})^{-1}, (\det(x_{ij}))^{-1}]\,. \quad (2)$$ Is this correct? 
 A: Edit As Tobias Kildetoft pointed out, I am only sure of the following argument when the characteristic of $k$ is zero. I have not ventured beyond that point much, but I would expect that the representation theory of $\operatorname{GL}_n$ is well-studied even in positive characteristic. You might still be able to make a similar argument.
By the algebraic Peter-Weyl theorem [27.3.9, TY], we can decompose the coordinate ring of $G$ as 
$$k[G]_d = \bigoplus_{\lambda\mathop{\vdash}_n d} V_\lambda \otimes V_\lambda^\ast$$
where $\lambda$ runs through all generalized partitions with at most $n$ rows and $d$ boxes, and $V_\lambda$ denotes the irreducible representation given by $\lambda$. Observe that this is the grading obtained by localizing the coordinate ring of $k^{n\times n}$ by $\det$, so $d$ may take negative values corresponding simply to taking inverse powers of the determinant.
Taking $U$-invariants, we get 
$$k[G]_d^U = \bigoplus_{\lambda\mathop{\vdash}_n d} V_\lambda $$
because $(V_\lambda^\ast)^U$ is one-dimensional.
Let $k[G]=k[x_{ij},\det^{-1}_n]$ and $x:=(x_{ij})\in k[G]^{n\times n}$. For subsets $I,J\subseteq \{1,\ldots,n\}=:[n]$, denote by $x_{IJ}$ the matrix obtained by choosing from $x$ the rows indexed by $I$ and the columns indexed by $J$. For $r:=|I|$, we write $x_I$ instead of $x_{I,[r]}$. We claim that 
$$ k[G/U] = k[G]^U = k[ \det(x_I) \mid I\subseteq \{1,\ldots,n\} ][\det\nolimits_n^{-1}] =: R.$$
First of all, we will show that $\det(x_I)$ is right-$U$-invariant. Let $r:=|I|$. For any $u\in U$ and $\newcommand{\of}[1]{\left(#1\right)}g\in G$, we calculate
\begin{align*}
((1,u).\det(x_I))(g) &=
 \det(x_I)(gu) = \det((gu)_I)
%\\ &= \det\of{ \of{ \sum\nolimits_{k=1}^n g_{ik} u_{kj} }_{i\in I,\,j\in\nums r}  }
%\\ &= \det\of{ \of{ \sum\nolimits_{k=1}^{j} g_{ik} u_{kj} }_{i\in I,\,j\in\nums r}  }
\\ &= \det\of{ g_I\cdot u_{[r]} } = \det(g_I) = \det(x_I)(g),
\end{align*}
because $u_{[r]}$ is the upper left $r\times r$ block of the unipotent matrix $u$, which has determinant $1$. Thus, $R\subseteq k[G]^U$. To see that they are equal, we will make use of the Peter Weyl theorem and the following:
Claim. $R$ is invariant under the action of $G$ from the left. 
Proof. Let $g,v\in G\newcommand{\nums}[1]{[#1]}$. Then,
\begin{align*}
\det( (gv)_I ) &= \sum_{\pi:I\cong\nums r} (-1)^\pi \prod_{i\in I} (gv)_{i,\pi(i)}
= \sum_{\pi:I\cong\nums r} (-1)^\pi \prod_{i\in I} \sum_{j=1}^n g_{ij} v_{j\pi_i}
\\ &= \sum_{\pi:I\cong\nums r} (-1)^\pi \sum_{j:I\to\nums n} \prod_{i\in I} g_{ij_i} v_{j_i\pi_i}
\\ &= \sum_{j:I\to\nums n} \sum_{\pi:I\cong\nums r} (-1)^\pi\prod_{i\in I} g_{ij_i} v_{j_i\pi_i}
\\ &= \sum_{\substack{J\subseteq\nums n\\|J|=r}}
  \sum_{\pi:I\cong\nums r} \det(g_{IJ})\cdot\det(v_J).
\end{align*}
Hence, $g.\det(x_I)\in \sum_{J\subseteq\nums n} k\cdot\det(x_J)\subseteq R$ for all $I$.
Since $R\subseteq k[G]^U$ and $R$ is a $G$-module, it will suffice to show that $R$ contains a highest weight vector of each weight $\lambda$.
Indeed, it is not hard to construct one highest weight vector of each of the generating weights:
\begin{align*}
\omega_1 &= (1,0,0,\ldots,0), \\
\omega_2 &= (1,1,0,\ldots,0), \\
\vdots & \\
\omega_n &= (1,1,1,\ldots,1), \\
\omega'_n &= (-1,\ldots,-1).
\end{align*}
The upper left minors $f_r:=\det(x_{\nums r})$ are also invariant under the action of $U$ from the left and an invertible diagonal matrix $t=\operatorname{diag}(t_1,\ldots,t_n)$ satisfies $t.f_r = t_1\cdots t_r\cdot f_r$. Hence, $f_r$ is a highest weight vector of weight $\omega_r$ and since $f_n=\det_n$ is invertible, we have also a highest weight vector $f_n^{-1}$ of weight $\omega'_n$. We can multiply the $f_r$ to construct any highest weight vector corresponding to a dominant weight.
[TY] Tauvel & Yu - Lie Algebras and Algebraic Groups
A: First we describe the variety $G/U$. We have
\begin{align}
\left( \begin{matrix} x_{11} & x_{12} & x_{13} \\
x_{21} & x_{22} & x_{23} \\
x_{31} & x_{32} & x_{33} \end{matrix} \right) \left( \begin{matrix} 1 & a & b \\
0 & 1 & c \\
0 & 0 & 1 \end{matrix} \right) = \left( \begin{matrix} x_{11} & x_{12} + a x_{11} & x_{13} + b x_{11} + cx_{12} \\
x_{21} & x_{22} + a x_{21} & x_{23} + b x_{21} + cx_{22} \\
x_{31} & x_{32} + a x_{31} & x_{33} + b x_{31} + cx_{32} \end{matrix} \right).
\end{align}
Therefore $g=\left( \begin{matrix} x_{11} & x_{12} & x_{13} \\
x_{21} & x_{22} & x_{23} \\
x_{31} & x_{32} & x_{33} \end{matrix} \right)$ and $g'=\left( \begin{matrix} x_{11}' & x_{12}' & x_{13}' \\
x_{21}' & x_{22}' & x_{23}' \\
x_{31}' & x_{32}' & x_{33}' \end{matrix} \right)$ if and only if the minors of $g$ and $g'$ which contains elements of the first column and with consecutive rows are the same. For example, $x_{11}=x_{11}', x_{11}x_{22}-x_{12}x_{21} = x_{11}'x_{22}'-x_{12}'x_{21}'$.
Therefore the coordinate ring of $G/U$ is $k[G/U] = k[x_{11}, x_{21}, x_{31}, x_{11}x_{22}-x_{12}x_{21}, x_{11}x_{32}-x_{12}x_{31}, x_{21} x_{32} - x_{22} x_{31}, \det(x_{ij})]$.
