TITLE: Left U_n-invariants of SL_n - an exercise in Kraft-Procesi
QUESTION [5 upvotes]: I am sorry for spamming MO with questions I have not thought about for more than 3 hours, but currently I am quite busy with preparing a talk on representations of $S_n$, and I don't want these to get lost. I hope this one is not quite as vague as the last one.
This here is an attempt to generalize Exercise I.21 in Kraft-Procesi, Classical Invariant Theory.
Let $K$ be a field - say, infinite, since we are going to do classical invariant theory. Let $K\left[\mathrm{SL}_n K\right]$ denote the $K$-algebra of polynomial functions on $\mathrm{SL}_n K$, which I define either as
$\left\lbrace f\mid_{\mathrm{SL}_n K} \ \mid \ f\in K\left[\mathrm{M}_n K\right]\right\rbrace$
or as $K\left[\mathrm{M}_n K\right]\diagup \left(\det-1\right)$ (proving the equivalence of these two definitions is not the matter, it's rather easy - even easier than Kraft and Procesi try to make one believe).
Now, the group $\mathrm{U}_n K$ of unipotent upper triangular matrices acts on $\mathrm{SL}_n K$ from the left. What is the invariant ring? It is easily seen that
$\det\left(\text{the submatrix formed by the intersection of the rows }i,i+1,...,n\text{ with the columns }j,i+1,i+2,...,n\right)$
is an invariant for any $i\geq j$. These generate the fraction field of the invariants, but do they also generate the ring of the invariants itself?
(The above-mentioned exercise is the above for $n=2$.)
Arguments using Victorian age methods (as opposed to Zariski-topological or other algebro-geometrical) would be particularly preferred.
EDIT: As Allen Knutson has pointed out, my question has a negative answer. However, the (larger) collection of determinants of the form
$\det\left(\text{the submatrix formed by the intersection of the rows }i,i+1,...,n\text{ with the columns }j_1, j_2, ..., j_{n-i+1}\right)$
for $1 < i \leq n$ and $1 \leq j_1 < j_2 < ... < j_{n-i+1} \leq n$ does generate the ring of invariants. When $K$ has characteristic $0$, this can be proven using the standard theory of highest-weight modules and multiplicity-free algebras explained in Kraft-Procesi (see my errata, "Page 9, Exercise 21" for a proof). I am still wondering whether it holds for arbitrary $K$ and has a more elementary or combinatorial proof.

REPLY [2 votes]: Your question (in char 0) has already been answered by Knutson and Victor Protsak. I just wanted to say that this holds in greater generality (char 0) but the method is not Victorian. 
So, consider the algebraically closed field $K$ of char 0, and $G=SL_n$. Given an irreducible $V_{\lambda}$ with highest weight $\lambda$ (relative to  the standard upper triangular subgroup $B=TU$where $T$ is the group of diagonals), we have the decomposition for the action of $G\times G$ on the coordinate ring $k[G]$
$$k[G]=\bigoplus V_{\lambda }^* \otimes V_{\lambda}.$$ Taking $U$ invariants on the left, we have 
$$k[U\backslash G]=\bigoplus V_{\lambda},$$ i.e. every irreducible representation $V_{\lambda}$  occurs exactly once and is generated as a $G$ module by the vector $v_{\lambda}$ which is invariant under the group $V$ of $lower \quad triangular$ unipotent matrices. If $\lambda_1, \cdots, \lambda_{n-1}$ are the highest weights of the fundamental representations, and $\lambda =\sum a_i\lambda _i$ with $a_i\geq 0$ then clearly (by multiplicity one) $v_{\lambda}=\prod v_i^{a_i}$ is a product of powers of $v_{\lambda_1}, \cdots ,v_{\lambda _{n-1}}$. Hence the ring of invariants is generated by the functions spanning $V_{\lambda_1},\cdots  V_{\lambda _{n-1}}$. Thus the ring of invariants is finitely generated.
The above proof works for any connected reductive group.