Let $k$ be a field and $k[{\bf x}] = k[x_{ij}: 1 \leq i, j \leq n]$ be a polynomial algebra that I can think of as the algebra of functions on $n \times n$ matrices that are polynomial in each coordinate. I'd like to understand the polynomials $f$ in $k[{\bf x}]$ that are invariant by left translations by the unipotent group $N$ of upper triangular matrices: that is, those $f \in k[{\bf x}]$ that satisfy $f(ng) = f(g)$ for all $g \in M_n(k)$ and all $n \in N$.
In this blog post on the representations of $GL_n$, I think that David Speyer claims that the algebra of $N$-invariants as described above is generated, as an algebra over $k$, by all the bottom-justified minors: that is, for every subset $\sigma =\{\sigma_1, \ldots, \sigma_r\} \subset \{1, \ldots, n\}$ let $f_{\sigma}$ be the polynomial that corresponds to the $r \times r$ minor given by the bottom $r$ rows and the columns $\sigma_1, \ldots, \sigma_r$. Then the algebra generated by all of the $f_\sigma$ (something like a Plücker algebra) is supposedly exactly the algebra of $N$-invariants.
It's not hard to convince yourself that all the $f_\sigma$s are $N$-invariant, but how can you show that there's nothing else? Speyer tantalizingly suggests that there are various arguments, and that Theorem 14.11 of Miller-Sturmfels provides one, but I can't figure out how it does that. For what it's worth, the theorem says that the $f_\sigma$s form a sagbi basis for the Plücker algebra with any term order where the leading term of each $f_\sigma$ is the diagonal (or antidiagonal) one. You can see the statement here; search inside for "diagonal or antidiagonal". (Keep in mind that Miller-Sturmfels uses top-justified minors, so the Plücker algebra there is invariant by left translation by the lower-triangular unipotents. but I don't think that changes things much.)
Any ideas would be much appreciated! I'd love to understand how Theorem 14.11 might be used to prove that the $f_\sigma$s generate the full algebra of $N$-invariants. I'm also terribly curious about other possible arguments.