**TL;DR version.**

Given a commutative ring $\mathbf{k}$ and $n+m$ "generic" vectors $\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_n, \mathbf{y}_1, \mathbf{y}_2, \ldots, \mathbf{y}_m$ in $\mathbf{k}^k$ (really over a polynomial ring, not over $\mathbf{k}$), how can we describe the polynomial relations that hold between the $nm$ scalars $\left(\mathbf{x}_i\right)^T \mathbf{y}_j$ ? Is the ideal of these relations spanned by the $\left(k+1\right)\times\left(k+1\right)$-determinants that one would expect?

**Long version.**

Here is a classical result.

Let $\mathbf{k}$ be a commutative ring with $1$. Let $n$ and $m$ be two nonnegative integers. Let $A$ be the polynomial ring $\mathbf{k}\left[z_{i,j} \mid 1\leq i\leq n, \ 1 \leq j \leq m\right]$, and let $B$ be the polynomial ring $\mathbf{k}\left[x_1,x_2,\ldots,x_n, y_1,y_2,\ldots,y_m\right]$. We define a $k$-algebra homomorphism $\phi : A \to B$ by sending each $z_{i, j}$ to $x_i y_j$. Then, $\operatorname{Ker}\phi$ is the ideal of $A$ generated by all elements of the form $z_{a, b} z_{c, d} - z_{a, d} z_{c, b}$ with $1 \leq a \leq n$, $1 \leq b \leq m$, $1 \leq c \leq n$ and $1 \leq d \leq m$.

A nice proof that Bjorn Poonen suggested to me in 2011 can be found on m.se (see also the original AoPS thread).

Now, let me attempt to generalize this, inspired by a suggestion from Alexander Chervov on a rather different problem in MO question #102874.

Let $\mathbf{k}$ be a commutative ring with $1$. Let $k$, $n$ and $m$ be three nonnegative integers. Let $A$ be the polynomial ring $\mathbf{k}\left[z_{i,j} \mid 1\leq i\leq n, \ 1 \leq j \leq m\right]$ (this is the same $A$ as before), and let $B_k$ be the polynomial ring $\mathbf{k}\left[\left(x_{i, \ell} \mid 1 \leq i \leq n,\ 1 \leq \ell \leq k \right) \cup \left(y_{j, \ell} \mid 1 \leq j \leq m, \ 1 \leq \ell \leq k\right)\right]$ (this is a polynomial ring in $nk+mk$ variables). For every $1 \leq i \leq n$, we let $\mathbf{x}_i$ be the vector $\left(x_{i,1}, x_{i,2}, \ldots, x_{i,k}\right)^T \in \left(B_k\right)^k$. For every $1 \leq j \leq m$, we let $\mathbf{y}_j$ be the vector $\left(y_{j,1}, y_{j,2}, \ldots, y_{j,k}\right)^T \in \left(B_k\right)^k$. We define a $k$-algebra homomorphism $\phi_k : A \to B_k$ by sending each $z_{i, j}$ to $\left(\mathbf{x}_i\right)^T \mathbf{y}_j$. The question is what $\operatorname{Ker}\left(\phi_k\right)$ is.

I have a guess: Let $W_k$ be the ideal of $A$ formed by all determinants of the form

$\det\left( \left( \left( \mathbf{x}_{i_a} \right)^T \mathbf{y}_{j_b} \right) _{1 \leq a \leq k+1, \ 1 \leq b \leq k+1} \right)$

ranging over all pairs $\left(\left(i_1,i_2,\ldots,i_{k+1}\right), \left(j_1,j_2,\ldots,j_{k+1}\right)\right) \in \left\{ 1, 2, \ldots, n \right\}^k \times \left\{ 1, 2, \ldots, m \right\}^k$. (Of course, the ideal does not change if we restrict these pairs by requiring $i_1 < i_2 < \cdots < i_{k+1}$ and $j_1 < j_2 < \cdots < j_{k+1}$.)

I suspect that $\operatorname{Ker}\left(\phi_k\right) = W_k$. At least $W_k \subseteq \operatorname{Ker}\left(\phi_k\right)$ is easy to prove, so the guess looks reasonable.

This seems to be related to semistandard Young tableaux -- after all, there is a bitableau basis of $A$, of which I know nothing beyound the definition; and there is a straightening rule that reduces a monomial in $A$ modulo the ideal $W_k$, giving a sum of monomials which contain no products $z_{i_1, j_1} z_{i_2, j_2} \cdots z_{i_k, j_k}$ with $i_1 < i_2 < \cdots < i_k$ and $j_1 < j_2 < \cdots < j_k$ (a condition that seems to make more sense in terms of the RSK tableau corresponding to the monomials). But I am not sure if this straightening rule is confluent and, generally, whether this is a good way to move forward. The whole thing could also be the second fundamental theorem for some algebraic group, whence the invariant-theory tag.

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