The combinatorics of the Nullstellensatz for the variety of nilpotent matrices

Question

Let $H_n$ denote the set of $n \times n$ nilpotent matrices with complex entries. The set $H_n$ may be regarded as an algebraic variety. Indeed, consider the polynomial ring $\mathbb{C}[A_{i,j} : 1 \leq i,j \leq n]$ in $n^2$ variables, and for $m \geq 0$, define the polynomial \begin{equation*} \phi_{m;i,j} := \sum_{ 1 \leq k_1,\ldots,k_{m-1} \leq n} A_{i,k_1}A_{k_1,k_2}\ldots A_{k_{m-1},j}, \end{equation*} with the understanding that $\phi_{1;i,j} = A_{i,j}$, and $\phi_{0;i,j} = \delta_{i,j}$.

Then $H_n$ is the zero set $V(I)$ of the ideal generated by the $n^2$ polynomials $\{\phi_{n;i,j} : 1 \leq i,j \leq n \}$.

Alternatively, for $0 \leq k \leq n$ define the polynomials \begin{equation} p_k := \sum_{S \subset [n], |S|=k} \det_{i,j \in S} A_{i,j}, \end{equation} with $p_0=1$. Then the characteristic polynomial of a matrix takes the form \begin{align*} \Delta_A(t) := \det(tI - A) = \sum_{k=0}^n t^{n-k}(-1)^k p_k. \end{align*} Since $A$ is nilpotent if and only if $\det(tI-A) = t^n$, it follows than $H_n$ may alternatively be characterised as the $V(J)$ of the ideal $J$ generated by the $n$ polynomials $\{p_1,\ldots,p_n\}$.

In particular, by Hilbert's Nullstellensatz, we have the equality of radicals, $\sqrt{I} = \sqrt{J}$. Clearly $J$ is not a subset of $I$, since $J$ contains polynomials of degree $1$, whereas $I$ contains only polynomials of degree $n$ and higher.

Conversely however, we can show that $I \subset J$. Indeed, by the Cayley–Hamilton theorem, $A$ is a root of its own characteristic polynomial, so that for each $1 \leq i,j \leq n$ we have the identity in $\mathbb{C}[A_{i,j} : 1 \leq i,j \leq n ]$ \begin{equation} 0 = \sum_{k=0}^n (-1)^k \phi_{n-k;i,j} p_k. \end{equation} Thus, since $p_0=1$, $\phi_{n;i,j}$ can be written as a linear combination of $p_k$, and hence $\phi_{n;i,j} \in J$.

Question: Have the combinatorics of the nullstellensatz for $H_n$ been studied? Here are some concrete examples of questions I'm interested in:

1. By the Nullstellensatz, for each $k$, there is some integer $r \geq 1$ such that $p_k^r$ can be written as a polynomial linear combination of the $\{\phi_{n;i,j} : 1 \leq i,j \leq n\}$. Can this linear combination be found explicitly?
2. If an $n \times n$ matrix $A$ satisfies $A^m = 0$ for some $m \geq n$, then it is guaranteed to satisfy $A^n = 0$. Thus, if we consider the ideal $I_m := \{ \phi_{m;i,j} : 1 \leq i,j \leq n \}$, then for $m \geq n$, $I_m$ and $I=I_n$ have the same zero set. It's easily verified $I_{m+1} \subset I_m$. However, by the nullstellensatz, for any $m' \geq m$, each $\phi_{m';i,j}$ has a power that can be written as a polynomial linear combination of the $\{ \phi_{m;i,j} : 1 \leq i,j \leq n \}$. Can this linear combination be found explicitly?
3. In the previous two questions, do we have a concept of how the exponent $r$ behaves as a function of $n,m,m'$?

Have this circle of questions been studied anywhere? There are some sort of related questions on this site and stackexchange, though without explicit reference to the combinatorics:

Here someone asks if the ideal $J$ above is prime. See also here.

Here someone asks if the geometry of $H_n$ has been studied. There are references to Jantzen's Lie Theory and Chriss-Ginzburg, as well as papers by Ness and by Richardson.

Here someone show that $H_n$ has dimensions $n^2-n$.

I should also mention the combinatorial proofs of the Cayley–Hamilton theorem: see Straubing - A combinatorial proof of the Cayley-Hamilton theorem.

Thank you for reading this far, and my apologies if I have used any unusual notation or terminology — I'm a beginner in algebraic geometry!

Answer 1

Thanks everyone for your replies! As Darij suggested in his answer, it appears the clever trick used in Gert Almkist's generalisation of a mistake of Bourbaki can be generalised to tackle the problem. Thanks Darij for the suggestion and for the reference.

If I'm not mistaken, here are the details. In the originally setting of Almkist's trick, we show that $(\sum_{i=1}T_{ii})^{n(n-1)+1}$ is a linear combination of the $\phi_{n;i,j}$ as follows: Note that if $e_1,\ldots,e_n$ is the standard basis of $\mathbb{C}^n$, then \begin{equation} (\sum_{i=1}^nT_{ii})\det(e_1,\ldots,e_n) = \sum_{k=1}^n \det(e_1,\ldots,Te_k,\ldots,e_n). \end{equation} By multilinearity, for any vectors $b_1,\ldots,b_n$ we have \begin{equation} (\sum_{i=1}^nT_{ii})\det(b_1,\ldots,b_n) = \sum_{k=1}^n \det(b_1,\ldots,Tb_k,\ldots,b_n). \end{equation} In particular, for any natural number $N$ we have \begin{equation} (\sum_{i=1}^nT_{ii})^N = \sum_{k_1 + \ldots + k_n=N} \frac{N!}{k_1!\ldots k_n!} \det(T^{k_1}e_1,\ldots,T^{k_n}e_n), \end{equation} where the sum is over nonnegative integers $k_1,\ldots,k_n$ adding to $N$. When $N \geq n(m-1)+1$, it is guaranteed that at least one of the $k_i$ will be $\geq m$, so that the quantity on the right-hand-side is a polynomial linear combination of $\phi_{m;i,j}$.

To generalise this method to the polynomials $p_k$, we begin by noting \begin{equation} p_k \det(e_1,\ldots,e_n) = \sum_{1 \leq i_1 < \ldots < i_k \leq n} \det(e_1,\ldots,Te_{i_1},\ldots,Te_{i_2},\ldots,Te_{i_k},\ldots,e_n), \end{equation} where we understand that between the dots we have $e_i$s. In particular, iterating like before, for any $N$ we have \begin{equation} p_k^N = \sum_{k_1+\ldots+k_N = kN} C_{N,k}(k_1,\ldots,k_n) \det(T^{k_1}e_1,\ldots,T^{k_n}e_n), \end{equation} where $C_{N,k}(k_1,\ldots,k_n)$ is the number of ways of putting balls into $n$ bins in $N$ rounds with $k_j$ balls in bin $j$ at time $N$, with the rule that on each round, we put $k$ balls into $k$ different bins. In any case, provided $kN \geq n(m-1)+1$, we are guaranteed to have some $k_i \geq m$ for each summand, and hence for such $N$, this formula expresses $p_k^N$ as a polynomial linear combination of $\{\phi_{n=m;i,j} \}$.

Note that as $k$ increases, we need only take smaller powers of $p_k$ (i.e. the $N_k^{\text{th}}$ power with $N_k$ the smallest integer greater than $(n(m-1)+1)/k$) to express $p_k^{N_k}$ in terms of $\phi_{m;i,j}$. This makes sense, as $p_k$ is a higher degree polynomial. In fact, $p_k^{N_k}$ has degree $n(m-1)+1+r(k)$, where $0 \leq r(k) < k$.

This method also gives us a (probably quite inefficient) way of expressing a power of $\phi_{m;i,j}$ in terms of the $\phi_{m';i',j'}$. Write $\phi_{m;i,j} = \sum_k \lambda_kp_k$, where $\lambda_k$ are polynomials. Then $\phi_{m;i,j}^{n(N-1)+1}$ is a polynomial in powers of $p_k$, which again by this pigeonhole style principle, has every monomial containing a power of a $p_k$ of at least $N$. Provided $kN \geq n(m'-1)+1$, this power can then be written as a polynomial linear combination of the $\phi_{m';i',j'}$.