Generalized complete-intersection conjecture for Jordan-type loci

The generalized complete-intersection conjecture for Jordan type

1. The theorem

Let $\mathsf k$ be an infinite field. For a partition $R$, write $\ell(R)$ for its number of parts and $\mathfrak D(R)$ for the Jordan type of the dense orbit in the nilpotent commutator of a nilpotent matrix of type $R$. Let

$$Q=(q_1>\cdots>q_\ell),\qquad q_a-q_{a+1}\ge2,$$

be stable, fix a Jordan matrix $J_Q$, and put

$$\mathcal N_{J_Q}=\{B\in M_n(\mathsf k):B^n=0,\ BJ_Q=J_QB\}.$$

For every partition $P$ satisfying $\mathfrak D(P)=Q$, put

$$W_P^Q=\{B\in\mathcal N_{J_Q}:P_B=P\}.$$

We prove that $\overline{W_P^Q}$ is an irreducible complete intersection of codimension $\ell(P)-\ell$, and that its defining ideal is generated by polynomials of degree at most $\ell$.

We use the Box Theorem of Irving--Kosir--Mastnak: on putting $d_a=q_a-q_{a+1}$, the partitions in $\mathfrak D^{-1}(Q)$ are indexed bijectively by

$$I=(i_1,\ldots,i_\ell),\qquad 1\le i_a\le d_a-1\ (a<\ell),\qquad 1\le i_\ell\le q_\ell, \tag{1.1}$$

and the corresponding partition $P_I^Q$ has $\ell(P_I^Q)=|I|:=\sum_ai_a$.

2. Multiplier coordinates and the proposed equations

Put $O=\mathsf k[[t]]$ and identify the underlying $\mathsf k[t]$-module with

$$V_Q=\bigoplus_{a=1}^{\ell}O/(t^{q_a}),$$

where $J_Q$ is multiplication by $t$. In row-vector convention a commuting endomorphism is represented by a multiplier matrix $M=(m_{ab}(t))$, with

$$\operatorname {ord}m_{ab}\ge(q_b-q_a)_+.$$

Each entry is represented by its unique polynomial in the appropriate truncation. Because the $q_a$ are distinct, the semisimple quotient of the centralizer is $\mathsf k^\ell$. Thus $M$ is nilpotent exactly when the constant term of every $m_{aa}$ is zero. These remaining coefficients are affine coordinates on $\mathcal N_{J_Q}$.

Fix a box index $I$ and set

$$z_r=d_{r-1}-i_{r-1}\quad(2\le r\le\ell).$$

For every $1\le j\le\ell$ and $2\le h\le i_j$ define one polynomial $f_{j,h}$. For $j=1$, set

$$f_{1,h}=[t^{h-1}]m_{11}.$$

For $j>1$, put $T_a(j,h)=\sum_{r=a}^{j-1}i_r+h$ for $1\le a\le j$, and, for $2\le a\le j$, put

$$W_a(j)=\sum_{r=a}^{j}z_r. \tag{2.1}$$

Choose the largest $a\in\{2,\ldots,j\}$ for which $W_a(j)\ge T_a(j,h)$, if such an $a$ exists, and otherwise put $a=1$. (When $h\le z_j$, this chooses $a=j$.) Define

$$f_{j,h}=[t^{T_a(j,h)-1}]\det M_{[a,j],[a,j]}. \tag{2.2}$$

For $a=j$, (2.2) is just $[t^{h-1}]m_{jj}$. Let

$$C_I=(f_{j,h}:1\le j\le\ell,\ 2\le h\le i_j),\qquad Y_I=V(C_I).$$

There are

$$m=\sum_j(i_j-1)=|I|-\ell \tag{2.3}$$

generators, and (2.2) has degree $j-a+1\le\ell$.

3. Connected permutation blocks and weighted carries

For consecutive indices write

$$D_{rj}(t)=\det M_{[r,j],[r,j]},\qquad D_{r,r-1}=1.$$

For $b\le c$, let $K_{bc}$ be the signed sum of the permutation monomials in $D_{bc}$ whose permutation has no break before $c$; a break at $u<c$ means $\pi([b,u])=[b,u]$. Put $K_{bb}=m_{bb}$.

Every permutation has a unique decomposition into consecutive indecomposable blocks. Splitting that decomposition at the cut $a-1\mid a$ gives the identity over $\mathbb Z$

$$D_{rj}=D_{r,a-1}D_{aj}+ \sum_{b=r}^{a-1}\sum_{c=a}^{j} D_{r,b-1}K_{bc}D_{c+1,j}. \tag{3.1}$$

If $b<c$, every indecomposable permutation crosses each cut $u\mid u+1$, $b\le u<c$, downward at least once. A downward entry crossing that cut contains $t^{d_u}$. Consequently

$$\operatorname {ord}K_{bc}\ge q_b-q_c =\sum_{u=b}^{c-1}d_u. \tag{3.2}$$

This bound is termwise and is therefore valid in every characteristic.

Lemma 3.1 (interval carry lemma)

If all equations in $C_I$ vanish at $M$, then, for every $1\le r\le j\le\ell$,

$$\operatorname {ord}D_{rj}\ge\nu^I_{rj}, \tag{3.3}$$

where

$$\nu^I_{1j}=\sum_{u=1}^{j}i_u, \tag{3.4}$$

and, for $r\ge2$,

$$\nu^I_{rj}= \min_{p=r-1,\ldots,j} \left(\sum_{u=r}^{p}z_u+\sum_{u=p+1}^{j}i_u\right). \tag{3.5}$$

For $p=r-1$ the first sum is empty; for $p=j$ the second is empty.

Proof

We prove a slightly sharper induction. While the equations at vertex $j$ have been processed only through $h$, replace the last $i_j$ in (3.4)--(3.5) by $h$; denote the resulting number by $\nu_{rj}(h)$. Thus, for $r\ge2$,

$$\nu_{rj}(h)=T_r(j,h)+ \min\left(0,\epsilon_r,\epsilon_r+\epsilon_{r+1},\ldots, \sum_{u=r}^{j}\epsilon_u\right), \tag{3.6}$$

where $\epsilon_u=z_u-i_u$ for $u<j$ and $\epsilon_j=z_j-h$.

Let $a$ be the endpoint selected in (2.1). Write $S_x=\sum_{u=x}^{j}\epsilon_u=W_x(j)-T_x(j,h)$. By maximality of $a$, $S_a\ge0$ when $a>1$, whereas $S_x<0$ for every $x>a$. For $x>a$, every proper prefix in (3.6) is $S_x-S_{p+1}>S_x$; hence the terminal sum is the minimum. For $x=a$, every proper prefix is nonnegative, so zero is the minimum. For $x<a$, the minimum occurs before $a$. It follows that

\begin{aligned} &\nu_{rj}(h)=\nu_{rj}(h-1)+1 &&\Longleftrightarrow r\le a,\tag{3.7}\\ &\nu_{aj}(h)=T_a(j,h) &&(a>1), \tag{3.8}\\ &\nu_{rj}(h)=W_r(j)=\nu_{rj}(h-1) &&(r>a), \tag{3.9}\\ &\nu_{rj}(h)=\nu^I_{r,a-1}+T_a(j,h) &&(r<a). \tag{3.10} \end{aligned}

We now use outer induction on $j$ and inner induction on $h$. The case $j=1$ is immediate from the diagonal equations. At $h=1$ for $j>1$, $z_j\ge1$, so $a=j$. Use (3.1) at the cut $j-1\mid j$:

$$D_{rj}=D_{r,j-1}m_{jj}+ \sum_{b=r}^{j-1}D_{r,b-1}K_{b,j}.$$

The first term has order at least $\nu^I_{r,j-1}+1=\nu_{rj}(1)$. In a crossing term, successively use $\nu^I_{r,b}\le\nu^I_{r,b-1}+d_b$; the remaining last cost is $d_{j-1}=i_{j-1}+z_j\ge i_{j-1}+1$. Hence every crossing term has the same lower bound. This proves the inner base. For $h\ge2$, inner induction and (3.7)--(3.8) give

$$\operatorname {ord}D_{aj}\ge T_a(j,h)-1.$$

Equation (2.2) raises this order to $T_a(j,h)$. Intervals starting to the right of $a$ require no increase by (3.9). If $r<a$, apply (3.1). Its noncrossing term has order at least the right side of (3.10). A crossing term has order at least

$$X_{b,c}=\nu^I_{r,b-1}+\sum_{u=b}^{c-1}d_u+\nu_{c+1,j}(h-1). \tag{3.11}$$

Here and below the $\nu$-value of an empty interval is zero.

The elementary choices in (3.5) give

$$\nu^I_{r,b}\le\nu^I_{r,b-1}+i_b\le\nu^I_{r,b-1}+d_b$$

and

$$\nu_{c,j}(h-1)\le z_c+\nu_{c+1,j}(h-1) \le d_{c-1}+\nu_{c+1,j}(h-1).$$

Absorbing the outer summands in (3.11) with these inequalities reduces its minimum to $b=a-1,c=a$. Using (3.9),

$$\begin{aligned} X_{b,c} &\ge \nu^I_{r,a-2}+d_{a-1}+W_{a+1}(j)\\ &=\nu^I_{r,a-2}+i_{a-1}+W_a(j)\\ &\ge\nu^I_{r,a-1}+T_a(j,h) =\nu_{rj}(h). \end{aligned} \tag{3.12}$$

Set $W_{j+1}(j)=0$, and give every empty interval determinant and its $\nu$-value order zero. If $a=1$, there is no propagation: (2.2) itself is the required full-prefix coefficient. This proves the induction. Taking $h=i_j$ proves (3.3). $\square$

In particular,

$$C_I(M)=0\quad\Longrightarrow\quad \operatorname {ord}\Delta_j\ge s_j:=\sum_{u=1}^{j}i_u, \qquad \Delta_j=D_{1j}. \tag{3.13}$$

Lemma 3.2 (Fitting consequence)

If (3.13) holds, then $\dim_{\mathsf k}\ker B\ge |I|$.

Proof

First, if $A$ is any set of $p$ rows, then

$$\operatorname {ord}\det M_{A,[1,p]}\ge s_p. \tag{3.14}$$

If $A=[1,p]$, this is exactly (3.13). Otherwise $p<\ell$. Group the determinant terms by the last break $v$ for which the first $v$ columns are matched to the first $v$ rows. The part before the break is $\Delta_v$. The remaining matching has no later internal break, and the nonprefix row set contains a row bigger than $p$. It therefore crosses every cut $(v+1)\mid(v+2),\ldots,p\mid(p+1)$ downward and costs at least $\sum_{r=v+1}^{p}d_r\ge\sum_{r=v+1}^{p}i_r$. This proves (3.14). If $|A|=|E|$ and $E$ contains $[1,p]$, Laplace expansion in those columns similarly gives

$$\operatorname {ord}\det M_{A,E}\ge s_p. \tag{3.15}$$

In row convention, $\operatorname {coker}B$ is presented over $O$ by the rows of

$$\begin{bmatrix}M\\D\end{bmatrix}, \qquad D=\operatorname {diag}(t^{q_1},\ldots,t^{q_\ell}). \tag{3.16}$$

A maximal minor of (3.16) is, up to sign,

$$t^{\sum_{r\in C}q_r}\det M_{A,E}, \qquad E=[1,\ell]\setminus C. \tag{3.17}$$

If $C=\varnothing$, (3.17) is $\Delta_\ell$. Otherwise put $p=\min C-1$. Then $E\supseteq[1,p]$, and (3.15), together with

$$\sum_{r=p+1}^{\ell}i_r \le\sum_{r=p+1}^{\ell-1}(d_r-1)+q_\ell \le q_{p+1}, \tag{3.18}$$

shows that (3.17) is divisible by $t^{|I|}$. The valuation of the zeroth Fitting ideal of a finite-length $O$-module is its length. Hence $\operatorname {length}_O\operatorname {coker}B\ge|I|$, which equals $\dim_{\mathsf k}\ker B$. $\square$

4. Gaussian chambers and separation

For a box index $I$, put $e_b=q_b-i_b$. Start with the following Gaussian parameters. The matrix $L$ is lower unitriangular, $U$ is upper triangular, and

$$\operatorname {ord}U_{aa}=i_a\quad(i_a<q_a),\qquad U_{\ell\ell}=0\quad(i_\ell=q_\ell),\qquad \operatorname {ord}L_{ab}\ge q_b-q_a-i_b\quad(a>b). \tag{4.1}$$

More explicitly, $U_{ab}\in O/(t^{q_b})$ for $a<b$, $U_{bb}\in t^{i_b}O/(t^{q_b})$ with nonzero leading coefficient, and $L_{ab}\in t^{e_b-q_a}O/(t^{e_b})$ for $a>b$. In the terminal possibility $i_\ell=q_\ell$, set $U_{\ell\ell}=0$; the condition on a nonzero leading coefficient is imposed only when $i_b<q_b$.

Form the ordinary power-series matrix $\widehat M=LU$, and let $M$ be its canonical multiplier matrix: reduce column $b$ modulo $t^{q_b}$. Let $\mathcal K_I$ be the image of this parameter space. Before and after reduction the product is tiled. Indeed, in a lower entry $(a,b)$, the term through $b$ has order at least $q_b-q_a$, while a term through $c<b$ has the additional positive amount $q_c-q_b-i_c$; diagonal cross terms have positive order. Thus $\mathcal K_I\subseteq\mathcal N_{J_Q}$.

The truncated parameter map is injective. This follows directly from Doolittle recursion. Assuming the preceding rows of $U$ and columns of $L$ have been recovered, row $b$ recovers every $U_{bc}$, $c\ge b$, from

$$M_{bc}-\sum_{s<b}L_{bs}U_{sc}\pmod {t^{q_c}}.$$

These products are independent of the representatives already recovered: for $s<c$, an ambiguity divisible by $t^{e_s}$ vanishes modulo $t^{q_c}$, since $e_s\ge q_{s+1}+1>q_c$. Then column $b$ recovers every $L_{ab}$, $a>b$, from

$$M_{ab}-\sum_{s<b}L_{as}U_{sb} =L_{ab}U_{bb}\pmod {t^{q_b}}.$$

Multiplication by $U_{bb}=t^{i_b}$ times a unit is injective on the displayed quotient $t^{e_b-q_a}O/(t^{e_b})$. If the last pivot is zero, it is prescribed and there are no parameters below it; the final residual entry is forced. Thus the recursion also covers the terminal case. The lower parameters contribute $q_a$ coefficients for every $a>b$, the strictly upper parameters contribute $q_b$ coefficients for every $a<b$, and the diagonal parameters contribute $q_b-i_b$. The ambient multiplier space has the same off-diagonal counts and $q_b-1$ diagonal coefficients. Multiplication is injective by the recursion above. Hence

$$\mathcal K_I\text{ is irreducible},\qquad \dim\mathcal K_I=\dim\mathcal N_{J_Q}-m. \tag{4.2}$$

We next account explicitly for the column truncations. Put $D_Q=\operatorname {diag}(t^{q_1},\ldots,t^{q_\ell})$. There is a power-series matrix $H$ such that

$$\widehat M=M+HD_Q. \tag{4.3a}$$

We first need a shifted flag estimate for the unreduced product. If $r\le p$ and $A\subseteq\{r,\ldots,\ell\}$ has $|A|=p-r+1$, then

$$\operatorname {ord}\det\widehat M_{A,[r,p]} \ge\nu^I_{r,p}. \tag{4.3b}$$

Indeed, apply ordinary Cauchy--Binet to $LU$. A middle set $S$ is contained in $[1,p]$. Let $u$ be its largest omitted element of $[r,p]$, with $u=r-1$ when none is omitted. Upper triangularity forces the pivots $u+1,\ldots,p$, at cost $i_{u+1}+\cdots+i_p$. For $r\le x\le u$,

$$\#(S\cap[1,x-1]) =(x-r)+\#\bigl(([r,p]\setminus S)\cap[x,p]\bigr) >x-r\ge\#(A\cap[1,x-1]).$$

Thus every nonzero matching in the lower minor crosses downward each cut $r-1\mid r,\ldots,u-1\mid u$. Moreover, for $a>s$,

$$\operatorname {ord}L_{as}\ge q_s-i_s-q_a \ge\sum_{x=s+1}^{a}z_x.$$

The lower cost is therefore at least $z_r+\cdots+z_u$. Together with the forced pivots this is the $u$-alternative in (3.5). When $r=1$, necessarily $S=[1,p]$, and its principal upper pivots instead give $\nu^I_{1p}=\sum_{x=1}^p i_x$ directly. This proves (4.3b).

Expand a canonical interval determinant multilinearly using (4.3a). In a term using a nonempty set $C\subseteq[r,j]$ of remainder columns, let $c=\min C$. It contains the factor $t^{\sum_{b\in C}q_b}$. Laplace expansion first in the $H$-columns and then in the untouched prefix $[r,c-1]$, followed by (4.3b), gives order at least $q_c+\nu^I_{r,c-1}$. Here an empty prefix has order zero. The definitions (3.4)--(3.5) and the box bounds give

$$\nu^I_{rj}\le\nu^I_{r,c-1}+\sum_{u=c}^ji_u,\qquad \sum_{u=c}^ji_u\le q_c. \tag{4.3c}$$

The last inequality is strict when $j<\ell$, by $\sum_{u=c}^j i_u\le\sum_{u=c}^j(d_u-1)<q_c$. When $j=\ell$ but $i_\ell<q_\ell$, it follows instead from

$$\sum_{u=c}^{\ell}i_u \le q_c-q_\ell-(\ell-c)+i_\ell<q_c.$$

Thus every remainder term has order at least $\nu^I_{rj}$, and the term with no remainder columns has the same bound by (4.3b). For the canonical multiplier matrix we have proved

$$\operatorname {ord}D_{rj}\ge\nu^I_{rj}. \tag{4.3}$$

The endpoint rule (2.1) says precisely that the right side is at least the target in (2.2), so $\mathcal K_I\subseteq Y_I$.

At the generic point of $\mathcal K_I$, equality holds in (4.3) whenever $j<\ell$ or $i_j<q_j$; this is the only form needed below. In these cases the second inequality in (4.3c) is strict, so every nonempty remainder term has order strictly greater than $\nu^I_{rj}$. For the raw product, the case $r=1$ is witnessed by the principal middle set and the monomial $\prod_{u=1}^jU_{uu}$. For $r\ge2$, besides the principal Cauchy--Binet term, each alternative indexed by $p=r,\ldots,j$ in (3.5) is witnessed by the middle set

$$\{r-1,r,\ldots,p-1,p+1,\ldots,j\}.$$

Match the lower rows $u=r,\ldots,p$ to $u-1$, then the upper rows $u-1$ to columns $u$; match the remaining rows diagonally. Its order is $\sum_{u=r}^{p}z_u+\sum_{u=p+1}^{j}i_u$, and the product of its independent leading Gaussian coefficients is a unique monomial with coefficient $\pm1$. Thus it cannot cancel in any characteristic.

There is also a direct kernel calculation which does not treat $L$ as an endomorphism of $V_Q$. For $x\in V_Q$, define the surjection

$$\Phi:V_Q\longrightarrow\bigoplus_b O/(t^{e_b}), \qquad \Phi(x)=((xL)_b\bmod t^{e_b})_b.$$

Condition (4.1) makes this well-defined, and triangular back-substitution makes it surjective. Since $\widehat M-M=HD_Q$, the two matrices induce the same endomorphism of $V_Q$. If $y=xL$, solve the equations $yU=0$ in increasing output column. Once $y_a\in t^{e_a}$ for $a<b$, its contribution to column $b$ vanishes because

$$e_a=q_a-i_a\ge q_{a+1}+1>q_b.$$

The $b$-th equation is therefore equivalent to $y_b\in t^{e_b}$. Consequently $xM=0$ if and only if $\Phi(x)=0$, and hence

$$\dim\ker B=\dim\ker\Phi =\sum_b(q_b-e_b)=\sum_b i_b=|I|. \tag{4.4}$$

We will need the following separation fact.

Lemma 4.1 (chamber separation)

If $|I|=|J|$ and all equations in $C_I$ vanish identically on $\mathcal K_J$, then $I=J$.

Proof

We first prove $J_j\ge I_j$ by induction on $j$. If $J_1<I_1$, the generic order of $m_{11}$ on $\mathcal K_J$ is $J_1$, contradicting Lemma 3.1 and (3.4); hence the case $j=1$ holds. Suppose it is known before $j$, but $J_j=I_j-\delta$ with $\delta>0$. In particular $J_j<q_j$, so the generic-order equality is valid for every interval ending at this $j$. Lemma 3.1, applied because $C_I$ vanishes on $\mathcal K_J$, therefore gives

$$\nu^J_{rj}\ge\nu^I_{rj}\qquad(1\le r\le j). \tag{4.5}$$

For $r\ge2$, (3.5) has the recursion

$$\nu^X_{rj}=\min\left\{A^X_{rj},\ z_r^X+\nu^X_{r+1,j}\right\}, \qquad A^X_{rj}=\sum_{u=r}^{j}X_u,\quad z_r^X=d_{r-1}-X_{r-1}. \tag{4.6}$$

Use the elementary fact that, if $A'<A$, $Z'\le Z$, and $\min(A',Z')\ge\min(A,Z)$, then $Z'=Z=\min(A,Z)$. At $r=j$, outer induction gives $z_j^J\le z_j^I$, while $A^J_{jj}=A^I_{jj}-\delta$. Equations (4.5)--(4.6) therefore force $z_j^J=z_j^I$, hence $J_{j-1}=I_{j-1}$, and also $\nu^J_{jj}=\nu^I_{jj}$. Descending in $r$, the same argument successively forces $J_{r-1}=I_{r-1}$ and $\nu^J_{rj}=\nu^I_{rj}$. At $r=1$, however,

$$\nu^J_{1j}=\sum_{u\le j}J_u =\sum_{u\le j}I_u-\delta<\nu^I_{1j},$$

contradicting (4.5). Thus $J\ge I$ coordinatewise. Equal total sums give $J=I$. $\square$

5. Incidence geometry

We record the dimension and irreducibility facts used below. Work first over an algebraic closure. For $R=(d^{m_d})_d$, let $A_R=C_{M_n}(J_R)$. There is a surjection

$$A_R\longrightarrow\prod_dM_{m_d}$$

with kernel $\operatorname {rad}A_R$, and an element of $A_R$ is nilpotent exactly when every component of its semisimple image is nilpotent. Thus $\mathcal N_{J_R}$ is the inverse image of a product of ordinary nilpotent cones. Consequently

$$\mathcal N_{J_R}\text{ is geometrically irreducible},\qquad \dim\mathcal N_{J_R}=\dim A_R-\ell(R). \tag{5.1}$$

If $\mathfrak D(P)=Q$, the commuting-pair incidence variety

$$\mathcal C_{P,Q} =\{(B,A)\in\mathcal O_P\times\mathcal O_Q:AB=BA\} \cong GL_n\times^{G_P}W_Q^P$$

is irreducible, because $W_Q^P$ is the dense open type locus in the irreducible variety $\mathcal N_{J_P}$. By (5.1),

$$\dim\mathcal C_{P,Q}=n^2-\ell(P).$$

Projection to $\mathcal O_Q$ gives

$$\dim W_P^Q=\dim A_Q-\ell(P),\qquad \operatorname {codim}_{\mathcal N_{J_Q}}W_P^Q=\ell(P)-\ell. \tag{5.2}$$

Also $\mathcal C_{P,Q}=GL_n\times^{G_Q}W_P^Q$. The group $G_Q=A_Q^\times$ is connected: it is an extension of $\prod_dGL_{m_d}$ by $1+\operatorname {rad}A_Q$. It therefore fixes each irreducible component of the fiber; more than one such component would make the associated bundle reducible. Hence $W_P^Q$ and its closure are geometrically irreducible.

For an arbitrary $R$ with $W_R^Q\ne\varnothing$, the same incidence calculation gives

$$\operatorname {codim}_{\mathcal N_{J_Q}}W_R^Q \ge\ell(R)-\ell, \tag{5.3}$$

and the inequality is strict if $\mathfrak D(R)\ne Q$: in that case the type-$Q$ locus is not dense in the irreducible $\mathcal N_{J_R}$.

6. Identification of the equation locus

For every box index $I$, let $R_I$ be the generic Jordan type on $\mathcal K_I$. From (4.4), $\ell(R_I)=|I|$. The generic-type open subset of $\mathcal K_I$ lies in $W_{R_I}^Q$, so (4.2) gives $\operatorname {codim}W_{R_I}^Q\le m$. On the other hand (5.3) gives $\operatorname {codim}W_{R_I}^Q\ge\ell(R_I)-\ell=m$. Equality holds, and the strict clause of (5.3) gives

$$\mathfrak D(R_I)=Q,\qquad \overline{\mathcal K_I}=\overline{W_{R_I}^Q}. \tag{6.1}$$

Indeed, both irreducible varieties in (6.1) have codimension $m$.

The assignment $\rho:I\mapsto R_I$ maps the finite Box index set into $\mathfrak D^{-1}(Q)$. It is injective. Indeed, if $R_I=R_J$, then $|I|=\ell(R_I)=\ell(R_J)=|J|$, and (6.1) gives $\overline{\mathcal K_I}=\overline{\mathcal K_J}$. Since $C_I$ vanishes on $\mathcal K_I$, it therefore vanishes on $\mathcal K_J$, and Lemma 4.1 gives $I=J$. By the Box Theorem, the domain and codomain of $\rho$ have the same finite cardinality. Thus $\rho$ is a bijection. In particular, every partition $P$ with $\mathfrak D(P)=Q$ is $R_I$ for a unique $I$, and

$$\overline{\mathcal K_I}=\overline{W_{R_I}^Q}\subseteq Y_I. \tag{6.2}$$

Now let $Z$ be any irreducible component of $Y_I$, and let $R_Z$ be its generic Jordan type. Since $C_I$ has $m$ generators, Krull's height theorem gives $\operatorname {codim}Z\le m$. The generic-type open part of $Z$ lies in $W_{R_Z}^Q$. Lemmas 3.1--3.2 give $\ell(R_Z)\ge|I|$, while (5.3) gives

$$\operatorname {codim}Z \ge\operatorname {codim}W_{R_Z}^Q \ge\ell(R_Z)-\ell \ge|I|-\ell=m. \tag{6.3}$$

Every inequality is equality. Thus $\mathfrak D(R_Z)=Q$ and $\ell(R_Z)=|I|$. By the bijection $\rho$, write $R_Z=R_J$; then $|J|=\ell(R_J)=|I|$. By (6.1), the irreducible closure $\overline{W_{R_J}^Q}$ is $\overline{\mathcal K_J}$. It has the same dimension as $Z$, hence equals $Z$. Therefore every polynomial in $C_I$ vanishes on $\mathcal K_J$. Lemma 4.1 gives $J=I$. Thus $Y_I$ has a unique irreducible component and

$$(Y_I)_{\mathrm{red}}=\overline{W_{R_I}^Q}. \tag{6.4}$$

It remains only to check the scheme structure. Let $M_0$ be the diagonal point with $m_{aa}=t^{i_a}$, using zero if $a=\ell,i_\ell=q_\ell$. It belongs to $Y_I$, because every coefficient selected in (2.2) is strictly below the order of the corresponding diagonal product. Order the equations by increasing $(j,h)$, and use the distinct ambient coordinates $y_{j,h}=[t^{h-1}]m_{jj}$. If $a=a(j,h)$ is the endpoint in (2.2), multilinearity gives

$$\frac{\partial f_{j,h}}{\partial y_{j,h}}(M_0) =[t^{\sum_{r=a}^{j-1}i_r}]D_{a,j-1}(M_0)=1. \tag{6.5}$$

An equation at an earlier vertex does not involve $m_{jj}$. At the same vertex, if $h'<h$, differentiating $f_{j,h'}$ with respect to $y_{j,h}$ selects a coefficient of the corresponding $D_{a',j-1}(M_0)$ strictly below its diagonal order, and hence gives zero. Thus this $m\times m$ Jacobian minor is triangular with diagonal one. The scheme $Y_I$ is smooth of codimension $m$ at $M_0$.

The ideal $C_I$ has $m$ generators and height $m$, so it is a complete intersection in the polynomial coordinate ring of $\mathcal N_{J_Q}$. The quotient is Cohen--Macaulay and has no embedded associated primes. By (6.4) there is only one minimal prime. Equation (6.5) gives a nonempty smooth open subset of its irreducible support, so that open contains the generic point and the generic localization is reduced. A Cohen--Macaulay ring is $S_1$; together with generic reducedness this implies reducedness. Thus $C_I$ is prime and is the defining ideal of $\overline{W_{R_I}^Q}$.

Given the original $P$, choose the unique $I$ with $P=R_I$. By (2.3), the codimension is $|I|-\ell=\ell(P)-\ell$, and every generator has degree at most $\ell$. This proves the theorem.

7. Ground field and references used

All equations and determinant identities above are integral, and all unique leading monomials have coefficient $\pm1$. The proof over an algebraic closure therefore gives geometric primeness and geometric irreducibility in every characteristic. Faithfully flat descent gives the assertion over the original infinite field $\mathsf k$.

The statement is Conjecture 4.2 of M. Boij, A. Iarrobino, and L. Khatami, Jordan type stratification of spaces of commuting nilpotent matrices, Linear Algebra Appl. 710 (2025), 183--202 (https://arxiv.org/abs/2409.13553). The Box Theorem used above is in J. Irving, T. Kosir, and M. Mastnak, A Proof of the Box Conjecture for Commuting Pairs of Matrices (https://arxiv.org/abs/2403.18574).