Straightening basis for generic symmetric determinantal rings

We first prove independence under the Gram substitution in every rank $r\ge 1$. Put $Z=X{X}^{\mathsf{T}}$, where $X=(x_{ic})$ is a generic $t\times r$ matrix, and order variables lexicographically by columns first: $x_{11}>\cdots >x_{t1}>x_{12}>\cdots >x_{tr}$. Cauchy–Binet says $p(A,B)(X{X}^{\mathsf{T}})={\sum }_{C\subseteq [r],\mathrm{\mid }C\mathrm{\mid }=s}\det X_{A,C}\det X_{B,C}$ for $\mathrm{\mid }A\mathrm{\mid }=\mathrm{\mid }B\mathrm{\mid }=s\le r$. For completeness, this identity follows by expanding each entry $z_{ij}={\sum }_c{x}_{ic}{x}_{jc}$ in the Leibniz determinant: group terms by the chosen column indices; repeated indices cancel by pairing permutations exchanging them, and the terms with one fixed set $C$ give the two displayed determinants. A determinant's leading monomial pairs the smallest available row with its first column, then the next row with its next column, recursively. Comparing column sets in this sum, the largest summand has $C=(1,\dots ,s)$. Thus the minor has leading monomial ${\prod }_{c=1}^s{x}_{a_{c}c}{x}_{b_{c}c}$ with coefficient one.

For a standard product, remove empty factors and reverse the sequence of its columns to obtain $B_q,A_q,\dots ,B_1,A_1$. Column lengths are nonincreasing, rows are weakly increasing by the standard-chain inequalities, and columns are strictly increasing. Each column length occurs an even number of times, so all row lengths are even. Conversely a semistandard tableau with even row lengths has an even number of columns of each length: those numbers are consecutive differences of even row lengths. Pairing consecutive equal-height columns and reversing their sequence gives a unique standard product. Finally its Gram leading monomial from step 1.1 records the entries in each tableau row, and weak increase uniquely reconstructs each row from those multiplicities. Distinct standard products with minor sizes at most $r$ therefore have distinct leading monomials and are linearly independent after the rank-$r$ Gram substitution.

Take $r=t$ in step 2.1. A standard product of polynomial degree $d$ corresponds exactly to an even-row tableau with $2d$ boxes and entries in $[t]$. By [F1], the number of these is $\left(\genfrac{}{}{0px}{}{N+d-1}{d}\right)$, where $N=t(t+1)\mathrm{/}2$. This equals the dimension of the degree-$d$ homogeneous part of the polynomial ring in the $N$ entries of $Z$: the monomials $z^e$ are indexed by $e\in {\mathbb{N}}^N$ with $\sum e_{ij}=d$. Step 2.1 proves independence of all these standard products already in this polynomial ring, because a linear relation there would remain a relation under Gram substitution. An independent set of the full finite dimension is a basis. Consequently the standard products of all minor sizes span the polynomial ring in every degree.

We make the resulting identities completely indexed and constructive. Let ${\mathcal{E}}_d=\{e=(e_{ij}{)}_{i\le j}:\sum e_{ij}=d\}$ and let ${\mathcal{T}}_d$ be the standard products of degree $d$, without empty factors; order both finite sets lexicographically in any fixed encoding. For a product $Q={\prod }_{h=1}^{q}p(A_h,B_h)$, with $s_h=\mathrm{\mid }{A}_h\mathrm{\mid }=\mathrm{\mid }{B}_h\mathrm{\mid }$ and ${\sum }_h{s}_h=d$, define $b_e(Q)={\sum }_{{\sigma }_h\in {\mathfrak{S}}_{s_h}}{\prod }_h\mathrm{sgn}({\sigma }_h)$, where the sum is restricted by the following explicit indexing condition: the multiset of unordered pairs $\{\min (a_{hc},b_{h,{\sigma }_h(c)}),\max (a_{hc},b_{h,{\sigma }_h(c)})\}$, over all $h$ and $1\le c\le s_h$, contains $(i,j)$ exactly $e_{ij}$ times. An empty restricted sum is zero. Leibniz expansion gives exactly $Q={\sum }_{e\in {\mathcal{E}}_d}{b}_e(Q)z^e$. Form the square integer matrix $M_d=(b_e(T){)}_{e\in {\mathcal{E}}_d,T\in {\mathcal{T}}_d}$. It is invertible over $\mathbb{Q}$ by step 3.1 applied over an algebraic closure of $\mathbb{Q}$. Since its nonzero determinant is an integer and $k$ has characteristic zero, it is invertible over $k$ as well.

If $M_d[T\leftarrow b(Q)]$ denotes the matrix with column $T$ replaced by $(b_e(Q){)}_e$, the explicit straightening identity is $Q={\sum }_{T\in {\mathcal{T}}_d}(\det M_d[T\leftarrow b(Q)]\mathrm{/}\det M_d)T$. Indeed multiplying the coefficient vector on the right by $M_d$ gives $b(Q)$ by [F2] applied to the finite square matrix $M_d$ over $k$, so its coefficient of each monomial $z^e$ is precisely the coefficient on the left. All sets, entries, signs, and denominators in this identity have been specified by finite ordered index data; no existence of a straightening law is assumed. Applying it once replaces any nonstandard product directly by standard products, so no further nonstandard term is produced and rewriting terminates. For $d=0$ both indexing sets consist of the unit and the matrix is $(1)$.

Return to $0<m<t$. Every minor of size $s\ge m+1$ lies in $I_{m+1}$. Explicitly, for increasing $A,B$ of size $s$, write $A^{\mathrm{\prime }}=\{a_1,\dots ,a_{m+1}\}$. Laplace expansion along those rows is $p(A,B)={\sum }_{C\subseteq B,\mathrm{\mid }C\mathrm{\mid }=m+1}(-1{)}^{(m+1)(m+2)\mathrm{/}2+{\sum }_{b_j\in C}j}p(A^{\mathrm{\prime }},C)p(A\setminus A^{\mathrm{\prime }},B\setminus C)$, with the zero-size determinant interpreted as one. Each summand contains a generating $(m+1)$-minor. Thus step 5.1, followed by discarding every standard product containing such a minor, spans the quotient by standard products with all sizes at most $m$. All generators of $I_{m+1}$ vanish under the rank-$m$ Gram substitution, so a relation among these surviving classes would contradict their Gram independence from step 2.1. They form a basis, and the same finite indexed identities, reduced modulo the ideal, supply terminating straightening there. For $m=0$, the quotient by the ideal of entries is $k$ and the only allowed product is the unit. For $m=t$, the ideal is zero and steps 3.1–5.1 give the polynomial-ring assertion.