Proof. We claim that if the inclusion $\varphi \colon K[YZ]\longrightarrow K[Y,Z]$ is pure for a fixed triple of positive integers $(m,n,t)$ , then purity holds as well for the inclusion of the K-algebras corresponding to a triple $(m',n',t)$ with $m'\leqslant m$ and $n'\leqslant n$ .

To see this, set $Y'$ to be the matrix consisting of the first $m'$ rows of Y, and $Z'$ to be the matrix consisting of the first $n'$ columns of Z, and consider the $\mathbb {N}$ -grading on $K[Y,Z]$ where the indeterminates from the submatrices $Y'$ and $Z'$ have degree $0$ , as does K, while the remaining indeterminates have degree $1$ so that ${K[Y,Z]}_0\ =\ K[Y',Z']$ . Then

$$\begin{align*}{K[YZ]}_0\ =\ K[Y'Z'], \end{align*}$$

so $K[Y'Z']$ is a pure subring of $K[YZ]$ . Since we are assuming $K[YZ]\longrightarrow K[Y,Z]$ is pure, it follows that the composition

$$\begin{align*}K[Y'Z']\ \subseteq\ K[YZ]\ \subseteq\ K[Y,Z] \end{align*}$$

is pure as well, but then so is $K[Y'Z']\subseteq K[Y',Z']$ . This proves the claim; similar reduction arguments will be used for other matrix families later in the paper.

Set

and

. We next prove that $\varphi $ is pure in the cases claimed in the theorem. When $t=1$ , the ring R coincides with the Segre product of the polynomial rings $K[Y]$ and $K[Z]$ , which is a pure subring of S. For the case $m\leqslant t$ , in light of the reduction step, it suffices to establish the purity when $m=t$ and $n\geqslant t$ . In this case, the ring R has dimension $mn$ , specifically the matrix entries

are algebraically independent over K and hence form a homogeneous system of parameters for R. By Theorem 2.1, it suffices to show that $H^{mn}_{\mathfrak {m}_R}(S)$ is nonzero; we show that

$$\begin{align*}\left[\frac{1}{\prod x_{ij}}\right]\ \in\ H^{mn}_{\mathfrak{m}_R}(S) \end{align*}$$

is a nonzero element, equivalently that for each $k\geqslant 1$ , one has

$$\begin{align*}\left(\prod x_{ij}\right)^{k-1}\ \notin\ (x_{11}^k,\dots,x_{mn}^k)S. \end{align*}$$

It is enough to show the above after specializing the entries of Y to the $t\times t$ identity matrix. This specialization maps $YZ$ to Z, with the image of S being the polynomial ring $K[Z]$ . The above display then takes the form

$$\begin{align*}\left(\prod z_{ij}\right)^{k-1}\ \notin\ (z_{11}^k,\dots,z_{mn}^k)K[Z], \end{align*}$$

which is immediately seen to hold. The case $n\leqslant t$ is much the same.

Next, suppose $t\geqslant 2$ . It remains to prove that $\varphi \colon K[YZ]\longrightarrow K[Y,Z]$ is not pure if $m>t$ and $n>t$ . By the reduction step at the beginning of the proof, it suffices to show that $\varphi $ is not pure in the case $m=t+1=n$ . In this case, the ring $R=K[YZ]$ is a hypersurface of dimension $t^2+2t$ , so it suffices by Theorem 2.1 to show that the local cohomology module

$$\begin{align*}H^{t^2+2t}_{\mathfrak{m}_R}(S) \end{align*}$$

is zero, where $\mathfrak {m}_R$ is the homogeneous maximal ideal of R. The minimal primes of the ideal $\mathfrak {m}_RS$ are described by Theorem 4.1; in the notation of that theorem, these are the primes $\mathfrak {p}_{0,t},\ \mathfrak {p}_{1,t-1},\ \dots ,\ \mathfrak {p}_{t,0}$ . With $\operatorname {cd}$ denoting the cohomological dimension, we shall prove that for each integer k with $0\leqslant k\leqslant t$ , one has

(4.2.1) $$ \begin{align} \operatorname{cd}\left(\mathfrak{p}_{0,t}\cap\mathfrak{p}_{1,t-1}\cap\dots\cap\mathfrak{p}_{k,t-k}\right)\ \leqslant\ t^2+t+1, \end{align} $$

from which it follows that $\operatorname {cd}(\mathfrak {m}_RS)\leqslant t^2+t+1$ ; since $t\geqslant 2$ , one has $t^2+t+1<t^2+2t$ .

We first claim that

(4.2.2) $$ \begin{align} \operatorname{cd}\left(\mathfrak{p}_{0,t}\cap\mathfrak{p}_{1,t-1}\cap\dots\cap\mathfrak{p}_{k,t-k}\right)\ &\leqslant\ \max\left\{\operatorname{cd}(\mathfrak{p}_{0,t}),\ \operatorname{cd}(\mathfrak{p}_{1,t-1}),\ \dots,\ \operatorname{cd}(\mathfrak{p}_{k,t-k}),\right. \notag \\&\quad \left. \operatorname{cd}(\mathfrak{p}_{0,t-1})-1,\ \operatorname{cd}(\mathfrak{p}_{1,t-2})-1,\ \dots,\ \operatorname{cd}(\mathfrak{p}_{k-1,t-k})-1\right\}. \end{align} $$

Quite generally, for ideals $\mathfrak {a}$ and $\mathfrak {b}$ of S, the Mayer–Vietoris sequence

shows that

$$\begin{align*}\operatorname{cd}(\mathfrak{a}\cap\mathfrak{b})\ \leqslant\ \max\left\{\operatorname{cd}(\mathfrak{a}),\ \operatorname{cd}(\mathfrak{b}),\ \operatorname{cd}(\mathfrak{a}+\mathfrak{b})-1 \right\}. \end{align*}$$

Using this for the ideals

and

, one has

$$ \begin{align*} &\operatorname{cd}\left([\mathfrak{p}_{0,t}\cap\mathfrak{p}_{1,t-1}\cap\dots\cap\mathfrak{p}_{k,t-k}]\cap\mathfrak{p}_{k+1,t-k-1}\right)\ \leqslant\ \max\left\{\operatorname{cd}\left(\mathfrak{p}_{0,t}\cap\mathfrak{p}_{1,t-1}\cap\dots\cap\mathfrak{p}_{k,t-k}\right),\right. \\&\qquad\quad\qquad\qquad\qquad\qquad\left. \operatorname{cd}(\mathfrak{p}_{k+1,t-k-1}),\\operatorname{cd}\left([\mathfrak{p}_{0,t}\cap\mathfrak{p}_{1,t-1}\cap\dots\cap\mathfrak{p}_{k,t-k}]+\mathfrak{p}_{k+1,t-k-1}\right)-1\right\}. \end{align*} $$

Up to taking radicals, the ideal

$$\begin{align*}[\mathfrak{p}_{0,t}\cap\mathfrak{p}_{1,t-1}\cap\dots\cap\mathfrak{p}_{k,t-k}]+\mathfrak{p}_{k+1,t-k-1} \end{align*}$$

coincides with

$$ \begin{align*} (\mathfrak{p}_{0,t}+\mathfrak{p}_{k+1,t-k-1})\cap(\mathfrak{p}_{1,t-1}+\mathfrak{p}_{k+1,t-k-1}) &\cap\dots\cap(\mathfrak{p}_{k,t-k}+\mathfrak{p}_{k+1,t-k-1})\\ &\quad \ =\ \mathfrak{p}_{0,t-k-1}\cap\mathfrak{p}_{1,t-k-1}\cap\dots\cap\mathfrak{p}_{k,t-k-1}\ =\ \mathfrak{p}_{k,t-k-1}, \end{align*} $$

since $\mathfrak {p}_{i_1,j_1}+\mathfrak {p}_{i_2,j_2}=\mathfrak {p}_{i,j}$ for

and

. It follows that

$$ \begin{align*} &\operatorname{cd}\left(\mathfrak{p}_{0,t}\cap\mathfrak{p}_{1,t-1}\cap\dots\cap\mathfrak{p}_{k+1,t-k-1}\right)\\&\qquad\qquad\qquad\qquad \leqslant\ \max\left\{\operatorname{cd}\left(\mathfrak{p}_{0,t}\cap\mathfrak{p}_{1,t-1}\cap\dots\cap\mathfrak{p}_{k,t-k}\right),\ \operatorname{cd}(\mathfrak{p}_{k+1,t-k-1}),\ \operatorname{cd}(\mathfrak{p}_{k,t-k-1})-1\right\}. \end{align*} $$

Using this inductively, one obtains the inequality (4.2.2).

Since the rings $S/\mathfrak {p}_{i,j}$ are Cohen–Macaulay for $i+j\leqslant t$ , Theorem 2.3 implies that $\operatorname {cd}(\mathfrak {p}_{i,j}) = \operatorname {ht}(\mathfrak {p}_{i,j})$ . Consequently, the inequality (4.2.2) gives

$$ \begin{align*} \operatorname{cd}\left(\mathfrak{p}_{0,t}\cap\mathfrak{p}_{1,t-1}\cap\dots\cap\mathfrak{p}_{k,t-k}\right)\ \leqslant\ \max &\left\{\operatorname{ht}(\mathfrak{p}_{0,t}),\ \operatorname{ht}(\mathfrak{p}_{1,t-1}),\ \dots,\ \operatorname{ht}(\mathfrak{p}_{k,t-k}),\right. \\&\quad \left. \operatorname{ht}(\mathfrak{p}_{0,t-1})-1,\ \operatorname{ht}(\mathfrak{p}_{1,t-2})-1,\ \dots,\ \operatorname{ht}(\mathfrak{p}_{k-1,t-k})-1\right\}. \end{align*} $$

Using the formula for $\operatorname {ht}(\mathfrak {p}_{i,j})$ from Theorem 4.1, it is readily verified that for each fixed integer $\ell $ with $0\leqslant \ell \leqslant t$ , one has

$$\begin{align*}\max\left\{\operatorname{ht}(\mathfrak{p}_{0,\ell}),\ \operatorname{ht}(\mathfrak{p}_{1,\ell-1}),\ \dots,\ \operatorname{ht}(\mathfrak{p}_{\ell,0})\right\}\ =\ \ell^2 -(2t+1)\ell +2t(t+1), \end{align*}$$

which then yields the inequality (4.2.1).

Proof. It suffices to consider the case where $m=t$ . Denote the symplectic form by B, and set , a codimension one subspace. We construct the subspaces $W_j$ by reverse induction on j. If $V_t$ has dimension t, simply choose $W_t$ to be $V_t$ itself. If $V_t$ has dimension less than t, then $\dim (V_t^\perp )>t$ , so $\dim (V_t^\perp \cap H) \geqslant t$ .

If $k<t$ , take $W_t$ to be a Lagrangian subspace of $K^{2t}$ containing $V_t$ . If $k=t$ , since ${V_t\subset H}$ , there exists a nonzero vector $x\in (V_t^\perp \cap H) \smallsetminus V$ . Then

$$\begin{align*}V_t +Kx\ \subseteq\ (V_t+Kx)^\perp \cap H. \end{align*}$$

Continuing in this manner, we can extend $V_t$ to a Lagrangian subspace of $K^{2t}$ on which L vanishes.

Assume that the vector spaces $W_{j+1}, W_{j+2}, \dots , W_t$ have been constructed satisfying the required conditions. There are two cases: If j is different from k, simply choose $W_j$ of dimension j such that $V_j \subseteq W_j \subseteq W_{j+1}$ . This can be done since $V_j$ has dimension at most j and $W_{j+1}$ has dimension $j+1$ . If j equals k, choose $W_k$ of dimension k such that ${V_k \subseteq W_k \subseteq H \cap W_{k+1}}$ ; this can indeed be done since $V_k$ has dimension at most k, and $H \cap W_{k+1}$ has dimension at least

$$\begin{align*}(2t-1)+(k+1)-2t\ =\ k. \end{align*}$$

Finally, since any subspace of an isotropic subspace is isotropic, we are done.

Proof. If $n\leqslant t$ , there is nothing to prove. If $Y'$ is a truncation of Y obtained by deleting certain columns, then the alternating matrix $Y^{\prime \operatorname {tr}}\Omega Y'$ is a truncation of the alternating matrix $Y^{\operatorname {tr}}\Omega Y$ obtained by deleting the corresponding columns and rows; thus, it suffices to prove the lemma when Y is size $2t\times (t+1)$ .

Next, note that any size $t+1$ minor of Y equals the determinants of a matrix of the form $Y\# Z$ , where Z is a suitable size $2t\times (t-1)$ matrix with entries $0$ and $1$ , and $\#$ denotes the concatenation of matrices; for example, for the upper size $t+1$ minor, one may take Z to be the block matrix .

Thus, it suffices to prove that for all matrices Z of size $2t\times (t-1)$ , one has

$$\begin{align*}\det(Y\# Z)\ \in\ (Y^{\operatorname{tr}}\Omega Y). \end{align*}$$

Since $\det (Y\# Z)=\operatorname {pf}((Y\# Z)^{\operatorname {tr}}\Omega (Y\# Z))$ , it suffices to prove that

$$\begin{align*}\operatorname{pf}((Y\# Z)^{\operatorname{tr}}\Omega(Y\# Z))\ \in\ (Y^{\operatorname{tr}}\Omega Y). \end{align*}$$

But $(Y\# Z)^{\operatorname {tr}}\Omega (Y\# Z)$ is a size $2t$ alternating matrix, and $Y^{\operatorname {tr}}\Omega Y$ its upper-left size $t+1$ submatrix. Working modulo the entries of $Y^{\operatorname {tr}}\Omega Y$ , it suffices to check that the Pfaffian of a size $2t$ alternating matrix of the form

$$\begin{align*}\begin{pmatrix} 0 & A\\ -A^{\operatorname{tr}} & B \end{pmatrix} \end{align*}$$

is zero, where A and B are size $t\times t$ , and the first column of A is zero. This is immediate, as the determinant of such a matrix is zero.

Proof. It suffices to prove that $K[Y]/\mathfrak {P}$ is a complete intersection ring after specializing the entries of the rows indexed

$$\begin{align*}1,\ 2,\ \dots,\ t+1-n,\qquad t+1,\ t+2,\ \dots,\ 2t+1-n \end{align*}$$

to zero since this leaves the number of defining equations unchanged. We may hence assume that the matrix Y has $2t-2(t+1-n)=2n-2$ rows, that is, Y is size $2(n-1)\times n$ , equivalently, size $2t\times (t+1)$ . Next, specialize the entries of Y to the corresponding entries of the matrix

This entails killing

$$\begin{align*}\binom{t+1}{2}+t(t+1)\ =\ 2t(t+1) -\binom{t+1}{2} \end{align*}$$

linear forms in $K[Y]$ . As $K[\overline {Y}]/(\overline {Y}^{\operatorname {tr}}\Omega \overline {Y})$ is Artinian, $K[Y]/\mathfrak {P}$ is a complete intersection ring.

Proof. As seen in the previous proof, the generators of the ideal $\mathfrak {a}$ form part of a system of parameters for the ring $S/\mathfrak {P}$ .

Proof. It suffices to consider the case where the field K is algebraically closed. Since $S/I$ is a complete intersection ring by the corollary above, we need to show that $\Delta $ does not belong to any minimal prime of the ideal I.

Let G be the subgroup of $\operatorname {Sp}_{2t}(K)$ consisting of matrices with

$$ \begin{align*} m_{11} &= 1 = m_{t+1,t+1} \\ m_{1i} &= 0 = m_{i1} \qquad \text{for}\ i\neq 1\\ m_{t+1,i} &= 0 = m_{i,t+1} \quad \text{for} \ i\neq t+1. \end{align*} $$

Deleting rows and columns $1$ and $t+1$ shows that G is isomorphic to $\operatorname {Sp}_{2t-2}(K)$ and is hence a connected algebraic group. The action of G on S via $M\colon Y\longmapsto MY$ induces an action on $S/I$ and thus on the (necessarily finite) set of minimal primes of $S/I$ . Since G is connected the action must be trivial, that is, G stabilizes each minimal prime of I.

Suppose a minimal prime P of I contains $\Delta $ . Using the fact that G stabilizes P, we shall first show that P contains each maximal minor of Y that involves the first row. Since row $t+1$ of Y is contained in I, hence in P, we need only consider maximal minors of Y that involve the first row and not row $t+1$ . We use to denote the image of $\Delta $ under an element M of G.

Let $\alpha $ be a size t subset of the row indices $\{1,\dots ,2t\}$ such that $1\in \alpha $ and $t+1\notin \alpha $ . We use $Y_{\alpha }$ for the square submatrix with rows $\alpha $ , and set $\ell (\alpha )$ to be the number of indices $a\in \alpha $ such that $a\leqslant t$ and $a+t\in \alpha $ . The proof that $\det (Y_{\alpha })\in P$ is by induction on $\ell (\alpha )$ .

For the case $\ell (\alpha )=0$ , proceed by induction on the number w of $a\in \alpha $ with $a>t$ . When $w=0$ , one has $\det (Y_{\alpha })=\Delta $ , which is an element of P. For the inductive step, consider the $2t\times 2t$ matrix M with

Observe that $M\in G$ , and that the matrix $MY$ is obtained from Y by the row operations where row $i+t$ is added to row i whenever $i\leqslant t$ and $i+t\in \alpha $ . It follows that

is the determinant of the $t\times t$ matrix whose i-th row is the sum of rows i and $i+t$ of Y if $i\leqslant t$ and $i+t\in \alpha $ , and is row i of Y otherwise. By the linearity of determinants along a row,

is the sum of $t\times t$ minors of Y, each of which is indexed by a set of rows $\beta $ with $\ell (\beta )=0$ . One of these is $\det (Y_{\alpha })$ , while the others have fewer indices greater than t. Using

and the inductive hypothesis, it follows that $\det (Y_{\alpha })\in P$ , settling the case $\ell (\alpha )=0$ .

Next, fix $\alpha $ with $\ell (\alpha )>0$ . Let $i,j\in \{1,\dots ,t\}$ be such that $i, i+t\in \alpha $ and $j,j+t\notin \alpha $ ; such a j exists by cardinality reasons. Let $\alpha ' = \alpha \smallsetminus \{i,i+t\}$ . Observe that each of

$$\begin{align*}\ell(\alpha'\cup\{i,j\}),\quad \ell(\alpha'\cup\{i,j+t\}),\quad \ell(\alpha'\cup\{i+t,j\}),\quad \ell(\alpha'\cup\{i+t,j+t\}) \end{align*}$$

is strictly less than $\ell (\alpha )$ . Let M be the $2t \times 2t$ matrix with

Note that $M\in G$ , and that the matrix $MY$ is obtained from Y by row operations where the $(j+t)$ -th row is added to the i-th row, and the $(i+t)$ -th row is added to the j-th row. Hence, up to choices of signs,

is the sum of

$$\begin{align*}\det(Y_{\alpha'\cup\{i,j\}}),\quad \det(Y_{\alpha'\cup\{i,i+t\}}),\quad \det(Y_{\alpha' \cup \{j,j+t\}}),\quad \det(Y_{\alpha' \cup \{i+t,j+t\}}). \end{align*}$$

By the inductive hypothesis, $\det (Y_{\alpha '\cup \{i,j\}})$ and $\det (Y_{\alpha '\cup \{i+t,j+t\}}$ are elements of the prime P, as is $\det (Y_{\alpha '\cup \{i,j\}})$ and hence

. It follows that, with a sign choice, one of

(6.5.1) $$ \begin{align} \det(Y_{\alpha'\cup\{i,i+t\}})\ \pm\ \det(Y_{\alpha'\cup\{j,j+t\}}) \end{align} $$

is an element of P. We claim that there exists a Plücker relation in $K[Y]$ of the form

(6.5.2) $$ \begin{align} \det(Y_{\alpha'\cup\{i,i+t\}}) \det(Y_{\alpha'\cup\{j,j+t\}}) \ \pm\ \det(Y_{\alpha'\cup\{i,j\}}) &\det(Y_{\alpha'\cup\{i+t,j+t\}}) \notag \\ &\ \ \pm\ \det(Y_{\alpha'\cup\{i,j+t\}}) \det(Y_{\alpha'\cup\{i+t,j\}})\ =\ 0. \end{align} $$

This may be verified, for example, by passing to a dense open subset of matrices where the rows $\alpha '\cup \{i,i+t\}$ form a basis for $K^t$ , and multiplying on the right by an invertible matrix so as reduce to the case where these rows are the standard basis for $K^t$ . The equality is now readily checked.

Since the other terms in (6.5.2) belong to P by the induction hypothesis, one obtains

(6.5.3) $$ \begin{align} \det(Y_{\alpha'\cup\{i,i+t\}})\det(Y_{\alpha'\cup\{j,j+t\}})\ \in\ P. \end{align} $$

Combining (6.5.1) and (6.5.3), bearing in mind that P is prime, it follows that

$$\begin{align*}\det(Y_{\alpha})\ =\ \det(Y_{\alpha'\cup\{i,i+t\}})\ \in\ P, \end{align*}$$

completing the proof that P contains each $t\times t$ minor of Y that involves the first row.

If P contains $y_{11}$ , Corollary 6.4 gives a contradiction. It follows that the prime ideal P must contain each size $t-1$ minor of the last $t-1$ columns of Y.

Let $Y'$ be the $2(t-1)\times t$ submatrix obtained by deleting rows $1$ and $t+1$ of Y, and $\Omega '$ be the size $2t-2$ standard symplectic block matrix. Set $I'$ to be the ideal of $K[Y']$ generated by the entries of $Y^{\prime \operatorname {tr}}\Omega 'Y'$ along with the size $t-1$ minors of the last $t-1$ columns of $Y'$ . On an open dense subset of $V(I')$ , the last column belongs to the span of colums $2,3,\dots ,t-1$ . Since the dimension of the Pfaffian nullcone corresponding to a $2(t-1)\times (t-1)$ matrix is

$$\begin{align*}2(t-1)^2-\binom{t-1}{2} \end{align*}$$

by Corollary 6.4, it follows that

$$\begin{align*}\dim V(I')\ \leqslant\ 2(t-1)^2-\binom{t-1}{2}+(t-2). \end{align*}$$

Accounting for the matrix entry $y_{11}$ , this implies

$$\begin{align*}\dim V(P)\ \leqslant\ 2(t-1)^2-\binom{t-1}{2}+(t-2)+1\ =\ 2t^2-\binom{t}{2}-2t. \end{align*}$$

But then

$$\begin{align*}\dim V(P)\ <\ \dim V(I)\ =\ 2t^2-\binom{t}{2}-(2t-1), \end{align*}$$

where the equality uses, again, Corollary 6.4. This is not possible since P is a minimal prime of I.

Proof. Since the projection map onto the first n columns provides a surjection of algebraic sets, it suffices to prove each result in the case $n=t$ . Let $\Delta $ be the upper $t\times t$ minor of Y.

We first consider $I_a$ . In this case, Corollary 6.4 and Lemma 6.5—after permuting columns—show that $\Delta $ is a nonzerodivisor modulo $I_a$ . Write Y as

$$\begin{align*}\begin{pmatrix}Y_1\\ Y_2\end{pmatrix}, \end{align*}$$

where $Y_1$ and $Y_2$ are size $t\times t$ . Since $Y_1$ is invertible over the ring $S_{\Delta }$ , one has $S_{\Delta }=K[Y_1,\ Z]_{\Delta }$ , where the entries of $Y_1$ and

are algebraically independent over K. Note that

so the ideal $(Y^{\operatorname {tr}}\Omega Y)S_{\Delta }$ is generated by the entries of

It follows that

$$\begin{align*}I_a S_{\Delta}\ =\ (Z-Z^{\operatorname{tr}})S_{\Delta}+(y_{11},\dots,y_{1a})S_{\Delta}. \end{align*}$$

Since $I_a S_{\Delta }$ is generated by linear forms belonging to the polynomial ring $K[Y_1,\ Z]$ , it is a prime ideal of $S_{\Delta }$ ; as $\Delta $ is a nonzerodivisor modulo $I_a$ , there is a bijection between the minimal primes of $I_a$ and those of $I_a S_{\Delta }$ . It follows that $I_a$ has a unique minimal prime, so $V(I_a)$ is irreducible.

In the case of $I^{\prime }_a$ , working again with $n=t$ , the ring $S/I^{\prime }_a$ is a polynomial extension of

$$\begin{align*}K[Y']/(Y^{\prime \operatorname{tr}}\Omega' Y'), \end{align*}$$

where $Y'$ is the $(2t-2)\times t$ matrix of indeterminates obtained by deleting rows $1$ and $t+1$ of Y, and

, and $\Omega '$ is the size $2t-2$ standard symplectic block matrix. It suffices to prove that the ring $S'/(Y^{\prime \operatorname {tr}}\Omega ' Y')$ has a unique minimal prime. Let $\Delta '$ be the upper left size $t-1$ minor of $Y'$ . Lemma 6.5 implies that $\Delta '$ is a nonzerodivisor on $S'/(Y^{\prime \operatorname {tr}}\Omega ' Y')$ . Writing the matrix $Y'$ as

$$\begin{align*}Y'\ =\ \begin{pmatrix}Y_1 & W_1\\ Y_2 & W_2\end{pmatrix}, \end{align*}$$

where $Y_1$ and $Y_2$ are square matrices of size $t-1$ , one has

The entries of $Y_1$ , $W_1$ ,

and

are algebraically independent over K, and $S^{\prime }_{\Delta '}$ may be viewed as $K[Y_1,\ W_1,\ Z_1,\ Z_2]_{\Delta '}$ . Since

it follows that

$$\begin{align*}S^{\prime}_{\Delta'}/(Y^{\prime \operatorname{tr}}\Omega' Y')_{\Delta'}\ =\ K[Y_1,\ W_1,\ Z_1,\ Z_2]_{\Delta'}/(Z_1-Z_1^{\operatorname{tr}},\ Z_2), \end{align*}$$

and is hence a domain. In particular, it has a unique minimal prime.

Proof. It suffices to prove the assertions when K is algebraically closed; we indeed work under this assumption. We begin by proving (1) for the algebraic set $V(I_{\sigma }+J_{s_k})$ . Consider matrices B of size $2t\times m$ for which the columns span an isotropic subspace, and the first k entries of the first row are zero. Since $m\leqslant t$ , Proposition 6.6 implies that the matrices B are the points of an irreducible algebraic set that we denote $\mathbb {V}_0$ .

For $1\leqslant j\leqslant m$ , let $C_j$ be a matrix of size $j\times (s_j-s_{j-1})$ , and set A to be the matrix

(6.7.1)

where $\#$ denotes the concatenation of matrices. It is readily seen that A is an element of the algebraic set $V(I_{\sigma }+J_{s_k})$ . The matrices $C_1,\dots ,C_m$ may be regarded as the points of an affine space $\mathbb {V}_1$ of dimension

$$\begin{align*}\sum_{j=1}^m j(s_j-s_{j-1}), \end{align*}$$

so that the construction (6.7.1) gives a map

$$\begin{align*}\mathbb{V}_0\times \mathbb{V}_1\longrightarrow V(I_{\sigma}+J_{s_k}). \end{align*}$$

Since the image of an irreducible algebraic set is irreducible, it suffices to verify that this map is surjective.

Let A be a matrix in the algebraic set $V(I_{\sigma }+J_{s_k})$ . For $1\leqslant j\leqslant m$ , let $V_j$ denote the span of the columns of the truncated matrix $A|_{s_j}$ . Consider the symplectic form (6.0.1) on $K^{2t}$ and the linear functional L that is projection to the first coordinate. By Lemma 6.1, there exist isotropic subspaces

$$\begin{align*}W_1 \subset W_2 \subset \dots \subset W_m \end{align*}$$

such that $V_j\subseteq W_j$ for each j, and $W_j$ has rank j. Consider a size $2t\times m$ matrix B such that $B|_j$ spans $W_j$ for each j. Then the columns of $A|_{s_j}$ belong to the column span of $B|_j$ for each j, so there exist matrices $C_j$ using which A may be obtained as in (6.7.1).

The proof that $V(I_{\sigma }+J^{\prime }_{s_k})$ is irreducible is similar: We consider instead matrices B of size $2t\times m$ , where the columns span an isotropic subspace, and for which the first row is zero, and the first k entries of row $t+1$ are zero. Proposition 6.6 implies that such matrices B are the points of an irreducible algebraic set. The linear functional used when applying Lemma 6.1 is now projection to the $t+1$ coordinate.

The proof of (2) is via induction, assuming the result for matrices Y of smaller size, as well as for larger ideals in the family, and applying Lemma 5.1. Set I to be either $I_{\sigma }+J_a$ or $I_{\sigma }+J^{\prime }_a$ . In the latter case, assume that $a<n$ since otherwise $K[Y]/(I_{\sigma }+J^{\prime }_a)$ arises from the smaller matrix obtained by deleting rows $1$ and $t+1$ of Y. To apply Lemma 5.1, choose

Specializing x to $1$ and each other entry to $0$ , we obtain a matrix in $V(I)\smallsetminus V(I+xS)$ , from which it follows that $I+xS$ is a larger ideal in the family, and hence radical by the inductive hypothesis. If $a=s_k$ for some k, then

is prime by (1); since $x\notin P$ , Lemma 5.1 implies that $I=P$ , so I is prime.

In the remaining cases, there exists an integer k with $s_k<a<s_{k+1}$ and the element x is either $y_{1,a+1}$ or $y_{t+1,a+1}$ . Set

and take P to be the prime ideal $I_{\sigma '}+J_a$ or $I_{\sigma '}+J^{\prime }_a$ in the respective cases; if $k=0$ , then

$$\begin{align*}\sigma'=(a,s_1,\dots,s_m) \end{align*}$$

is not standard, but the primality follows nonetheless from the case of a matrix of size $2t\times (n-a)$ . The specialization used earlier shows that $x\notin P$ . Using Lemma 5.3, one has

$$\begin{align*}y_{1,a+1}I_{k+1}(Y|_a)\ \subseteq\ I_{k+2}(Y|_{a+1})+J_a\ \subseteq\ I_{k+2}(Y|_{s_{k+1}})+J_a\ \subseteq\ I_{\sigma}+J_a \end{align*}$$

and

$$\begin{align*}y_{t+1,a+1}I_{k+1}(Y|_a)\ \subseteq\ I_{k+2}(Y|_{a+1})+J^{\prime}_a\ \subseteq\ I_{k+2}(Y|_{s_{k+1}})+J^{\prime}_a\ \subseteq\ I_{\sigma}+J^{\prime}_a, \end{align*}$$

so $xP\subseteq I$ in either case. It follows that I is radical by Lemma 5.1.

For (3), let V denote the algebraic set $V(I_{\sigma }+J_a)$ or $V(I_{\sigma }+J^{\prime }_a)$ . We first compute the dimension of V. In each case, V has an open subset U in which each matrix has the property that the submatrix consisting of the columns indexed

(6.7.2) $$ \begin{align} s_0+1,\ s_1+1,\ \dots,\ s_{m-1}+1 \end{align} $$

has rank exactly m. Note that $m\leqslant t$ and that $m\leqslant n$ . This open set U is nonempty, hence dense, for it contains the matrix in which the columns indexed (6.7.2) are, respectively, the standard basis vectors

$$\begin{align*}e_{t+2},\ e_{t+3},\ \dots,\ e_{t+m},\ e_{t+1}, \end{align*}$$

and all other columns are zero. The order of the standard basis vectors above accounts for the possibility that V may be $V(I_{\sigma }+J^{\prime }_{n-1})$ , though it cannot be $V(I_{\sigma }+J^{\prime }_n)$ , given our hypotheses. It suffices to compute the dimension of U.

Given a matrix A in the open set U, let B denote the $2t \times m$ submatrix consisting of the columns indexed (6.7.2). For each j with $1\leqslant j\leqslant m$ , the submatrix $D_j$ of A consisting of the columns indexed $s_{j-1}+1,\dots ,s_j$ can be uniquely written as a linear combination of the columns of $B|_j$ . The coefficients needed comprise the columns of a size $j\times (s_j-s_{j-1})$ matrix that we denote $C_j$ . The first column of $C_j$ is

$$\begin{align*}(0,\ 0,\ \dots,\ 0,\ 1)^{\operatorname{tr}} \end{align*}$$

while the other $j(s_j-s_{j-1}-1)$ entries are arbitrary scalars. In the case $V(I_{\sigma }+J_a)$ , the matrices B vary in a space of dimension

$$\begin{align*}2mt-\binom{m}{2}-k \end{align*}$$

by Corollary 6.4, and it follows that U has dimension

$$ \begin{align*} 2mt-\binom{m}{2}-k+1(s_1-s_0-1)+2(s_2-s_1-1)+\dots+m &(s_m-s_{m-1}-1) \\ &\qquad\quad =\ m(2t+n-m)-k-\sum_{j=1}^{m-1}s_j. \end{align*} $$

The dimension count for $V(I_{\sigma }+J^{\prime }_a)$ is similar, bearing in mind that in this case the matrices B vary in a space of dimension

$$\begin{align*}2mt-\binom{m}{2}-m-k. \end{align*}$$

The proof of the Cohen–Macaulay property is again via induction, assuming the result for matrices Y of smaller size, as well as for larger ideals in the family. Consider a prime of the form $I_{\sigma }+J_{s_k}$ , where $k\leqslant m-1$ . Since $y_{1,s_k+1}$ is a nonzerodivisor on $S/(I_{\sigma }+J_{s_k})$ , it suffices to prove that

$$\begin{align*}S/(I_{\sigma}+J_{s_k}+y_{1,s_k+1}S)\ =\ S/(I_{\sigma}+J_{s_k+1}) \end{align*}$$

is Cohen–Macaulay. If $s_k+1=s_{k+1}$ , then this is immediate from the inductive hypothesis. Else, $s_k+1<s_{k+1}$ , and we claim that $I_{\sigma }+J_{s_k+1}$ has minimal primes

where

if $k=0$ , then $\sigma '=(1,s_1,\dots ,s_m)$ is not standard, but $Q_2$ is prime by the case of a matrix of size $2t\times (n-1)$ , and the dimension of $S/Q_2$ is readily computed. Since $I_{\sigma }+J_{s_k+1}$ is radical and contained in each $Q_i$ , it suffices to verify that

$$\begin{align*}Q_1Q_2\ \subseteq\ I_{\sigma}+J_{s_k+1}. \end{align*}$$

This is straightforward since

$$\begin{align*}y_{1b}I_{k+1}(Y|_{s_k+1})\ \subseteq\ I_{k+2}(Y|_{s_{k+1}}) + J_{s_k+1}\ \subseteq\ I_{\sigma}+J_{s_k+1} \end{align*}$$

for each b with $b\leqslant s_{k+1}$ by Lemma 5.3. By the inductive hypothesis, each $Q_i$ is prime, defining a Cohen–Macaulay ring $S/Q_i$ . Moreover,

$$\begin{align*}Q_1+Q_2\ =\ I_{\sigma'}+J_{s_{k+1}} \end{align*}$$

is prime, and Lemma 5.2 applies since

$$\begin{align*}\dim S/Q_1\ =\ \dim S/Q_2\ =\ m(2t+n-m)-k-1-\sum_{j=1}^{m-1}s_j\ =\ \dim S/(Q_1+Q_2)+1. \end{align*}$$

This concludes the argument that

$$\begin{align*}S/(I_{\sigma}+J_{s_k+1})\ =\ S/(Q_1\cap Q_2) \end{align*}$$

is Cohen–Macaulay. The proof for a prime ideal of the form $I_{\sigma }+J^{\prime }_{s_k}$ , with $k\leqslant m-1$ , is similar.

The remaining case is a prime of the form $I_{\sigma }+J_n$ , where it suffices to prove that

$$\begin{align*}S/(I_{\sigma}+J_n+y_{t+1,1}S)\ =\ S/(I_{\sigma}+J^{\prime}_1) \end{align*}$$

is Cohen–Macaulay. This follows from the inductive hypothesis if $s_1=1$ . If $s_1>1$ , we claim that $I_{\sigma }+J^{\prime }_1$ has minimal primes

For this, it suffices to verify that $Q_1Q_2\subseteq I_{\sigma }+J^{\prime }_1$ , which follows using $I_2(Y|_{s_1})\subseteq I_{\sigma }$ . Note that $S/Q_2$ and $S/(Q_1+Q_2)$ are Cohen–Macaulay using the case of a smaller matrix, namely the matrix with the first column of Y deleted. Since

$$\begin{align*}\dim S/Q_1\ =\ \dim S/Q_2\ =\ m(2t+n-m-1)-1-\sum_{j=1}^{m-1}s_j\ =\ \dim S/(Q_1+Q_2)+1, \end{align*}$$

Lemma 5.2 allows us to conclude that

$$\begin{align*}S/(I_{\sigma}+J^{\prime}_1)\ =\ S/(Q_1\cap Q_2) \end{align*}$$

is Cohen–Macaulay.

Proof. The formulae for the dimension coincide when n equals t or $t+1$ .

If $n\leqslant t$ , take $\sigma =(0,1,2,\dots ,n-1,n)$ in Theorem 6.7.3, to obtain

$$\begin{align*}\dim S/\mathfrak{P}\ =\ n(2t+n-n)-(1+2+\dots+(n-1))\ =\ 2nt -\displaystyle{\binom{n}{2}}, \end{align*}$$

while if $n>t$ , take $\sigma =(0,1,2,\dots ,t-1,n)$ , in which case the theorem gives

$$\begin{align*}\dim S/\mathfrak{P}\ =\ t(2t+n-t)-(1+2+\dots+t-1)\ =\ nt+\binom{t+1}{2}, \end{align*}$$

completing the proof.

Proof. We claim that if the inclusion $\varphi \colon K[Y^{\operatorname {tr}}\Omega Y]\longrightarrow K[Y]$ is pure for fixed $(n,t)$ , then purity holds as well for the inclusion of the K-algebras corresponding to $(n',t)$ with $n'\leqslant n$ .

Set , that is, $Y'$ is the submatrix consisting of the first $n'$ columns of Y. Consider the $\mathbb {N}$ -grading on $K[Y]$ , where the indeterminates from $Y'$ have degree $0$ , as does K, while the remaining indeterminates have degree $1$ . Then

$$\begin{align*}{K[Y^{\operatorname{tr}}\Omega Y]}_0\ =\ K[Y^{\prime \operatorname{tr}}\Omega Y'], \end{align*}$$

so $K[Y^{\prime \operatorname {tr}}\Omega Y']$ is a pure subring of $K[Y]$ . It follows that the composition

$$\begin{align*}K[Y^{\prime \operatorname{tr}}\Omega Y']\ \subseteq\ K[Y^{\operatorname{tr}}\Omega Y]\ \subseteq\ K[Y] \end{align*}$$

is pure as well, but then so is $K[Y^{\prime \operatorname {tr}}\Omega Y']\subseteq K[Y']$ .

Set and . We next prove that $\varphi $ is pure in the case $n=t+1$ . In this case, the ring R is regular, with the upper triangular entries of $Y^{\operatorname {tr}}\Omega Y$ forming a regular homogeneous system of parameters for R. As $\dim R=\binom {n}{2}$ , it suffices by Theorem 2.1 to verify that the local cohomology module

$$\begin{align*}H^{\binom{n}{2}}_{\mathfrak{m}_R}(S) \end{align*}$$

is nonzero, where $\mathfrak {m}_R$ is the homogeneous maximal ideal of R. This is immediate from Theorem 6.8, which implies that $\mathfrak {m}_RS$ is an ideal of height $\binom {n}{2}$ .

It remains to prove that $\varphi \colon R\longrightarrow S$ is not pure if $n\geqslant t+2$ . By the reduction step, this comes down to the case $n=t+2$ . In this case, the ring $R=K[Y^{\operatorname {tr}}\Omega Y]$ is again regular, of dimension $\binom {n}{2}$ , so by Theorem 2.1 it suffices to verify the vanishing of $H^{\binom {n}{2}}_{\mathfrak {m}_R}(S)$ . This follows from Theorem 6.8, which implies that $\mathfrak {m}_RS$ is an ideal of height

$$\begin{align*}\binom{n}{2}-1, \end{align*}$$

defining a Cohen–Macaulay ring $S/\mathfrak {m}_RS$ .

Proof. It suffices to prove the assertion after specializing the entries of the last $d-2n+1$ rows to zero. We may hence assume that $n=(d+1)/2$ , that is, Y is size $(2n-1)\times n$ .

First, suppose $k=0$ . Specialize the entries of Y to the corresponding entries of the matrix

A routine—albeit tedious—count shows that this specialization entails killing

$$\begin{align*}3n(n-1)/2 \end{align*}$$

linear forms in $K[Y]$ . The ideal $(Y^{\operatorname {tr}} Y)$ has $\binom {n+1}{2}$ minimal generators; since

$$\begin{align*}\dim K[Y]\ =\ (2n-1)n\ =\ \binom{n+1}{2} + \frac{3}{2}n(n-1), \end{align*}$$

it suffices to verify that

$$\begin{align*}K[\overline{Y}]/(\overline{Y}^{\operatorname{tr}}\overline{Y}) \end{align*}$$

has dimension zero. The $(1,n)$ entry of the matrix $\overline {Y}^{\operatorname {tr}}\overline {Y}$ is $y_{n1}^2$ . Modulo $y_{n1}$ , the $(2,n)$ entry is $y_{n+1,2}^2$ . Proceeding in this order, examining the last column of $\overline {Y}^{\operatorname {tr}}\overline {Y}$ , we see that

$$\begin{align*}y_{n1},\ y_{n+1,2},\ y_{n+2,3},\ \dots,\ y_{2n-2,n-1},\ y_{2n-1,n} \end{align*}$$

are nilpotent in $K[\overline {Y}]/(\overline {Y}^{\operatorname {tr}}\overline {Y})$ . Modulo these elements, the last column and the last two rows of $\overline {Y}$ are zero; proceed inductively.

Since the displayed specialization $\overline {Y}$ entails killing $n-1$ entries from the first row, the case $0<k<n$ follows as well.

Proof. Let $\widetilde {Y}$ denote the upper $(d-1)\times n$ submatrix of Y. In the ring $S/\mathfrak {S}$ , one has

$$\begin{align*}y_{di}\ =\ y_{1i}+\dots+y_{d-1,i} \end{align*}$$

for each i, so $S/\mathfrak {S}$ is a homomorphic image of $K[\widetilde {Y}]$ . Making the substitutions using the equation displayed above, one sees that

$$\begin{align*}S/\mathfrak{S}\ \cong\ K[\widetilde{Y}]/(\widetilde{Y}^{\operatorname{tr}}\Psi\widetilde{Y}), \end{align*}$$

where $\Psi $ is the $(d-1)\times (d-1)$ alternating matrix

$$\begin{align*}\Psi\ =\ \begin{pmatrix} 0 & 1 & 1 & \ldots & 1\\ 1 & 0 & 1 & \ldots & 1\\ 1 & 1 & 0 & & 1\\ \vdots & \vdots & & \ddots & \\ 1 & 1 & 1 & & 0 \end{pmatrix}. \end{align*}$$

It is readily checked that $\Psi $ is invertible if $d-1$ is even and that it has rank $d-2$ otherwise. Since alternating matrices of the same size are cogredient as in (6.0.2) precisely if they have the same rank, if $d-1$ is even, then $\Psi $ is cogredient to the standard symplectic block matrix $\Omega $ , whereas, if $d-1$ is odd, then $\Psi $ is cogredient to

$$\begin{align*}\begin{pmatrix} \Omega & 0\\ 0 & 0 \end{pmatrix}, \end{align*}$$

where $\Omega $ is size $d-2$ . This largely reduces the proof to an application of Theorem 6.8:

If $d=2t+1$ , the ring $S/\mathfrak {S}$ is isomorphic to $K[Z]/(Z^{\operatorname {tr}}\Omega Z)$ , where Z is a $2t\times n$ matrix of indeterminates. It follows that $S/\mathfrak {S}$ is a Cohen–Macaulay integral domain, with

$$\begin{align*}\dim S/\mathfrak{S}=\begin{cases} 2nt -\displaystyle{\binom{n}{2}} \ =\ nd -\displaystyle{\binom{n+1}{2}}& \text{if }\ n\leqslant t+1,\\ nt+\displaystyle{\binom{t+1}{2}}& \text{if }\ n\geqslant t. \end{cases} \end{align*}$$

If $d=2t+2$ , then $S/\mathfrak {S}$ is isomorphic to a polynomial ring in n indeterminates over the ring $K[Z]/(Z^{\operatorname {tr}}\Omega Z)$ , where Z is a matrix of indeterminates of size $2t\times n$ . It follows that $S/\mathfrak {S}$ is again a Cohen–Macaulay integral domain and that

$$\begin{align*}\dim S/\mathfrak{S}=\begin{cases} n+2nt-\displaystyle{\binom{n}{2}}\ =\ nd -\displaystyle{\binom{n+1}{2}}& \text{if }\ n\leqslant t+1,\\ n+nt+\displaystyle{\binom{t+1}{2}}& \text{if }\ n\geqslant t, \end{cases} \end{align*}$$

which completes the proof.

Proof. First, consider the case $\alpha =\{1,\dots ,k\}=\beta $ . Let

$$\begin{align*}Q\ =\ \begin{pmatrix} A & B\\ C & D \end{pmatrix}, \end{align*}$$

where A is a square matrix of size k, and D is a square matrix of size $n-k$ . Then

using which one has

Taking determinants gives

$$\begin{align*}\det Q\det D\ =\ \det A, \end{align*}$$

which is precisely the assertion of the lemma in this case.

For arbitrary $\alpha ,\beta $ , permute the rows of Q by sending the rows indexed $(\alpha ,\alpha ^{\mathrm {c}})$ to the rows indexed $(1,\dots ,n)$ , and the columns indexed $(\beta ,\beta ^{\mathrm {c}})$ to the columns indexed $(1,\dots ,n)$ . This yields an orthogonal matrix with determinant $\operatorname {sgn}(\alpha )\operatorname {sgn}(\beta )\det (Q)$ . The result now follows from the previous case.