Proof. By induction on $n\geq 2$ . The base case, where $n=2$ and $k=d=n-k=1$ is trivial. For the inductive step, we assume that for every choice of $k'$ and $d'$ satisfying $1\leq k'\leq d'\leq (n-1)-k'$ , the set $\left \{F_{T'}(d')|T'\in \operatorname {STab}(n-1,k',d')\right \}$ is linearly independent. Fix any integers $1\leq k\leq d\leq n-k$ and suppose we have a dependence relation

(7) $$ \begin{align} \sum_{T\in\operatorname{STab}(n,k,d)}c_TF_T(d)=0. \end{align} $$

Note that for every d-standard tableau $T\in \operatorname {STab}(n,k,d)$ as in (5), we must have either $n=i_d$ or $n=j_k$ ; let $I_d\subset \operatorname {STab}(n,k,d)$ denote the set of d-standard tableaux with $n=i_d$ and let $J_d\subset \operatorname {STab}(n,k,d)$ be the ones with $n=j_k$ . Let $\pi \colon R\rightarrow S={\mathbb F}[x_1,\ldots ,x_{n-1}]$ be the projection sending $x_n$ to $0$ , and note that for every $T\in I_d$ we have $\pi (F_T(d))=0$ . Moreover, for every $T\in J_d$ , we have $\pi (F_T(d))=F_{T'}(d)$ where

We see that $T'\in \operatorname {STab}(n-1,k-1,d)$ , hence the induction hypothesis applies. Note that since the map $J_d\rightarrow \operatorname {STab}(n-1,k-1,d)$ , sending $T\mapsto T'$ is one-to-one, and since

$$ \begin{align*}\pi\left(\sum_{T\in \operatorname{STab}(n,k,d)=I_d\sqcup J_d}c_TF_T(d)\right)=\sum_{T \in J_d}c_TF_{T'}(d)=0,\end{align*} $$

we deduce, by the induction hypothesis, that $c_T=0$ for all $T\in J_d$ . Then our dependence relation (7) becomes

$$ \begin{align*}\sum_{T \in I_d}c_TF_T(d)=0=x_n\cdot\left(\sum_{T \in I_d}c_TF_{T''}(d)\right),\end{align*} $$

where

In this case, we see that $T''\in \operatorname {STab}(n-1,k,d-1)$ , and again our induction hypothesis applies. Since the map $I_d\rightarrow \operatorname {STab}(n-1,k,d-1)$ sending $T\mapsto T''$ is one-to-one, and since

$$ \begin{align*}\sum_{T \in I_d}c_TF_{T''}(d)=0,\end{align*} $$

the induction hypothesis implies that $c_T=0$ for every $T\in I_d$ too, and therefore the dependence relation (7) must be trivial.

Proof. Note that for $1\leq i\leq a-1$ and for $b\leq i\leq n$ we have $\gamma ^i(T)=\gamma ^i((a,b)\cdot T)$ . Assume then that $a\leq i\leq b-1$ , and suppose that a and b belong to columns r and q in T, respectively. Then

$$ \begin{align*}\gamma^i((a,b)\cdot T) =\gamma^i(T)\ \begin{array}{c} \text{with }r\text{th part decreased by }1\\ \text{and }q\text{th part increased by }1,\\ \end{array}\end{align*} $$

and since we are assuming that $q<r$ , it follows that $\gamma ^i(T)\triangleleft \gamma ^i((a,b)\cdot T)$ , and the result follows.

Proof. We show by downward induction on the composition-dominance order on $\operatorname {Tab}(\lambda (d))$ that for every $T\in \operatorname {Tab}(\lambda (d))$ , $F_T(d)$ can be written as a linear combination of shifted Specht polynomials associated to d-standard tableaux. For the base case, note, as above, that the largest tableau of shifted shape $\lambda (d)$ is already standard. Inductively, fix a shifted tableau $T\in \operatorname {Tab}(\lambda (d))$ as in (5), and assume that for every tableau $T'$ that dominates T, $F_{T'}(d)$ can be written as a linear combination of shifted Specht polynomials corresponding to standard tableaux. We may assume that the columns of T are increasing. If T has no row descents, then T must be standard and we are done. Otherwise T has some row descent, say between the $a^{th}$ and $a+1^{st}$ column. There are several cases to consider and we claim that in all cases we can write $F_T(d)$ as a linear combination

$$ \begin{align*}F_T(d)=\sum_{T \triangleleft T'}c_{T'}F_{T'}(d).\end{align*} $$

Case 1: $1\leq a\leq n-k-d-1$ . Set $b=d+a$ ; in this case, we can merely swap $i_b$ and $i_{b+1}$ without affecting $F_T(d)$ ; in other words setting $T'=(i_b,i_{b+1})\cdot T$ we have

$$ \begin{align*}F_T(d)=F_{T'}(d),\end{align*} $$

and since $T\triangleleft (i_b,i_{b+1})\cdot T$ , it follows from our inductive hypothesis that $F_T(d)$ can be written as a linear combination of d-standard Specht polynomials on the shifted shape $\lambda (d)$ .

Case 2: $a=n-k-d$ . Then we have

and we have the descent $i_{n-k}>j_1>i_1$ . Then by Lemma 2.4, we see that

$$ \begin{align*}T\triangleleft T'=\left(i_{n-k},j_{1}\right)\cdot T \ \text{and} \ T\triangleleft T''=\left(i_1,i_{n-k}\right)\cdot T.\end{align*} $$

Moreover, one can easily check that

$$ \begin{align*}F_T(d)=F_{T'}(d)-F_{T''}(d)=G\left((x_{i_1}-x_{i_{n-k}})-(x_{j_1}-x_{i_{n-k}})\right),\end{align*} $$

and inductive hypothesis applies.

Case 3: $n-k-d+1\leq a\leq n-d-1$ . Set $b=a-n+k+d$ so we have

Sub-Case 3a: $i_{b+1}<i_b<j_b$ . Then

$$ \begin{align*}T\triangleleft T'=\left(i_{b+1},j_b\right)\cdot T \ \text{and} \ T\triangleleft T''=\left(i_{b+1},i_b\right)\cdot T,\end{align*} $$

and again one can easily check that

$$ \begin{align*}F_T(d)=F_{T'}(d)+F_{T''}(d)= G\cdot\left((x_{i_b}-x_{i_{b+1}})(x_{j_{b}}-x_{j_{b+1}})+(x_{i_{b+1}}-x_{j_b})(x_{i_b}-x_{j_{b+1}})\right),\end{align*} $$

and our inductive hypothesis applies.

Sub-Case 3b: $i_b<i_{b+1}<j_{b+1}<j_b$ . Then

$$ \begin{align*}T\triangleleft T'=\left(i_{b+1},j_b\right)\cdot T \ \text{and} \ T\triangleleft T''=\left(j_{b+1},j_b\right)\cdot T,\end{align*} $$

and again one can easily check that

$$ \begin{align*}F_T(d)=F_{T'}(d)+F_{T''}(d)= G\cdot\left((x_{i_b}-x_{i_{b+1}})(x_{j_{b}}-x_{j_{b+1}})+(x_{i_{b}}-x_{j_{b+1}})(x_{i_{b+1}}-x_{j_{b}})\right),\end{align*} $$

and our inductive hypothesis applies.

Case 4: $a=n-d$ . Then

and we have the descent $j_{k}>i_k>i_{k+1}$ . Then by Lemma 2.4, we have

$$ \begin{align*}T\triangleleft T'=\left(i_{k+1},i_k\right)\cdot T \ \text{and} \ T\triangleleft T''=\left(i_{k+1},j_k\right)\cdot T.\end{align*} $$

and again one can check that

$$ \begin{align*}F_T(d)=F_{T'}(d)+F_{T''}(d)=G\left(x_{i_k}\left(x_{i_{k+1}} -x_{j_k}\right) +x_{j_k}\left(x_{i_k}-x_{i_{k+1}}\right)\right),\end{align*} $$

to which the inductive hypothesis once again implies.

Case 5: $n-d+1\leq a\leq n-k-1$ . In this case, set $b=a-(n-k-d)$ so that we have

with the descent $i_b>i_{b+1}$ . Hence by Lemma 2.4, we have

$$ \begin{align*}T\triangleleft T'=\left(i_{b+1},i_b\right)\cdot T,\end{align*} $$

and in this case we clearly have

$$ \begin{align*}F_T(d)=F_{T'}(d),\end{align*} $$

to which the induction hypothesis applies again.

Therefore, in all cases, we have shown that $F_T(d)$ is a linear combination of shifted Specht polynomials indexed by tableaux on the shifted shape $\lambda (d)$ which dominate T. Therefore, by induction, the d-standard shifted Specht polynomials

$$ \begin{align*}\left\{F_T(d)|T\in\operatorname{STab}(\lambda(d))\right\},\end{align*} $$

must span $V(n,k,d)$ .

Proof of Theorem 2.2. By Lemma 2.3, the polynomials in the set

$$ \begin{align*}\left\{F_T(d) \ | \ T\in\operatorname{STab}(\lambda(d))\right\},\end{align*} $$

are linearly independent, and by Lemma 2.5, they generate the shifted Specht module $V(n,k,d)$ , and hence they must form a basis.

Proof. With $T\in \operatorname {STab}(n,k,k)$ and $T'\in \operatorname {STab}(n-1,k-1,k)$ as above, we compute

$$ \begin{align*}F_T=\prod_{t=1}^k(x_{i_t}-x_{j_t})=\prod_{t=1}^{k-1}(x_{i_t}-x_{j_t})\cdot (x_{i_k}-x_n)=\prod_{t=1}^{k-1}(y_{j_t}-y_{i_t})\cdot (-y_{i_k})=(-1)^k\cdot \Phi\left(F_{T'}\right),\end{align*} $$

and hence by Theorem 2.2, the first statement follows. To see the second statement note that the ring automorphism $\Phi \colon R\rightarrow R$ becomes the negative identity map on the quotient $\bar {\Phi }=-I\colon R/(x_n)\rightarrow R/(x_n)$ , and hence

$$ \begin{align*}\mathfrak{a}(n,k,k)+(x_n)=\mathfrak{a}(n-1,k-1,k)+(x_n),\end{align*} $$

which is the second statement.

Proof. We prove the formula by induction on n, the base case $n=1$ ( $k=0$ , $d=1$ ) being trivial. For the inductive step, assume that the formula holds for $n-1$ , that is,

$$ \begin{align*}\dim_{{\mathbb F}}\left(V(n-1,j,e)\right)=\binom{n-1}{e}-\binom{n-1}{j},\end{align*} $$

for all integers $j,e$ satisfying $1\leq j\leq e\leq n-1-j$ . Then fix integers $k,d$ satisfying $1\leq k\leq d\leq n-k$ . If $d>k$ , then by removing the box containing n, we get a bijection of indexing sets $\operatorname {STab}(n,k,d)\rightarrow \operatorname {STab}(n-1,k,d-1)\sqcap \operatorname {STab}(n-1,k-1,d)$ , and hence we find that

$$ \begin{align*} \dim_{{\mathbb F}}\left(V(n,k,d)\right)&= \dim_{{\mathbb F}}\left(V(n-1,k,d-1)\right)+\dim_{{\mathbb F}}\left(V(n-1,k-1,d)\right)\\ &= \left(\binom{n-1}{d-1}-\binom{n-1}{k-1}\right)+\left(\binom{n-1}{d}-\binom{n-1}{k-2}\right)\\ &= \binom{n}{d}-\binom{n}{k-1}, \end{align*} $$

where the last equality is Pascal’s identity. If $d=k$ , then by Corollary 2.7, we have

$$ \begin{align*} \dim_{{\mathbb F}}\left(V(n,k,k)\right)&= \dim_{{\mathbb F}}\left(V(n-1,k-1,k)\right)=\binom{n-1}{k}-\binom{n-1}{k-2}\\ &= \left(\binom{n-1}{k}+\binom{n-1}{k-1}-\binom{n-1}{k-1}-\binom{n-1}{k-2}\right)=\binom{n}{k}-\binom{n}{k-1}, \end{align*} $$

again by Pascal’s identity.

Proof. (1) $\Rightarrow $ (2). Assume that $p=0$ or $p\geq k+1$ . For each index j, $2k\leq j\leq n$ , define the nested chain of monomial complete intersections

$$ \begin{align*}C_{2k}\subset \cdots \subset C_n, \ \ \text{where} \ \ C_j=\frac{{\mathbb F}[x_1,\ldots,x_j]}{(x_1^2,\ldots,x_j^2)}. \end{align*} $$

By Lemma 3.3, the first monomial complete intersection $C_{2k}$ has the weak Lefschetz property, and in particular the derivative map

$$ \begin{align*}D_{C_{2k}}\colon \left(C_{2k}\right)_i\rightarrow \left(C_{2k}\right)_{i-1},\end{align*} $$

is surjective for all $1\leq i\leq k$ . Inductively for $j>2k$ , note that for each $0\leq i\leq k$ , we have direct sum decomposition of graded vector spaces

$$ \begin{align*}\left(C_{j}\right)_i\cong \left(C_{j-1}\right)_i\oplus x_j\cdot \left(C_{j-1}\right)_{i-1},\end{align*} $$

and in particular the derivative map in degree k decomposes into block triangular form

Since the maps

$$ \begin{align*}D_{C_{j-1}}\colon\left(C_{j-1}\right)_i\rightarrow\left(C_{j-1}\right)_{i-1}\end{align*} $$

is surjective for all $1\leq i\leq k$ , it follows that the maps

$$ \begin{align*}D_{C_j}\colon\left(C_j\right)_i\rightarrow \left(C_j\right)_{i-1}\end{align*} $$

is also surjective for all $1\leq i\leq k$ . In particular, this argument shows that the derivative maps $D\colon A_i\rightarrow A_{i-1}$ must be surjective for all $1\leq i\leq k$ as well.

(2) $\Rightarrow $ (3) Obvious.

(3) $\Rightarrow $ (1) Assume that $0<p<k+1$ . Then setting $i=p\leq k$ , we claim that $D\colon A_p\colon A_{p-1}$ cannot be surjective. Indeed, set $\alpha \in A_p$ to be the pth-elementary symmetric polynomial in the first $2p-1\leq n-1$ variables:

$$ \begin{align*}\alpha=e_p(x_1,\ldots,x_{2p-1})\in A_p.\end{align*} $$

Note first that over a field of characteristic p, we have

$$ \begin{align*}D(\alpha)=p\cdot e_{p-1}(x_1,\ldots,x_{2p-1})\equiv 0.\end{align*} $$

On the other hand, we have

$$ \begin{align*}\alpha(1,\ldots,1)=\binom{2p-1}{p}=\frac{(2p-1)\cdots(p+1)}{(p-1)!}\neq 0.\end{align*} $$

This shows that $\alpha \in P_p=\ker (D)\cap A_p$ , but $\alpha \not \in V(n,p,p)$ (since every polynomial in $V(n,p,p)$ necessarily vanishes at any point in which at least $p+1$ -entries are equal). Therefore, by Lemma 3.2, $D\colon A_p\rightarrow A_{p-1}$ is not surjective, and hence by [Reference Migliore, Miró-Roig and Nagel11, Prop. 2.1], $D\colon A_k\rightarrow A_{k-1}$ cannot be surjective either.

Proof. Containment $V(j,k,k)\subseteq V(n,k,k)\cap B_k$ is clear. For the reverse containment, suppose that $\alpha \in V(n,k,k)\cap B_k$ . The key observation here is that the restriction of the derivative map on A, call it $D_A$ , to the subspace $B\subset A$ is the same as $D_B$ . Since $\alpha \in V(n,k,k)\subseteq P_{A,k}=\ker (D_A)\cap A_k$ , it follows that $D_A(\alpha )=0$ , and hence also $D_B(\alpha )=0$ . This means that $\alpha \in P_{B,k}$ . From the proof of Lemma 3.4, we deduce that $D_B\colon B_k\rightarrow B_{k-1}$ is surjective, which by Lemma 3.2 implies that $\alpha \in V(j,k,k)$ , as desired.

Proof. (1) $\Rightarrow $ (2). This is due to Ikeda; see [Reference Hochster and Eagon5, Prop. 3.66].

(2) $\Rightarrow $ (1). Note that if $p\leq n$ , then we must have $L^p=0$ , and $(A,L)$ cannot be strong Lefschetz.

(2) $\Rightarrow $ (3). If $(A,L)$ is strong Lefschetz then the map

$$ \begin{align*}L^{n-k-d+1}\colon A_d\rightarrow A_{n-k+1},\end{align*} $$

must have maximal rank, and hence we must have

$$ \begin{align*}\dim(\ker(L^{n-k-d+1})\cap A_d)=\dim(A_d)-\dim(A_{n-k+1})=\binom{n}{d}-\binom{n}{k-1}=\dim(V(n,k,d)).\end{align*} $$

But since we already have containment $V(n,k,d)\subseteq \ker (L^{n-k-d+1})\cap A_d$ it must be equality. It follows that $P_i=\ker (L^{n-2i+1})\cap A_i$ , and hence we have the containment

$$ \begin{align*}\bigoplus_{i=k}^dL^{d-i}(P_i)\subseteq \ker(L^{n-k-d+1})\cap A_d.\end{align*} $$

Since $(A,L)$ is strong Lefschetz it follows that for each i, $L^{d-i}(P_i)\cong P_i$ , and hence a simple dimension count reveals this containment must also be equality.

(3) $\Rightarrow $ (2). Assume (3) holds, and assume that (2) does not. Fix integer k satisfying $1\leq k\leq \left \lfloor \frac {n}{2}\right \rfloor $ and suppose that $\alpha \in \ker (L^{n-2k})\cap A_k$ . Then certainly $\alpha \in \ker (A^{n-2k+1})\cap A_k$ hence $\alpha \in P_k$ by (3). But also according to (3) we have

$$ \begin{align*}A_{n-k}=V(n,0,n-k)=\bigoplus_{i=0}^{n-k}L^{n-k-i}(P_i).\end{align*} $$

On the other hand, if $L^{n-2k}(\alpha )=0$ , then $\dim (L^{n-2k}(P_k))<P_k$ , and hence we must have

$$ \begin{align*}\dim\left(A_{n-k}\right)=\sum_{i=0}^{n-k}\dim\left(L^{n-k-i}(P_i)\right)<\sum_{i=0}^{n-k}P_i=\binom{n}{k},\end{align*} $$

which is a contradiction. Therefore, (2) must hold after all.

Proof of Theorem 4.4. Assuming that Theorem 4.3 holds, we prove Theorem 4.4 by induction on $n\geq 2$ . For the base case $n=2$ , the only possibilities for integers $k,d$ are $k=0$ and $d=1,2$ and $k=1=d$ . In the case $k=0$ , we have $\mathfrak {a}(2,0,d)=(x_1,x_2)^{\langle d\rangle }$ which is radical by Lemma 4.5. In the case, $k=d=1$ we have $\mathfrak {a}(2,1,1)=((x_1-x_2))$ is prime, and therefore radical. For the inductive step, assume that $\mathfrak {a}(n-1,j,e)$ is radical for all integers satisfying $0\leq j\leq e\leq n-1-j$ . Fix integers $k,d$ satisfying $1\leq k\leq d\leq n-k$ . First, we argue for the case $d=k$ . Let $I=\mathfrak {a}(n,k,k)$ and $x=x_n$ . Then by Corollary 2.7 it follows that we have

$$ \begin{align*}I+(x)=\mathfrak{a}(n,k,k)+(x_n)=\mathfrak{a}(n-1,k-1,k)+(x_n).\end{align*} $$

By induction hypothesis, $\mathfrak {a}(n-1,k-1,k)$ is radical, and since its generators are polynomials which are independent of $x_n$ , it follows that the sum $I+(x)$ is radical. Also using the change of coordinates map $\Phi \colon x_i\mapsto y_i$ in Corollary 2.7 we find that

$$ \begin{align*} \Phi\left(I:x\right)=\left(\Phi\left(I\right):\Phi(x_n)\right)=(\mathfrak{a}(n-1,k-1,k):x_n)=\mathfrak{a}(n-1,k-1,k)=\Phi\left(I\right), \end{align*} $$

from which it follows that $(I:x)=I$ . Therefore, it follows from Lemma 4.1 that $I=\mathfrak {a}(n,k,k)$ itself must be radical.

For $d>k$ , we appeal again to Theorem 4.3:

$$ \begin{align*}\mathfrak{a}(n,k,d)=\mathfrak{a}(n,k,d-1)\cap (x_1,\ldots,x_n)^{\langle d\rangle}=\mathfrak{a}(n,k,k)\cap (x_1,\ldots,x_n)^{\langle d\rangle}.\end{align*} $$

Since $\mathfrak {a}(n,k,k)$ is radical, and $(x_1,\ldots ,x_n)^{\langle d\rangle }$ is radical, it follows that $\mathfrak {a}(n,k,d)$ is radical too.

Proof. It is not difficult to see that every polynomial $P\in \mathfrak {a}(n,k,d-1)$ decomposes into a sum of terms of the form

$$ \begin{align*}P=\sum_{\mathbf{a}\in{\mathbb N}^{n}}\mathbf{x}^{\mathbf{a}}\cdot \ \nu_{\mathbf{a}},\end{align*} $$

where $\nu _{\mathbf {a}}\in V(n,k,d-1)$ . We want to show that if $P\in (x_1,\ldots ,x_n)^{\langle d\rangle }$ , then each of its summands are too, that is, $\mathbf {x}^{\mathbf {a}}\cdot \nu _{\mathbf {a}}\in (x_1,\ldots ,x_n)^{\langle d\rangle }$ for all $\mathbf {a}\in {\mathbb N}^n$ . Suppose by way of contradiction that for some $\mathbf {a}\in {\mathbb N}^n$ and some $\nu _{\mathbf {a}}\in V(n,k,d-1)$ that $\mathbf {x}^{\mathbf {a}}\cdot \nu _{\mathbf {a}}\notin (x_1,\ldots ,x_n)^{\langle d\rangle }$ . Since $(x_1,\ldots ,x_n)^{\langle d\rangle }$ is a monomial ideal, there must be some monomial $\mathbf {x}^{\mathbf {b}}$ which appears in the monomial expansion of $\mathbf {x}^{\mathbf {a}}\cdot \nu _{\mathbf {a}}$ such that $\mathbf {x}^{\mathbf {b}}\notin (x_1,\ldots ,x_n)^{\langle d\rangle }$ . Define the weight of monomial $\mathbf {x}^{\mathbf {b}}$ as $\operatorname {wt}(\mathbf {x}^{\mathbf {b}})=\#\{b_i>0\}$ . Since $(x_1,\ldots ,x_n)^{\langle d\rangle }$ consists of all monomials of weight at least d, it follows that $\operatorname {wt}(\mathbf {x}^{\mathbf {b}})\leq d-1$ . On the other hand, since $\nu _{\mathbf {a}}$ is a linear combination of shifted Specht polynomials of type $\lambda (d)$ , it follows that every monomial in the monomial expansion of $\nu _{\mathbf {a}}$ also has weight $d-1$ . This implies that $\operatorname {wt}(\mathbf {x}^{\mathbf {b}})=d-1$ , and therefore that

$$ \begin{align*}\frac{\mathbf{x}^{\mathbf{b}}}{\sqrt{\mathbf{x}^{\mathbf{b}}}} = \mathbf{x}^{\mathbf{a}}.\end{align*} $$

In particular, we see that the monomial $\mathbf {x}^{\mathbf {b}}$ is unique to the term $\mathbf {x}^{\mathbf {a}}\cdot \nu _{\mathbf {a}}$ , and hence must occur with the same coefficient in the monomial expansion of $\mathbf {x}^{\mathbf {a}}\cdot \nu _{\mathbf {a}}$ as it does in the monomial expansion of $P=\sum _{\mathbf {a}\in {\mathbb N}^n}\mathbf {x}^{\mathbf {a}}\cdot \nu _{\mathbf {a}}$ . Therefore $P\notin (x_1,\ldots ,x_n)^{\langle d\rangle }$ , as desired.

Proof. If there is a variable $x_i\in \operatorname {supp}(\mathbf {x}^{\mathbf {a}})$ which is not in $\operatorname {supp}(F_T(d-1))$ , then certainly $\mathbf {x}^{\mathbf {a}}\cdot F_T(d-1)\in \mathfrak {a}(n,k,d)$ . Otherwise, there must be two indices $i\neq j$ such that $x_i,x_j\in \operatorname {supp}(\mathbf {x}^{\mathbf {a}})$ and $i,j$ in the same column of T. Choose any index $r\neq i,j$ such that $x_r\notin \operatorname {supp}(F_T(d-1))$ , which exists since $0\leq k\leq d-1<d\leq n-k$ , and let $(i,r), (j,r)\in \mathfrak {S}_n$ be the transpositions swapping $i,r$ and $j,r$ , respectively. Then we have

$$ \begin{align*}F_T(d-1)=F_{(i,r).T}(d-1)+F_{(j,r).T}(d-1),\end{align*} $$

and since $x_i\cdot F_{(i,r).T}(d-1), x_j\cdot F_{(j,r).T}(d-1)\in \mathfrak {a}(n,k,d)$ , it follows that

$$ \begin{align*}\mathbf{x}^{\mathbf{a}}\cdot F_T(d-1)\in\mathfrak{a}(n,k,d),\end{align*} $$

as desired.

Proof. We observe that

$$ \begin{align*}\mathbf{x}^{\mathbf{m}}\cdot \left(\nu-\mathbf{x}^{\mathbf{m}}\cdot \nu'\right)\in (x_1,\ldots,x_n)^{\langle d\rangle},\end{align*} $$

where $\nu '\in V([n]_m,k-m,d-1-m)$ is the image of $\nu $ in the map above. Indeed note that for each standard tableau $T\in \operatorname {STab}_m(n,k,d-1)$ as in (12), the support of the difference $F_T(d-1)-\mathbf {x}^{\mathbf {m}}\cdot F_{T'}(d-1-m)$ does not contain the monomial $\mathbf {x}^{\mathbf {m}}$ , hence the product $\mathbf {x}^{\mathbf {m}}\cdot \left (F_T(d)-\mathbf {x}^{\mathbf {m}}\cdot F_{T'}(d-1-m)\right )\in (x_1,\ldots ,x_n)^{\langle d\rangle }$ . Hence, if $\mathbf {x}^{\mathbf {m}}\cdot \nu \in (x_1,\ldots ,x_n)^{\langle d\rangle }$ , it follows that $\left (\mathbf {x}^{\mathbf {m}}\right )^2\cdot \nu '\in (x_1,\ldots ,x_n)^{\langle d\rangle }$ , and hence that $\nu '\in \left ((x_1,\ldots ,x_n)^{\langle d\rangle }:\mathbf {x}^{\mathbf {m}}\right )=(x_{m+1},\ldots ,x_n)^{(d-m)}$ . For degree reasons this implies that $\nu '=0$ , and hence by Corollary 2.8, $\nu =0$ as well.

Proof of Theorem 4.3. The containment $\mathfrak {a}(n,k,d)\subset \mathfrak {a}(n,k,d-1)\cap (x_1,\ldots ,x_n)^{\langle d\rangle }$ is clear. For the reverse containment, Lemma 4.5 implies that it suffices to check it on products of monomials with polynomials in $V(n,k,d-1)$ . Since $(x_1,\ldots ,x_n)^{\langle d\rangle }$ is generated by square-free monomials we may assume that our monomials are square-free, and by symmetry, we may assume that our monomial has the form $\mathbf {x}^{\mathbf {m}}=x_1\cdots x_m$ for some integer $1\leq m\leq d$ . Then as above we have

$$ \begin{align*}V(n,k,d-1)=V_m(n,k,d-1)\oplus V^m(n,k,d-1),\end{align*} $$

and by Lemma 4.7, $\mathbf {x}^{\mathbf {m}}\cdot \nu \in \mathfrak {a}(n,k,d)$ if $\nu \in V_m(n,k,d-1)$ . Furthermore, Lemma 4.6 implies that for $\nu \in V^m(n,k,d-1)$ if $\mathbf {x}^{\mathbf {m}}\cdot \nu \in (x_1,\ldots ,x_n)^{\langle d\rangle }$ then $\mathbf {x}^{\mathbf {m}}\cdot \nu \in \mathfrak {a}(n,k,d)$ , and the result follows.

Proof. By linearity of D it suffices to assume that $P^j$ is a square-free monomial, and by symmetry, we may assume is $P^j(x_1,\ldots ,x_{n-1})=x_1\cdots x_j$ . Then we have

$$ \begin{align*} P^j(x_1,\ldots,x_{n-1})&= x_1\cdots x_j= (x_n-y_1)\cdots(x_n-y_j)\\ &= x_n^2\cdot \left(\text{stuff}\right)+(-1)^{j-1}\cdot x_n\left(\sum_{i=1}^j y_1\cdots \hat{y_i}\cdots y_j\right)+(-1)^j\cdot y_1\cdots y_j\\ &\quad (\text{where }\hat{y}\text{ means omission})\\ &= (-1)^j\left(y_1\cdots y_j - x_nD\left(y_1\cdots y_j\right)\right)+x_n^2\cdot \left(\text{stuff}\right)\\ &\equiv (-1)^j\left(P^j(y_1,\ldots,y_{n-1})-x_nD\left(P^j(y_1,\ldots,y_{n-1})\right)\right) & \!\!\!\! \text{mod} \ (x_n^2), \end{align*} $$

as claimed. The second statement follows from the first since $D(\alpha )=0$ .

Proof. We have

$$ \begin{align*} J(n,k)+(x_n)&= I(n-1,k-1)+(y_1,\ldots,y_{n-1})^{\langle k\rangle}+(x_{n}^2)+(x_n)\\ &= \mathfrak{a}(n-1,k-1,k-1)+(x_1,\ldots,x_{n-1})^{\langle k\rangle}+(y_1,\ldots,y_{n-1})^{\langle k\rangle}+(x_n)\\ &= \mathfrak{a}(n-1,k-1,k-1)+(x_1,\ldots,x_{n-1})^{\langle k\rangle}+(x_n)\\ &= I(n-1,k-1)+(x_n). \end{align*} $$

For the other equality, note first that containment $(J(n,k):x_n)\supseteq (x_1,\ldots ,x_{n-1})^{\langle k-1\rangle }+(x_n)=(y_1,\ldots ,y_{n-1})^{\langle k-1\rangle }+(x_n)$ follows from Lemma 5.7. Indeed identifying the space of square-free polynomials with the monomial complete intersection $B={\mathbb F}[x_1,\ldots ,x_{n-1}]/(y_1^2,\ldots ,y_{n-1}^2)$ , Lemma 3.4 implies that the derivative map $D\colon B_{k}\rightarrow B_{k-1}$ is surjective, and hence for any square-free monomial of degree $k-1$ in variables $y_1,\ldots ,y_{n-1}$ , say $P=y_1\cdots y_{k-1}$ , we know by Lemma 3.4 there is square-free polynomial of degree k for which

$$ \begin{align*}D(Q(y_1,\ldots,y_{n-1}))=P(y_1,\ldots,y_{n-1})=y_1\cdots y_{k-1}.\end{align*} $$

Then Lemma 5.7 implies that

$$ \begin{align*} P\cdot x_n&= y_1\cdots y_{k-1}\cdot x_n=x_n\cdot D\left(Q(y_1,\ldots,y_{n-1})\right)\\ &\equiv Q(x_1,\ldots,x_{n-1})\pm Q(y_1,\ldots,y_{n-1}) & \text{mod} \ (x_n^2). \end{align*} $$

Since $Q(x_1,\ldots ,x_{n-1})\in I(n-1,k-1)$ and $Q(y_1,\ldots ,y_{n-1})\in (y_1,\ldots ,y_{n-1})^{\langle k\rangle }$ , it follows that $x_n\cdot P=x_n\cdot D(Q(y_1,\ldots ,y_{n-1}))\in J(n,k)$ , and hence $P=y_1\cdots y_{k-1}\in (J(n,k):x_n)$ . For the reverse containment, fix $G\in (J(n,k):x_n)$ . Then for each $S\in \operatorname {STab}(n-1,0,k)$ there exists polynomials $d_S\in R$ for which

$$ \begin{align*}x_n G-\sum_{S\in \operatorname{STab}(n-1,0,k)}d_SM_S\in I(n-1,k-1)+(x_n^2),\end{align*} $$

where $M_S=M_S(y_1,\ldots ,y_{n-1})\in (y_1,\ldots ,y_{n-1})^{\langle k\rangle }$ are square-free monomials of degree k in the y-variables. By Lemma 5.7, we have for each $S\in \operatorname {STab}(n-1,0,k)$

$$ \begin{align*} M_S(y_1,\ldots,y_{n-1})\equiv & \pm \left(M_S(x_1,\ldots,x_{n-1})-x_n D(M_S(x_1,\ldots,x_{n-1}))\right) & \text{mod} \ (x_n^2), \end{align*} $$

and since $M_S(x_1,\ldots ,x_{n-1})\in (x_1,\ldots ,x_{n-1})^{\langle k\rangle }\subset I(n-1,k-1)$ , we see that

$$ \begin{align*} x_nG-\sum_{S\in \operatorname{STab}(n-1,0,k)}d_SM_S(y_1,\ldots,y_{n-1}) &\equiv x_nG- \sum_{S\in \operatorname{STab}(n-1,0,k)}d_Sx_nD(M_S(y_1,\ldots,y_{n-1}))\\ &\equiv 0 \ \ \text{mod} \ I(n-1,k-1)+(x_n^2). \end{align*} $$

Since $x_n$ is a nonzero divisor for $I(n-1,k-1)$ , it follows that

$$ \begin{align*} &G-\sum_{S\in \operatorname{STab}(n-1,0,k)}d_SD(M_S(x_1,\ldots,x_{n-1}))\in I(n-1,k-1) \\ &\quad +(x_n)\subset (x_1,\ldots,x_{n-1})^{\langle k-1\rangle}+(x_n),\end{align*} $$

and since $D(M_S(x_1,\ldots ,x_{n-1}))\in (x_1,\ldots ,x_{n-1})^{\langle k-1\rangle }$ for all $S\in \operatorname {STab}(n-1,0,k)$ , we see also that $G\in (x_1,\ldots ,x_{n-1})^{\langle k-1\rangle }+(x_n)$ , as desired.

Proof of (1) $\Rightarrow $ (2) in Theorem 5.5(4)

Assume that $p=0$ or $p\geq k+1$ . We prove by induction on $n\geq 3$ for each integer k satisfying $1\leq k<k+1\leq n-k$ then the ideal $I(n,k)$ is perfect. The base case is $n=3$ where the only possible k value is $k=1$ . Here, the assumption on p is vacuous, and we have

$$ \begin{align*} I(3,1)= & \mathfrak{a}(3,1,1)+(x_1,x_2,x_3)^{\langle 2\rangle}=(x_1-x_3,x_2-x_3,x_1x_2,x_1x_3,x_2x_3). \end{align*} $$

Note that ${\mathbb F}[x_1,x_2,x_3]/I(3,1)\cong {\mathbb F}[z]/(z^2)$ is Cohen–Macaulay, which implies that $I(3,1)$ is perfect of grade $n-k+1=3$ .

For the inductive step, assume that $I(n-1,j)$ is perfect for all integers j satisfying $1\leq j<j+1\leq n-1-j$ . Fix k satisfying $1\leq k<k+1\leq n-k$ , and consider the Specht-monomial ideal

$$ \begin{align*} I(n,k)= \mathfrak{a}(n,k,k)+(x_1,\ldots,x_n)^{\langle k+1\rangle}. \end{align*} $$

Consider the sum $J(n,k)$ as in (16), that is,

$$ \begin{align*}J(n,k)=I(n-1,k-1)+(y_1,\ldots,y_{n-1})^{\langle k\rangle}+(x_n^2).\end{align*} $$

Note that $I(n-1,k-1)$ is perfect by the induction hypothesis. Also $(y_1,\ldots ,y_{n-1})^{\langle k\rangle }+(x_n^2)$ is perfect since $(y_1,\ldots ,y_{n-1})^{\langle k\rangle }$ is perfect (by Lemma 5.2), and $x_n^2$ is $(y_1,\ldots ,y_{n-1})^{\langle k\rangle }$ -regular. Therefore, by Lemma 5.3, $I(n,k)$ is perfect if and only if $J(n,k)$ is perfect. But according to Lemma 5.8 $J(n,k)+(x_n)=I(n-1,k-1)+(x_n)$ and $(J(n,k):x_n)=(x_1,\ldots ,x_{n-1})^{\langle k-1\rangle }+(x_n)$ which are both perfect of the same grade $g=n-k+2$ , and hence by Lemma 5.2, it follows that $J(n,k)$ is also perfect (of grade $g=n-k+2$ ). Thus it follows that $I(n,k)$ is perfect completing the induction step, and hence the proof.

Proof. Assume that $I(n,k)$ is perfect, and consider the intersection of primary ideals as in (17), that is,

$$ \begin{align*}I'(n,k)=\bigcap_{\sigma\in \mathfrak{S}_n}Q_{\sigma},\end{align*} $$

with minimal associated prime divisors given by

$$ \begin{align*}P_\sigma=\sqrt{Q_{\sigma}}=\sigma.\left(x_1,\ldots,x_{n-k+1}\right).\end{align*} $$

We would like to show that $I(n,k)=I'(n,k)$ . Note first that they have the same radical:

(18) $$ \begin{align} \sqrt{I'(n,k)}=\bigcap_{\sigma\in\mathfrak{S}_n}P_{\sigma}=(x_1,\ldots,x_n)^{\langle k\rangle}=\sqrt{I(n,k)}. \end{align} $$

For the last equality, the containment $I(n,k)\subseteq (x_1,\ldots ,x_n)^{\langle k\rangle }$ is clear, and the other containment follows from the containment:

$$ \begin{align*}x_{i_1}^2\cdots x_{i_k}^2=x_{i_1}\cdots x_{i_k}\cdot F_T+\left(\text{stuff in} \ (x_1,\ldots,x_n)^{\langle k+1\rangle}\right)\in I(n,k),\end{align*} $$

where

It follows that $\{P_\sigma \ | \ \sigma \in \mathfrak {S}_n\}$ is a complete list of minimal prime divisors of $I(n,k)$ , which implies Equation (18).

Set $I'=I'(n,k)$ and $I=I(n,k)$ . Suppose that a primary decomposition of I is given by $I=U_1\cap \cdots \cap U_m$ . Since I is perfect, all of its associated prime divisors must be minimal, and hence the primary components must be indexed by the symmetric group and we can write

$$ \begin{align*}I=\bigcap_{\sigma\in \mathfrak{S}_n}U_{\sigma},\end{align*} $$

where $\sqrt {U_\sigma }=P_{\sigma }$ . Fix $\sigma \in \mathfrak {S}_n$ , and set $U=U_\sigma $ , $P=P_\sigma $ , and $Q=Q_\sigma $ . We want to show that $U_\sigma =Q_\sigma $ . Let $R_P$ be the polynomial ring $R={\mathbb F}[x_1,\ldots ,x_n]$ localized at the prime ideal P. By a theorem of Nagata [Reference Nagata12, Th. 8.7] we have

$$ \begin{align*}U=IR_P\cap R \quad \text{and} \quad Q=JR_P\cap R.\end{align*} $$

We prove that in the local ring $R_P$ , the ideals are equal $IR_P=JR_P$ . Certainly because of the containment $I\subseteq J$ , we have also $IR_P\subseteq JR_P$ . In the other direction, we observe that $JR_P=QR_P$ . Without loss of generality we may assume that $\sigma =e$ , and $Q=(x_1-x_2,\ldots ,x_{1}-x_{n-k+1},x_1^2)$ . For each pair $1\leq r<s\leq n-k+1$ we may choose $1\leq i_1<\cdots <i_{k-1}\leq n-k+1$ such that $r,s\notin \left \{i_1,\ldots ,i_{k-1}\right \}$ and setting $j_i=n-k+1+i$ define the tableau

Then in the local ring $R_P$ the Specht polynomial $F_T\in I$ has the form $F_T=u\cdot (x_r-x_s)$ where $u\in R_P$ is a unit. Also consider the monomial

$$ \begin{align*}M_S=x_{n-k+2}\cdots x_n\cdot x_r\cdot x_s\in (x_1,\ldots,x_n)^{\langle k\rangle}\subset I.\end{align*} $$

Then in the local ring $R_P$ it has the form $M=w\cdot x_r\cdot x_s$ where $w\in R_P$ is a unit. Therefore, the generators of $QR_P$ satisfy $(x_r-x_s)\in IR_P$ and $x_r\cdot x_s\in IR_P$ it follows that $QR_P\subset IR_P$ and hence that $JR_P\subset IR_P$ , as desired.

Proof of (2) $\Rightarrow $ (1) in Theorem 5.5(4)

Assume that p satisfies $0<p<k+1$ . We show that $I(n,k)$ is not perfect by showing it does not satisfy Decomposition (17), or equivalently that

(19) $$ \begin{align} & \mathop{\underbrace{\mathfrak{a}(n,k,k)+(x_1,\ldots,x_n)^{\langle k+1\rangle}}}_{I(n,k)} \nonumber \\ &\quad \neq \mathop{\underbrace{\mathfrak{a}(n-1,k-1,k-1)+(x_1,\ldots,x_{n-1})^{\langle k\rangle}}}_{I(n-1,k-1)}\cap\left((y_1,\ldots,y_{n-1})^{\langle k\rangle}+(x_{n}^2)\right). \end{align} $$

We identify the space of square-free x-monomials with the monomial complete intersection $B={\mathbb F}[x_1,\ldots ,x_{n-1}]/(x_1^2,\ldots ,x_{n-1}^2)$ and $D=D_B=\frac {\partial }{\partial x_1}+\cdots +\frac {\partial }{\partial x_{n-1}}$ the associated lowering operator. From our assumptions on p, Lemma 3.4 implies that the derivative map $D\colon B_k\rightarrow B_{k-1}$ is not surjective, and hence (by Lemma 3.2), $V(n-1,k,k)=\ker (D)\cap B_k$ . In particular, there must exist a square-free polynomial of degree k, say $f=f(x_1,\ldots ,x_{n-1})\in (x_1,\ldots ,x_{n-1})^{\langle k\rangle }$ with the property that $D(f)=0$ but $f\notin V(n-1,k,k)$ . Since $D(f)=0$ , we deduce from Lemma 5.7 that

$$ \begin{align*}f(x_1,\ldots,x_{n-1})\equiv f(y_1,\ldots,y_{n-1}) \equiv 0\ \ \text{mod} \ (y_1,\ldots,y_{n-1})^{\langle k\rangle}+ (x_n^2).\end{align*} $$

Therefore, $f\in I(n-1,k-1)\cap \left ((y_1,\ldots ,y_{n-1})^{\langle k\rangle }+(x_{n}^2)\right )$ . On the other hand, since $f\notin V(n-1,k,k)$ it follows from Lemma 3.5 that $f\notin V(n,k,k)$ either, and it follows that $f\notin I(n,k)$ . This shows that Inequality (19) holds, and hence by Lemma 5.9, the Specht-monomial ideal $I(n,k)$ is not perfect.

Proof of Theorem 5.6. First assume that $p=0$ or $p\geq k+1$ . We prove by induction on $n\geq 3$ that for each integer k satisfying $1\leq k<k+1\leq n-k$ , and each i satisfying $1\leq i\leq k$ the Specht ideal $\mathfrak {a}(n+1,i+1,i+1)$ is perfect. First, recall that Corollary 2.7 says the change of coordinates map $\Phi $ gives a ring isomorphism

$$ \begin{align*}\mathfrak{a}(n+1,i+1,i+1)\cong \mathfrak{a}(n,i,i+1)=\mathfrak{a}(n,i,i)\cap (x_1,\ldots,x_n)^{\langle i+1\rangle},\end{align*} $$

where the second equality follows from Theorem 4.3.

For the base case $n=3$ and the only possible $k=1$ gives

$$ \begin{align*}\mathfrak{a}(4,2,2)\cong \mathfrak{a}(3,1,2)=\mathfrak{a}(3,1,1)\cap (x_1,x_2,x_3)^{\langle 2\rangle}.\end{align*} $$

Note that $\mathfrak {a}(3,1,1)$ and $(x_1,x_2,x_3)^{\langle 2\rangle }$ are both perfect of grade $g=2$ Also note that the Specht-monomial ideal $I(3,1)=\mathfrak {a}(3,1,1)+(x_1,x_2,x_3)^{\langle 2\rangle }$ has grade $g+1=3$ and its quotient

$$ \begin{align*}\frac{{\mathbb F}[x_1,x_2,x_3]}{I(3,1)}\cong\frac{{\mathbb F}[z]}{(z^2)}\end{align*} $$

is Cohen–Macaulay. Therefore, $I(3,1)$ is perfect and hence by Lemma 5.3, so is $\mathfrak {a}(4,2,2)$ .

For the inductive step, assume that $\mathfrak {a}(n,i,i)$ is perfect for every $1\leq i\leq k$ . We have

$$ \begin{align*}\mathfrak{a}(n+1,i+1,i+1)\cong \mathfrak{a}(n,i,i)\cap (x_1,\ldots,x_n)^{\langle i+1\rangle}.\end{align*} $$

Also note that since $1\leq i<i+1\leq k+1\leq n-k\leq n-i$ , Theorem 5.5(4) implies that the Specht-monomial ideal

$$ \begin{align*}I(n,i)=\mathfrak{a}(n,i,i)+(x_1,\ldots,x_n)^{\langle i+1\rangle}\end{align*} $$

is perfect of grade $g+1=n-i+1$ . Since $\mathfrak {a}(n,i,i)$ and $(x_1,\ldots ,x_n)^{\langle i+1\rangle }$ are both perfect of grade $g=n-i$ . Then it follows from Lemma 5.3 that $\mathfrak {a}(n+1,i+1,i+1)\cong \mathfrak {a}(n,i,i)\cap (x_1,\ldots ,x_n)^{\langle i+1\rangle }$ is also perfect. This proves implication (1).

For (2), assume that $n\geq 2p+1$ . By (1), we see that $\mathfrak {a}(n,p,p)$ is perfect. Also $(x_1,\ldots ,x_n)^{\langle p\rangle }$ is perfect by Lemma 5.2. Therefore, by Lemma 5.3, it follows that the Specht ideal $\mathfrak {a}(n+1,p+1,p+1)$ and the Specht-monomial ideal $I(n,p)$ are perfect or not, alike. But Theorem 5.5(4) implies that the Specht-monomial ideal

$$ \begin{align*}I(n,p)=\mathfrak{a}(n,p,p)+(x_1,\ldots,x_n)^{\langle p+1 \rangle}\end{align*} $$

is not perfect, and hence the Specht-monomial ideal $\mathfrak {a}(n+1,p+1,p+1)$ is not perfect either.

Proof. Certainly, we have that $\mathfrak {a}(n,k,k)\subset \mathfrak {a}(n-1,k,k)\subset \mathfrak {a}(n-1,k-1,k-1)$ , and also $(x_1,\ldots x_n)^{\langle k+1\rangle }\subset (x_1,\ldots ,x_{n-1})^{\langle k\rangle }$ , hence $I(n,k)\subset I(n-1,k-1)$ (if $k=1$ , we should regard $I(n-1,0)$ as R). It remains to see why $I(n,k)\subset (y_1,\ldots ,y_{n-1})^{\langle k\rangle }+(x_{n}^2)$ . For $T\in \operatorname {STab}(n,k,k)$ we have

$$ \begin{align*}F^k_T(x_1,\ldots,x_n) &=(x_{i_1}-x_{j_1})\cdots(x_{i_k}-x_n) \\ & =(y_{j_1}-y_{i_1})\cdots (y_{j_{k-1}}-y_{i_{k-1}})\cdot (-y_{i})\in (y_1,\ldots,y_{n-1})^{\langle k\rangle}+(x_{n}^2).\end{align*} $$

For $S\in \operatorname {STab}(n,0,k+1)$ , if $x_n\in \operatorname {supp}(M_S^{k+1})$ then $M_S^{k+1}=x_n\cdot M_{S'}^k$ for $S'\in \operatorname {STab}(n-1,0,k)$ and by Lemma 5.7, we have

$$ \begin{align*}M_{S'}^k(x_1,\ldots,x_{n-1})=M_{S'}^k(y_1,\ldots,y_{n-1})+x_n\cdot D(M_{S'}^k(y_1,\ldots,y_{n-1})) \ \ \text{mod} \ (y_1,\ldots,y_{n-1})^{\langle k\rangle}.\end{align*} $$

It follows that $M_S^{k+1}=x_n\cdot M_{S'}^k\in (y_1,\ldots ,y_{n-1})^{\langle k\rangle }+(x_{n}^2)$ . If $x_n\notin \operatorname {supp}(M_S^{k+1})$ , then it is obvious that $M_S^{k+1}\in (y_1,\ldots ,y_{n-1})^{\langle k\rangle }+(x_{n}^2)$ . Hence $I(n,k)\subseteq I(n-1,k-1)\cap \left ((y_1,\ldots ,y_{n-1})^{\langle k\rangle }+(x_{n}^2)\right )$ , as desired.

Proof. (1) is obvious from the definitions. For (2), note that for $S\in \operatorname {STab}(n-1,0,k-1)$ , $x_n\cdot M_S^k\in V(n,0,k+1)\subset I(n,k)$ . Also for $T\in \operatorname {STab}(n-1,k-1,k-1)$ and for index $1\leq i\leq n-1$ such that $i\notin \operatorname {supp}(T)$ , which exists because $n-1\geq 2k>2(k-1)$ , we have

$$ \begin{align*}x_n^2\cdot F_T^{k-1}=x_n(x_n-x_i)\cdot F_T^{k-1}+x_nx_i\cdot F_T^{k-1},\end{align*} $$

and since $(x_n-x_i)\cdot F_T^{k-1}\in V(n,k,k)$ and $x_nx_i\cdot F_T^{k-1}\in V(n,0,k+1)$ , it follows that $x_n^2\cdot F_T^{k-1}\in I(n,k)$ , and (2) follows. Finally fix a homogeneous polynomial $P\in I(n-1,k-1)$ . Then for each $T\in \operatorname {STab}(n-1,k-1,k-1)$ and each $S\in \operatorname {STab}(n-1,0,k)$ , there exist polynomials, which we may take in the y-variables, $g_T(\mathbf {y})$ of degree m and $h_S(\mathbf {y})$ of degree $m-1$ for which

$$ \begin{align*}P=\sum_{T \in\operatorname{STab}(n-1,k-1,k-1)}g_T(\mathbf{y})\cdot F_T^{k-1}+\sum_{S\in \operatorname{STab}(n-1,0,k)}h_S(\mathbf{y})\cdot M_S.\end{align*} $$

Writing $g_T(\mathbf {y})=g^0_T(y_1,\ldots ,y_{n-1})+x_ng^1_T(y_1,\ldots ,y_{n-1})+x_n^2g^2_T(\mathbf {y})$ and also $h_S(\mathbf {y})=h^0_S(y_1,\ldots ,y_{n-1})+x_nh^1_S(\mathbf {y})$ , it follows from (2) that $x_n^2g^2_T(\mathbf {y})\cdot F_T^{k-1}\in I(n,k)$ , $x_nh^1_S(\mathbf {y})\cdot M_S^k\in I(n,k)$ , and also that $x_ng^1_T(y_1,\ldots ,y_{n-1})\cdot F_T^{k-1}\in U$ . Then taking monomial expansions, reversing orders of summations, and grouping like monomial terms, we get

$$ \begin{align*} P&\equiv \sum_{T \in\operatorname{STab}(n-1,k-1,k-1)}\sum_{\mathbf{a}\in{\mathbb N}^{n-1}(m)}c_T^0(\mathbf{a})\mathbf{y}^{\mathbf{a}}\cdot F_T^{k-1}\\ &\quad +\sum_{T \in \operatorname{STab}(n-1,k-1,k-1)}\sum_{\mathbf{b}\in{\mathbb N}^n(m-1)}c_T^1(\mathbf{b})\mathbf{y}^{\mathbf{b}}\cdot x_nF_T^{k-1}\\ &\quad +\sum_{S\in \operatorname{STab}(n-1,0,k)}\sum_{\mathbf{b}\in{\mathbb N}^n(m-1)}d_S^0(\mathbf{b})\mathbf{y}^{\mathbf{b}}\cdot M_S\\ &= \sum_{\mathbf{a}\in{\mathbb N}^n(m)}\mathbf{y}^{\mathbf{a}}\cdot\left(\sum_{T \in \operatorname{STab}(n-1,k-1,k-1)}c_T^0(\mathbf{a})F_T^{k-1}\right) + \sum_{\mathbf{b}\in{\mathbb N}^{n-1}(m-1)}\mathbf{y}^{\mathbf{b}}\\ &\quad \cdot \left(\sum_{T \in \operatorname{STab}(n-1,k-1,k-1)}c_T^1(\mathbf{b})F_T^{k-1}+\sum_{S\in \operatorname{STab}(n-1,0,k)}d_S^0(\mathbf{b})M_S^k\right)\quad \text{mod} \ I(n,k), \end{align*} $$

and (3) follows.

Proof. By Lemma 5.14, we may write

$$ \begin{align*}P=\mathop{\underbrace{\sum_{\mathbf{a}\in{\mathbb N}^{n-1}(m)}\mathbf{y}^{\mathbf{a}}\nu_{\mathbf{a}}}}_{P_1} +\mathop{\underbrace{\sum_{\mathbf{b}\in{\mathbb N}^{n-1}(m-1)}\mathbf{y}^{\mathbf{b}}\mu_{\mathbf{b}}}}_{P_2}+P_3\end{align*} $$

for some $\nu _{\mathbf {a}}\in V$ , some $\mu _{\mathbf {b}}\in U$ , and some $P_3\in I(n,k)$ . Next note that if $P\in (y_1,\ldots ,y_{n-1})^{\langle k\rangle }+(x_{n}^2)$ then both $P_1\in (y_1,\ldots ,y_{n-1})^{\langle k\rangle }+(x_{n}^2)$ and $P_2\in (y_1,\ldots ,y_{n-1})^{\langle k\rangle }+(x_{n}^2)$ . Indeed, note that in their respective y-monomial expansions, those monomials in $P_1$ are all independent of $y_n$ , whereas all monomials in the y-monomial expansion of $P_2$ are either divisible by $y_n=x_n$ , or in $(y_1,\ldots ,y_{n-1})^{\langle k\rangle }+(x_{n}^2)$ already, by Lemma 5.7.

For the monomial products in $P_1$ we assume by way of contradiction that $\mathbf {y}^{\mathbf {a}}\cdot \nu _{\mathbf {a}}\notin (y_1,\ldots ,y_{n-1})^{\langle k\rangle }+(x_{n}^2)$ for some $\mathbf {a}\in {\mathbb N}^{n-1}$ . Then since $(y_1,\ldots ,y_{n-1})^{\langle k\rangle }+(x_{n}^2)$ is a monomial ideal, it follows that in the y-monomial expansion of $\mathbf {y}^{\mathbf {a}}\nu _{\mathbf {a}}$ , there must be some monomial say $\mathbf {y}^{\mathbf {d}}$ which is not in $(y_1,\ldots ,y_{n-1})^{\langle k\rangle }+(x_{n}^2)$ . Since $\mathbf {y}^{\mathbf {a}}\nu _{\mathbf {a}}$ is independent of $x_n$ , so is $\mathbf {y}^{\mathbf {d}}$ . Since $(y_1,\ldots ,y_{n-1})^{\langle k\rangle }$ consists of all y-monomials of weight at least k, it follows that $\operatorname {wt}(\mathbf {y}^{\mathbf {d}})\leq k-1$ . On the other hand, every y-monomial in the monomial expansion of $\nu _{\mathbf {a}}$ has weight equal to $k-1$ , hence it follows that $\operatorname {wt}(\mathbf {y}^{\mathbf {d}})=k-1$ . But then we can deduce, as in the proof of Lemma 4.5, that

$$ \begin{align*}\frac{\mathbf{y}^{\mathbf{d}}}{\sqrt{\mathbf{y}^{\mathbf{d}}}}=\mathbf{y}^{\mathbf{a}},\end{align*} $$

and hence the exponent vector $\mathbf {d}$ uniquely determines the exponent vector $\mathbf {a}$ . This implies that $\mathbf {y}^{\mathbf {d}}$ occurs with the same coefficient in the monomial expansion of the term $\mathbf {y}^{\mathbf {a}}\nu _{\mathbf {a}}$ as it does in the entire sum

$$ \begin{align*}P_1=\sum_{\mathbf{a}\in{\mathbb N}^{n-1}(m)}\mathbf{y}^{\mathbf{a}}\nu_{\mathbf{a}},\end{align*} $$

contradicting the fact that $P_1\in (y_1,\ldots ,y_{n-1})^{\langle k\rangle }+(x_{n}^2)$ . The argument for $P_2$ is similar.

Proof. The first statement for $\nu $ is obvious for degree reasons. For the second statement, we can write $\mu =\mu (x_1,\ldots ,x_n)=x_{n}\alpha +\beta $ for polynomials $\alpha =\alpha (x_1,\ldots ,x_{n-1})\in V_{\mathbf {x}}(n-1,k-1,k-1)$ and $\beta =\beta (x_1,\ldots ,x_{n-1})\in V_{\mathbf {x}}(n-1,0,k)$ . Here, it is important to distinguish between polynomials in x-variables and those in the y-variables, hence we adopt the notation $V_{\mathbf {x}}(m,j,j)$ and $V_{\mathbf {y}}(m,j,j)$ to denote the ${\mathbb F}$ -span of Specht polynomials in the x-variables and y-variables, respectively; note that if $m\leq n-1$ these two subspaces coincide.

Consider inclusions of monomial complete intersections, $B\subset A$ defined in the y-variables by:

$$ \begin{align*}B=\frac{{\mathbb F}[y_1,\ldots,y_{n-1}]}{(y_1^2,\ldots,y_{n-1}^2)}\hookrightarrow A= \frac{{\mathbb F}[y_1,\ldots,y_{n}]}{(y_1^2,\ldots,y_n^2)},\end{align*} $$

and let $L_B,D_B,H_B$ , and $L_A,D_A,H_A$ be their respective raising, lowering, and semi-simple operators, respectively, as in §3. Then since $p=0$ or $p\geq k+1$ , Lemma 3.4 implies that the lowering maps for B and A,

$$ \begin{align*}D_B\colon B_k\rightarrow B_{k-1}, \ \ \text{and} \ \ D_A\colon A_k\rightarrow A_{k-1}\end{align*} $$

are both surjective, which by Lemma 3.2 is equivalent to their primitive subspaces $P_{B,k}=\ker (D_B)\cap B_k$ and $P_{A,k}=\ker (D_A)\cap A_k$ satisfying

$$ \begin{align*}P_{B,k}=V_{\mathbf{y}}(n-1,k,k), \ \ \text{and} \ \ P_{A,k}=V_{\mathbf{y}}(n,k,k).\end{align*} $$

We identify B (resp. A) with the subspace spanned by square-free monomials in variables $y_1,\ldots ,y_{n-1}$ (resp. $y_1,\ldots ,y_n$ ).

Then if $\mu =x_n\alpha (x_1,\ldots ,x_{n-1})+\beta (x_1,\ldots ,x_{n-1})\in (y_1,\ldots ,y_{n-1})^{\langle k\rangle }+(x_{n}^2)$ , by Lemma 5.7, we have

(24) $$ \begin{align} \mu\equiv & (-1)^{k-1}x_n\left(\alpha(y_1,\ldots,y_{n-1})+D_B(\beta(y_1,\ldots,y_{n-1}))\right)\equiv 0 & \text{mod} \ (y_1,\ldots,y_{n-1})^{\langle k\rangle}+(x_{n}^2), \end{align} $$

(note that $D_B(\alpha (y_1,\ldots ,y_{n-1}))=0$ and $\beta (y_1,\ldots ,y_{n-1})\in (y_1,\ldots ,y_{n-1})^{\langle k\rangle }+(x_{n}^2)$ automatically). Dividing by $x_n$ in (24), we see that

$$ \begin{align*}\alpha(y_1,\ldots,y_{n-1}\kern-1pt)+D_B(\beta(y_1,\ldots,y_{n-1}\kern-1pt))\in ((y_1,\ldots,y_{n-1})^{\langle k\rangle}\kern-1pt+x_n^2):x_n=(y_1,\ldots,y_{n-1})^{\langle k\rangle}\kern-1pt+(x_n\kern-1pt),\end{align*} $$

which, for degree reasons, implies that

$$ \begin{align*}\alpha(y_1,\ldots,y_{n-1})+D_B(\beta(y_1,\ldots,y_{n-1}))=0.\end{align*} $$

Therefore, $\beta _{\mathbf {y}}=\beta (y_1,\ldots ,y_{n-1})\in B_k$ is a square-free polynomial in $y_1,\ldots ,y_{n-1}$ such that $D_B(\beta _{\mathbf {y}})=-\alpha _{\mathbf {y}}\in V(n-1,k-1,k-1)=P_{B,k-1}$ . Applying the commutator relation

$$ \begin{align*}\left[D_B,L_B\right]=H_B,\end{align*} $$

to $\alpha _{\mathbf {y}}=\alpha (y_1,\ldots ,y_{n-1})$ , we find that

$$ \begin{align*}D_B\circ L_B\left(\alpha_{\mathbf{y}}\right)=H\left(\alpha_{\mathbf{y}}\right)=\left(n-1-2(k-1)\right)\cdot \alpha_{\mathbf{y}},\end{align*} $$

which implies that

(25) $$ \begin{align} \mu_B(y_1,\ldots,y_{n-1}):= L_B\left(\alpha_{\mathbf{y}}\right)+(n-2k+1)\cdot \beta_{\mathbf{y}}\in P_{B,k}=V_{\mathbf{y}}(n-1,k,k). \end{align} $$

Note that Equation (25) also implies that $\mu _B:= \mu _B(x_1,\ldots ,x_{n-1})\in V_{\mathbf {x}}(n-1,k,k)$ . We can also apply the commutator relations for those operators on A. Note that the restriction of $D_A=D_B+\partial /\partial y_n$ to B is $D_B$ , and that by Lemma 3.5, we have

$$ \begin{align*}V_{\mathbf{y}}(n-1,k,k)=V_{\mathbf{y}}(n,k,k)\cap B_k.\end{align*} $$

It follows that $D_A(\beta _{\mathbf {y}})=D_B(\beta _{\mathbf {y}})=-\alpha _{\mathbf {y}}\in V_{\mathbf {y}}(n-1,k-1,k-1)\subset V_{\mathbf {y}}(n,k-1,k-1)$ , and hence applying the commutator relation

$$ \begin{align*}\left[D_A,L_A\right]=H_A,\end{align*} $$

to $D_A(\beta _{\mathbf {y}})=-\alpha _{\mathbf {y}}$ we also find that

(26) $$ \begin{align} \mu_A(y_1,\ldots,y_{n-1},y_n):= L_A(\alpha_{\mathbf{y}})+(n-2k+2)\cdot\beta_{\mathbf{y}}\in P_{A,k}=V_{\mathbf{y}}(n,k,k), \end{align} $$

and hence $\mu _A=\mu _A(x_1,\ldots ,x_n)\in V_{\mathbf {x}}(n,k,k)$ . Noting that $L_A=L_B+x_n$ we see that adding $\mu $ to $\mu _B$ yields $\mu _A$ , that is,

$$ \begin{align*} \mathop{\underbrace{\left(x_n\alpha+\beta\right)}}_{\mu}+\mathop{\underbrace{\left((x_1+\cdots+x_{n-1})\cdot \alpha+(n-2k+1)\beta\right)}}_{\mu_B}=\mathop{\underbrace{(x_1+\cdots+x_n)\cdot\alpha+(n-2k+2)\beta}}_{\mu_A}. \end{align*} $$

It follows from (25) and (26) that $\mu =x_n\cdot \alpha +\beta =\mu _A-\mu _B\in V_{\mathbf {x}}(n,k,k)=V(n,k,k)\subset \mathfrak {a}(n,k,k)$ , as claimed.

Proof. For (1) assume $\sqrt {\mathbf {x}^{\mathbf {a}}}\notin \operatorname {supp}(F_T^{k-1})$ . If some variable $x_i\in \operatorname {supp}(\mathbf {y}^{\mathbf {a}})$ but $x_i\notin \operatorname {supp}(F_T^{k-1})$ , then clearly

$$ \begin{align*}\mathbf{y}^{\mathbf{a}}\cdot F_T^{k-1}=\mathbf{y}^{\mathbf{a}'}\cdot y_i\cdot F_T^{k-1}=\mathbf{y}^{\mathbf{a}'}\cdot (x_n-x_i)\cdot F_{T}^{k-1}\in V(n,k,k).\end{align*} $$

We may therefore assume that every variable of $\mathbf {x}^{\mathbf {a}}$ lies in the support of $F_T^{k-1}$ . But since $\sqrt {\mathbf {x}^{\mathbf {a}}}\notin \operatorname {supp}(F_T^{k-1})$ there must be two distinct indices $i,j$ for which $y_i,y_j\in \operatorname {supp}(\mathbf {y}^{\mathbf {a}})$ and $i,j$ lie in the same column in T. Choose any index $r\neq i,j$ such that $x_r\notin \operatorname {supp}(F_T^{k-1})$ –presumably there is such an r since we are assuming that $k+1\leq n-k$ –and let $(i,r), (j,r)\in \mathfrak {S}_n$ be the transpositions swapping $i,r$ and $j,r$ , respectively. Then we have

$$ \begin{align*}\mathbf{y}^{\mathbf{a}}\cdot F_T^{k-1}=\mathbf{y}^{\mathbf{a}}\cdot\left(F_{(i,r).T}^{k-1}+F_{(j,r).T}^{k-1}\right),\end{align*} $$

and since $x_i\notin \operatorname {supp}(F_{(i,r).T}^{k-1})$ , we must have

$$ \begin{align*}\mathbf{y}^{\mathbf{a}}\cdot F_{(i,r).T}^{k-1}=\mathbf{y}^{\mathbf{a}'}\cdot y_i\cdot F_{(i,r).T}^{k-1}=\mathbf{y}^{\mathbf{a}'}\cdot (x_n-x_i)\cdot F_{(i,r).T}^{k-1}\in \mathfrak{a}(n,k,k)\subset I(n,k),\end{align*} $$

and similarly for $F^{k-1}_{(j,r).T}$ . This proves that $\mathbf {y}^{\mathbf {a}}\cdot F_T^{k-1}\in I(n,k)$ . Item (2) is easier and left to the reader.