2.1 Functions as polynomials

We introduce the ring of functions on a vector space; we will also call this the coordinate ring.

Definition 2.1.1. Given $V \in \mathbf{Vec}$ , we write K[V] for the K-algebra of functions $V\to K$ . This has a natural algebra filtration

\[ \{0\}=K[V]_{\leq -1} \subseteq K[V]_{\leq 0} \subseteq K[V]_{\leq1} \subseteq K[V]_{\leq 2} \subseteq \cdots \]

where $K[V]_{\leq d}$ is the set of functions $f:V \to K$ for which there exists an element of $\bigoplus_{e=0}^d S^e V^*$ that defines the function f.

We stress that K[V] is an algebra of functions, not of polynomials. More precisely, K[V] is the quotient of the symmetric algebra $S V^*$ by the ideal generated by the polynomials $x^q-x$ as x runs through (a basis of) $V^*$ . Since these polynomials are not homogeneous, K[V] has no natural grading; however, as seen above, it does have a natural filtration.

Note further that K[V] is a finite-dimensional K-vector space, of dimension $q^{\dim(V)}$ , the number of elements of V.

The following lemma is immediate; the natural isomorphisms in it will be interpreted as equalities throughout the paper.

Similarly, the set of arbitrary maps $V \to W$ is canonically isomorphic to $K[V]\otimes W$ via the K-linear map from right to left that sends $f\otimes w$ to the function $v \mapsto f(v) \cdot w$ .

Furthermore, we write $K[V]_0=K$ for the sub-K-algebra of constant functions, and $K[V]_{>0}$ for the K-vector space spanned by all functions that vanish at zero.

2.2 Polynomial generic representations over K

Recall Theorem 1.5.3, which characterises polynomial representations among all generic representations. We will use the following alternative characterisation instead.

Polynomial representations form an abelian category, in which the morphisms are natural transformations.

Any polynomial representation in the sense of Theorem 1.5.3 is a subquotient of a direct sum $T^{d_1} \oplus \cdots \oplus T^{d_k}$ , and this implies that it is polynomial in the sense of the definition above, of degree at most the maximum of the $d_i$ . In Remark 2.4.2, we will see that, conversely, any generic representation that is polynomial in the sense of the definition above is polynomial in the sense of Theorem 1.5.3.

Every finite-degree strict polynomial functor $\mathbf{Vec} \to \mathbf{Vec}$ in the sense of Friedlander-Suslin [Reference Friedlander and SuslinFS97] gives rise to a polynomial representation. But this forgetful functor is not an equivalence of abelian categories. For instance, if Q is a strict polynomial functor of degree d over K, then its q-Frobenius twist is a polynomial functor of degree dq over K and hence not isomorphic to Q. However, Q and its q-Frobenius twist give rise to the same generic representation.

2.3 Schur algebras over K

In spite of the discrepancy between strict polynomial functors and polynomial generic representations, a version of the theorem by Friedlander and Suslin that relates polynomial functors to representations of the Schur algebra, does hold.

Fix a natural number d and a $U \in \mathbf{Vec}$ . The composition map $\operatorname{End}\nolimits(U)\times \operatorname{End}\nolimits(U) \to \operatorname{End}\nolimits(U)$ gives rise, via pullback of functions, to a K-linear map

\[ K[\operatorname{End}\nolimits(U)]_{\leq d} \to K[\operatorname{End}\nolimits(U)]_{\leq d} \otimes K[\operatorname{End}\nolimits(U)]_{\leq d}. \]

We write $A_{\leq d}(U):=K[\operatorname{End}\nolimits(U)]_{\leq d}^*$ . Dualising the map above, we obtain a K-bilinear map

\[ A_{\leq d}(U) \times A_{\leq d}(U) \to A_{\leq d}(U). \]

A straightforward computation, using the associativity of composition of linear maps, shows that this turns $A_{\leq d}(U)$ into a unital, associative algebra, with unit element $f \mapsto f(\operatorname{id}_U)$ .

We remark that this is in fact a subalgebra of the Schur algebra in [Reference Friedlander and SuslinFS97], which is the dual space to the space of polynomials of degree at most d on $\operatorname{End}\nolimits(U)$ . Our Schur algebra consists of only those linear functions that vanish on the ideal of polynomials that define the zero function on $\operatorname{End}\nolimits(U)$ .

The Schur algebra comes with a homomorphism of monoids (not of K-algebras) $\operatorname{End}\nolimits(U) \to A_{\leq d}(U)$ defined by $\varphi \mapsto(f \mapsto f(\varphi))$ . This homomorphism is an embedding if $d \geq 1$ .

2.4 A finite-field analogue of the Friedlander–Suslin lemma

Given a polynomial generic representation P of degree at most d and a vector space U, we turn P(U) into an $A_{\leq d}(U)$ -module by a construction very similar to the construction of $A_{\leq d}(U)$ : first, the map

\[ \operatorname{End}\nolimits(U) \times P(U) \to P(U), (\varphi,p) \mapsto P(\varphi)(p) \]

gives rise, via pullback, to a K-linear map

\[ P(U)^* \to K[\operatorname{End}\nolimits(U)]_{\leq d} \otimes P(U)^*. \]

Dualising, we obtain a K-bilinear map

\[ A_{\leq d}(U) \times P(U) \to P(U) \]

that turns P(U) into a (unital) $A_{\leq d}(U)$ -module.

The following proposition is proved exactly as Friedlander and Suslin’s corresponding theorem, and it is also almost equivalent to [Reference KuhnKuh94a, Proposition 4.10].

Remark 2.4.3. If P is an irreducible polynomial generic representation, then for each $U \in \mathbf{Vec}$ , P(U) is (zero or) an irreducible $\operatorname{End}\nolimits(U)$ -module. Indeed, if M were a nonzero proper submodule, then, for varying V,

\[ Q(V):=\{p\in P(V) \mid \forall \varphi \in \operatorname{Hom}\nolimits_\mathbf{Vec}(V,U)\ P(\varphi)p \in M\} \]

would define a nonzero proper subrepresentation.

2.5 Filtering a polynomial representation by degree

A strict polynomial functor in the sense of Friedlander and Suslin has a grading by degree. In contrast, we will see that a polynomial generic representation only has a filtration by degree. One notable exception is the degree-zero part of a polynomial representation.

Definition 2.5.1. Let $P:\mathbf{Vec} \to \mathbf{Vec}$ be a polynomial generic representation. Then we define $P_0:\mathbf{Vec} \to \mathbf{Vec}$ by $P_0(V):=P(0)=:U$ for all $V \in \mathbf{Vec}$ and $P_0(\varphi):=\operatorname{id}_U$ for all $\varphi \in \operatorname{Hom}\nolimits_\mathbf{Vec}(V,W)$ . This is a direct summand of P in the abelian category of polynomial generic representations, called the degree-zero part or constant part of P; we will also informally say that U is the constant part of P. The constant part $P_0$ has a unique complement in P, namely,

\begin{align*}P_{>0}(V):=\{p \in P(V) \mid & P(0\cdot \operatorname{id}_V)p=0\}.\end{align*}

A polynomial representation P of degree at most some $e \leq d$ is in particular a polynomial representation of degree at most d. On the Schur algebra side, this inclusion of abelian categories is made explicit as follows. Take a vector space U of dimension at least d. Then the inclusion $K[\operatorname{End}\nolimits(U)]_{\leq e} \to K[\operatorname{End}\nolimits(U)]_{\leq d}$ dualises to a linear surjection $A_{\leq d}(U) \to A_{\leq e}(U)$ . This surjection is an algebra homomorphism, and hence if M is a module over the latter algebra, then it is also naturally a module over the former algebra.

This interpretation also shows which $A_{\leq d}(U)$ -modules are also $A_{\leq e}(U)$ -modules, namely, those for which the kernel I of the surjection acts as zero. Furthermore, if M is an $A_{\leq d}(U)$ -module, and N is an $A_{\leq d}(U)$ -submodule of M, then $M/N$ is an $A_{\leq e}(U)$ -module if and only if $I \cdot (M/N)=0$ , i.e., if and only if N contains the $A_{\leq d}(U)$ -submodule $I \cdot M$ . We conclude that there is a unique inclusionwise minimal $A_{\leq d}(U)$ -submodule N of M such that $M/N$ is an $A_{\leq e}(U)$ -module, namely, $N=I \cdot M$ .

By Proposition 2.4.1 we may translate this back to polynomial representations.

We clearly have

\[ P=P_{>-1} \supseteq P_{> 0} \supseteq \cdots \supseteq P_{>d}=\{0\}\]

where $d=\deg(P)$ ; and a straightforward check shows that $P_{>0}$ in this definition agrees with the direct complement $P_{>0}$ of $P_0$ in Definition 2.5.1.

Proof. We compute

\[ P/\alpha^{-1}(Q_{>e}) \cong \operatorname{im}(\alpha)/(\operatorname{im}(\alpha) \cap Q_{>e})\subseteq Q/Q_{>e}. \]

Since the latter representation has degree at most e, so does the first. The defining property of $P_{>e}$ then implies that $P_{>e} \subseteq \alpha^{-1}(Q_{>e})$ . This is equivalent to the statement in the lemma.

Proof. By Lemma 2.5.5, the morphism $P \to P/R$ maps $P_{>e}$ into $(P/R)_{>e}$ , and its kernel on $P_{>e}$ is R, so that $P_{>e}/R$ maps injectively into $(P/R)_{>e}$ . To see that it also maps surjectively, we note that

\[ (P/R)/(P_{>e}/R) \cong P/P_{>e} \]

has degree at most e. Hence $P_{>e}/R$ contains $(P/R)_{>e}$ by definition of the latter object.

2.6 Shifting

Just as a univariate polynomial can be shifted over a constant, and then its leading term does not change, a polynomial representation can be shifted over a constant vector space, and we will see that its top-degree part does not change.

If P is polynomial of degree at most d, then $\mathrm{Sh}_U P$ is also polynomial of degree at most d; below we will prove a more precise statement.

We have a morphism $\alpha:P \to \mathrm{Sh}_U P$ in the abelian category of polynomial generic representations defined by $\alpha_V=P(\iota_V):P(V) \to P(U \oplus V)$ , where $\iota_V:V \to U\oplus V$ is the inclusion $v \mapsto 0 + v$ . Indeed, that $(\alpha_V)_V$ is a morphism follows from the commutativity of the following diagram, for any $\varphi \in\operatorname{Hom}\nolimits_\mathbf{Vec}(V,W)$ .

This in turn follows from the fact that P is a representation and that

\[ (\operatorname{id}_U \oplus \varphi) \circ \iota_V = \iota_W \circ \varphi. \]

Similarly, we have a morphism $\beta:\mathrm{Sh}_U P\to P$ defined by $\beta_V=P(\pi_V):P(U \oplus V) \to P(V)$ , where $\pi:U \oplus V \to V$ is the projection $u+v \mapsto v$ . The relation $\pi \circ \iota=\operatorname{id}_V$ translates to $\beta \circ \alpha=\operatorname{id}_P$ . This implies that $\mathrm{Sh}_U P$ is the direct sum of $\operatorname{im}(\alpha) \cong P$ and the polynomial representation $Q:=\ker(\beta)$ .

The following lemma says, informally, that the top-degree part of a polynomial representation is invariant under shifting.

To see this, we recall that $\mathrm{Sh}_U P=\operatorname{im}(\alpha) \oplus Q$ , where $Q=\ker(\beta)$ . Accordingly,

\[ (\mathrm{Sh}_U P)/\alpha(P_{>d-1}) \cong (\alpha(P)/\alpha(P_{>d-1})) \oplus Q. \]

Here the first summand on the right is isomorphic to $P/P_{>d-1}$ , hence of degree at most $d-1$ . So it suffices to show that Q has degree at most $d-1$ as well. Consider a vector $q \in Q(V)$ and a linear map $\varphi \in\operatorname{Hom}\nolimits(V,W)$ . Then we have

\begin{align*}Q(\varphi)(q)&=P(\operatorname{id}_U \oplus \varphi)(q)=P(\operatorname{id}_U \oplus \varphi)(q) -P(\operatorname{id}_U \oplus \varphi)(\alpha_V(\beta_V(q))) \\&= (P(\operatorname{id}_U \oplus \varphi)-P(0_U \oplus \varphi))(q)\end{align*}

where the second equality follows from $\beta_V(q)=0$ and the last equality follows from the definition of $\alpha$ and $\beta$ . Now, for $\psi$ running through $\operatorname{Hom}\nolimits(U \oplus V,U \oplus W)$ , $P(\psi)$ can be described by a polynomial map of degree at most d. If in this map we substitute for $\psi$ the maps $\operatorname{id}_U \oplus \varphi$ and $0_U \oplus\varphi$ respectively, we obtain the same degree-d parts in $\varphi$ . Hence the map $\varphi \mapsto P(\operatorname{id}_U \oplus \varphi)-P(0_U \oplus \varphi)$ is given by a polynomial map of degree at most $d-1$ in the entries of $\varphi$ . This shows that Q has degree at most $d-1$ , as desired.

2.7 A well-founded order on polynomial representations

To see that $\preceq$ is well-founded, suppose we had an infinite chain

\[ P_1 \succ P_2 \succ \cdots. \]

To each $P_i$ we associate a length sequence $\ell(P_i) \in {\mathbb Z}_{\geq0}^{\{-1,0,1,2,\ldots\}}$ , where $\ell(P_i)(e)$ is the length of any composition chain of $(P_i)_{>e}$ in the abelian category of polynomial representations; by Proposition 2.4.1 this length is finite.

Note that $\ell(P_i)(e)=0$ for $e \geq \deg(P_i)$ , i.e., $\ell(P_i)$ has finite support. Now $P_i \succ P_{i+1}$ implies that $\ell(P_{i+1})$ is lexicographically strictly smaller than $\ell(P_i)$ . Since the lexicographic order on sequences with finite support is a well-order, we arrive at a contradiction.

The following construction will play a crucial role in our proof.

2.8 The coordinate ring of a polynomial representation

Note that $P(\varphi)^\#$ is an algebra homomorphism; this is going to be of crucial importance in § 4.6. The coordinate ring comes with a natural ring filtration,

\[ \{0\}=K[P]_{\leq -1} \subseteq K[P]_{\leq 0} \subseteq K[P]_{\leq1} \subseteq K[P]_{\leq 2} \subseteq \cdots, \]

where $K[P]_{\leq e}$ assigns to V the space $K[P(V)]_{\leq e}$ .

Example 2.8.3. Let $P=S^2$ and assume $|K|>2$ . Take $V=K^n$ with basis $e_1,\ldots,e_n$ , so that P(V) has basis $e_i e_j$ with $i \leq j$ . For $k>l$ , let $g_{kl}(s) \in \operatorname{End}\nolimits(V)$ be the matrix with 1s on the diagonal, an s in position (k,l), and 0s elsewhere. We have

\begin{align*}P(g_{kl}(s)) \sum_{i \leq j} a_{ij} e_i e_j &= \sum_{i \leq j} a_{ij}(g_{kl}(s) e_i)(g_{kl}(s) e_j)\\&= \sum_{i \leq j} a_{ij} (e_i + \delta_{il} s e_k)(e_j + \delta_{jl} s e_k) \\&= \sum_{i \leq j} a_{ij} (e_i e_j + s(\delta_{jl}e_i e_k + \delta_{il} e_k e_j) + s^2 \delta_{il} \delta_{jl} e_k^2)\\&= \Bigg(\sum_{i \leq j} a_{ij} e_i e_j \Bigg)+ s \Bigg(\sum_{i \leq l} a_{il} e_i e_k+\sum_{j \geq l} a_{lj} e_k e_j\Bigg)+ s^2 a_{ll} e_k^2.\end{align*}

Observe that by acting with $g_{kl}$ on (linear combinations of) the basis vectors $e_i e_j$ , indices l either remain the same or turn into indices k.

We now look at the dual. Let $\{x_{ij} \mid i \leq j\}$ be the basis of $P(V)^*$ dual to the given basis of P(V). Then, for instance, for $l< i < k$ we have

\[ P(g_{kl}(s))^\# x_{ik}=x_{ik} + s x_{li}, \]

as can be seen by taking the coefficient of $e_i e_k$ in the expression above. We observe here that indices k either remain the same or turn into indices l. We can also write the above as

\[ P(g_{lk}(s)^T)^\# x_{ik}=x_{ik} + s x_{li}. \]

Note that $P(g)^\#$ is contravariant in g, and hence $P(g^T)^\#$ is again covariant. This explains the $V^*$ in Lemma 2.8.2.

3.1 Multiplicative monoid homomorphisms $K \to K$

A monoid homomorphism $(K, \cdot) \to (K,\cdot)$ is a map $\varphi:K \to K$ with $\varphi(1)=1$ and $\varphi(ab)=\varphi(a) \varphi(b)$ for all $a,b \in K$ . In particular, $\varphi$ restricts to a group homomorphism from the multiplicative group $K^\times:=K \setminus \{0\}$ to itself. Since $K^\times$ is cyclic, say with generator g, the monoid homomorphism $\varphi$ is uniquely determined by its values on g and on 0. Write $\varphi(g)=g^e$ for a unique exponent $e \in \{1,\ldots,q-1\}$ . If $e\neq q-1$ , so that $\varphi(g) \neq 1$ , then $\varphi(0)$ is forced to be 0, since otherwise $\varphi(g) \cdot \varphi(0)$ does not equal $\varphi(g\cdot 0)=\varphi(0)$ . If $e = q-1$ , then there are two possibilities for $\varphi(0)$ , namely, $\varphi(0)=1$ and $\varphi(0)=0$ . In the first case we will denote $\varphi$ by $c \mapsto c^0$ , and in the second case we denote by $c \mapsto c^{q-1}$ . The following lemma is now straightforward.

Note that this monoid is not cancellative, since $0 \oplus j = (q-1)\oplus j$ for all $j \in \{1,\ldots,q-1\}$ . Nevertheless, it will be convenient to have a notation for subtracting elements in the following sense: for $i \in \{1,\ldots,q-1\}$ and $j \in \{0,\ldots,q-1\}$ we write $i \ominus j$ for the unique element in $\{1,\ldots,q-1\}$ that equals $i-j$ modulo $q-1$ .

3.2 Acting with diagonal matrices

Let $P:\mathbf{Vec} \to \mathbf{Vec}$ be a polynomial representation and set $V:=K^n$ , so that we may identify $\operatorname{End}\nolimits(V)$ with the space of $n \times n$ matrices. Then the monoid $\operatorname{End}\nolimits(V)$ acts linearly on P(V) via the map $\operatorname{End}\nolimits(V)\to \operatorname{End}\nolimits(P(V)), \varphi \mapsto P(\varphi)$ , and hence so does its submonoid $D_n \subseteq \operatorname{End}\nolimits(V)$ of diagonal matrices.

Lemma 3.2.1. We have

\[ P(V)=\bigoplus_{\chi:D_n \to K} P(V)_\chi \]

where $\chi$ runs over all monoid homomorphisms $(D_n,\cdot) \to(K,\cdot)$ and where

\[ P(V)_\chi:=\{p \in P(V) \mid \forall \varphi \in D_n:P(\varphi)p=\chi(\varphi)p \}. \]

Proof. Each element $\varphi \in D_n$ satisfies $\varphi^q=\varphi$ , and therefore also $P(\varphi)^q=P(\varphi^q)=P(\varphi)$ . Consequently, $P(\varphi)$ is a root of the polynomial $h=T\cdot (T^{q-1}-1) \in K[T]$ . This polynomial is square-free, so that $P(\varphi)$ is diagonalisable over a separable closure of K. But also the eigenvalues of $P(\varphi)$ are roots of h, i.e., elements of K, so $P(\varphi)$ is diagonalisable over K. Moreover, all elements of $D_n$ commute, and therefore so do all elements of $P(D_n)$ . Hence the latter are all simultaneously diagonalisable. We therefore have

\[ P(V)=\bigoplus_{\chi:D_n \to K} P(V)_\chi \]

where, a priori, $\chi$ runs through all maps $D_n \to K$ .

Now if $P(V)_\chi \neq 0$ , then it follows that $\chi(\operatorname{diag}\nolimits(1,\ldots,1))=1$ and $\chi(\varphi\psi)=\chi(\varphi) \chi(\psi)$ , i.e., $\chi$ is a monoid homomorphism $D_n \to K$ .

Note that monoid homomorphisms $D_n \to K$ can be naturally identified with n-tuples of monoid homomorphisms $K \to K$ , and hence, by Lemma 3.1.1, with elements of $\{0,\ldots,q-1\}^n$ . Explicitly, $\chi$ is identified with the tuple $(a_1,\ldots,a_n)$ if $\chi(\operatorname{diag}\nolimits(t_1,\ldots,t_n))=t_1^{a_1} \cdots t_n^{a_n}$ for all $(t_1,\ldots,t_n) \in K^n$ .

In analogy with the theory of representations of algebraic groups, we will use the word weight for monoid homomorphisms $\chi:D_n \to K$ , and we call a vector in $P(V)_\chi$ a weight vector of weight $\chi$ .

We use the notation $\oplus$ also in this context: if $\chi,\mu \in \{0,\ldots,q-1\}^n$ are weights, then $\chi \oplus\mu$ is their componentwise sum with respect to $\oplus$ . Note that

\[ (\chi \oplus\mu)(\operatorname{diag}\nolimits(t_1,\ldots,t_n))=\chi(\operatorname{diag}\nolimits(t_1,\ldots,t_n)) \cdot\mu(\operatorname{diag}\nolimits(t_1,\ldots,t_n)).\]

3.3 Acting with additive one-parameter subgroups

Let $P:\mathbf{Vec} \to \mathbf{Vec}$ be a polynomial representation, $n \in {\mathbb Z}_{\geq 2}$ , and $i,j \in [n]$ distinct. Then we have a one-parameter subgroup

\[ g_{ij}:(K,+) \to \operatorname{GL}\nolimits_n(K),\quad g_{ij}(s):=I+s E_{ij}, \]

where $E_{ij}$ is the matrix with zeros everywhere except for a 1 in position (i,j). For $b=0,\ldots,q-1$ we define the linear map $F_{ij}[b]:P(K^n) \to P(K^n)$ by

\[ F_{ij}[b]p{\kern.7pt}=\text{the coefficient of $s^b$ in}P(g_{ij}(s))p,\]

where we write $P(g_{ij}(s))p$ as a reduced polynomial in s with coefficients in $P(K^n)$ .

Proof. Let $p \in Q(K^n)$ . Then for all $s \in K$ the element

\[ P(g_{ij}(s))p=F_{ij}[0]p + s F_{ij}[1]p + \cdots +s^{q-1} F_{ij}[q-1]p \]

lies in $Q(K^n)$ . The Vandermonde matrix $(s^e)_{s \in K, e \in\{0,\ldots,q-1\}}$ is invertible, and this implies that each of the $F_{ij}[e]p$ above are linear combinations of the $P(g_{ij}(s))p$ , and therefore in $Q(K^n)$ .

1. (1) We have $\tilde{p}=p$ for $b=0$ .

2. (2) If $a_j=0$ , then $\tilde{p}=0$ for $b \neq 0$ .

3. (3) If $0<a_j \neq b$ , then $\tilde{p}$ is a weight vector of weight $a \ominus (b e_j) \oplus (b e_i)$ .

4. (4) If $0<a_j = b$ , then $\tilde{p}$ is a sum of a weight vector of weight

   \[ a \ominus b e_j \oplus b e_i= (a_1,\ldots,a_i \oplus b,\ldots,q-1,\ldots,a_n) \]
   and a weight vector of weight
   \[ a - b e_j \oplus b e_i= (a_1,\ldots,a_i \oplus b,\ldots,0,\ldots,a_n). \]

Proof. We write

\[ P(g_{ij}(s))p=p_0 + sp_1 + \cdots + s^{q-1}p_{q-1}, \]

where the $p_b \in P(K^n)$ are uniquely determined by the fact that the identity above holds for all $s \in K$ .

By setting s equal to zero we obtain $P(g_{ij}(0))p=P(\operatorname{id}_{K^n})p=p$ on the left-hand side, and $p_0$ on the right-hand side. This proves the first item.

If $a_j=0$ , then

\[ P(\operatorname{diag}\nolimits(1,\ldots,1,0,1\ldots,1)) p = p \]

where the 0 is in position j. Therefore

\[ P(g_{ij}(s)) p = P(g_{ij}(s)\operatorname{diag}\nolimits(1,\ldots,1,0,1,\ldots,1)) p =P(\operatorname{diag}\nolimits(1,\ldots,1,0,1,\ldots,1)) p \]

does not depend on s and hence $F_{ij}[b] p =0$ for $b\neq 0$ .

We now assume $a_j>0$ . We have $F_{ij}[b] p=p_b$ . To determine the weight(s) appearing in $p_b$ , we act on $p_b$ with diagonal matrices. For $t=(t_1,\ldots,t_n) \in K^n$ and $t_j \neq 0$ we have

\[\operatorname{diag}\nolimits(t_1,\ldots,t_n) \cdot g_{ij}(s)= g_{ij}(t_i s t_j^{-1}) \cdot \operatorname{diag}\nolimits(t_1,\ldots,t_n) \]

and therefore

\begin{align*}\sum_{d=0}^{q-1} s^d P(\operatorname{diag}\nolimits(t_1,\ldots,t_n)) p_d&= P(\operatorname{diag}\nolimits(t_1,\ldots,t_n) g_{ij}(s)) p \\&= P(g_{ij}(t_ist_j^{-1}) \operatorname{diag}\nolimits(t_1,\ldots,t_n)) p \\&= t_1^{a_1} \cdots t_n^{a_n} \cdot P(g_{ij}(t_i st_j^{-1})) p \\&= t_1^{a_1} \cdots t_n^{a_n} \cdot \sum_{d=0}^{q-1} (t_i s t_j^{-1})^d p_d.\end{align*}

Comparing coefficients of $s^b$ , we find

\[ P(\operatorname{diag}\nolimits(t))p_b=t^{a-be_j+be_i} p_b \]

for all $t \in K^{j-1} \times K^\times \times K^{n-j}=:D$ . Hence $p_b$ is a linear combination of weight vectors with weights that on D agree with the weight $a \ominus b e_j \oplus be_i$ . If $a_j \neq b$ , then there is only one such weight, namely, $a \ominus b e_j \oplus b e_i$ . If $a_j=b$ , then there are two such weights, namely, $a \ominus b e_j \oplus b e_i$ and $a - be_j \oplus b e_i$ .

3.4 Spreading out weight

Retaining the notation from § 3.3, suppose we are given a nonzero weight vector $p \in P(K^n)$ of weight $(a_1,\ldots,a_n) \in \{0,\ldots,q-1\}^n$ and a $j \in [n]$ with $a_j >0$ . We construct vectors $\tilde{p} \in P(K^{n+1})$ by identifying p with $P(\iota)p$ , where $\iota:K^n \to K^{n+1}$ is the embedding adding a 0 in the last position. Then p is a vector of weight $a=(a_1,\ldots,a_n,0)$ in $P(K^{n+1})$ , and we compute

\[ \tilde{p}:=F_{n+1,j}[b] p \]

for various b. The vector $\tilde{p}$ is guaranteed to be nonzero for at least two values of b, namely, for $b=0$ (in which case $\tilde{p}=p$ ), and, as we will now see, for $b=a_j$ . Indeed, in the latter case, by Lemma 3.3.2, $\tilde{p}$ is the sum of a weight vector $\tilde{p}_0$ of weight $a-a_je_j + a_j e_{n+1}$ and a weight vector $\tilde{p}_1$ of weight $a + (q-1-a_j) e_j + a_j e_{n+1}$ .

Proof. The vector $\tilde{p}_0$ is obtained by applying $P(\pi_j)$ to $\tilde{p}$ , where $\pi_j$ is the projection $K^{n+1} \to K^{n+1}$ that sets the jth coordinate to zero. Furthermore, we have $P(\pi_{n+1})p=p$ , where $\pi_{n+1}$ sets the $(n+1)$ th coordinate to zero. We can then compute $\tilde{p}_0$ as the coefficient of $s^{a_j}$ in

\begin{align*}P(\pi_j) P(g_{n+1,j}(s)) p &= P(\pi_j g_{n+1,j}(s)\pi_{n+1}) p\\&= P((j,n+1))P(\operatorname{diag}\nolimits(1,\ldots,1,s,1,\ldots,1,0)) p \\&= P((j,n+1))s^{a_j} p.\\[-30pt]\end{align*}

If $F_{n+1,j}[b] p \neq 0$ for some $b \neq 0,a_j$ or if $F_{n+1,j}[a_j]p \neq P((j,n+1)) p$ , then we find a new vector p’ in the subrepresentation of P generated by p whose weight has strictly more nonzero entries; we have spread out the weight of p.

Definition 3.4.2. A nonzero weight vector $p \in P(K^n) \subseteq P(K^{n+1})$ of weight $a \in \{0,\ldots,q-1\}^n $ is called maximally spread out if for all $j \in [n]$ with $a_j>0$ we have

\[ P(g_{n+1,j}(s)) p = p + s^{a_j} P((j,n+1)) p. \]

3.5 The prime field case

In this section we assume that q is a prime, so that K is a prime field. We retain the notation from above.

Define $\varphi:K^{n+1} \to K^{n}$ by

\[\varphi(c_1,\dots,c_{n+1}):=(c_1,\ldots,c_j+c_{n+1},\ldots,c_n).\]

We then have

\[ \varphi \circ g_{n+1,j}(s) \circ \iota=\operatorname{diag}\nolimits(1,\dots,1+s,\dots,1) \]

and therefore

\[ P(\varphi) P(g_{n+1,j}(s)) P(\iota) p = (1+s)^{a_j} \cdot p. \]

The coefficient of $s^1$ in the latter expression is $a_j \cdot p$ , which is nonzero since $a_j<q$ and q is prime. That coefficient is also equal to $P(\varphi)\tilde{p}$ , where $\tilde{p}:=F_{n+1,j}p$ . Hence $\tilde{p} \neq 0$ .

By Lemma 3.5.2, if $\chi=(a_1,\ldots,a_n)$ with $a_j>1$ , then $F_{n+1,j}$ maps $P(K^n)_\chi$ injectively into $P(K^{n+1})_{\chi'}$ , where $\chi'=\chi-e_j+e_{n+1}$ . On the other hand, if $a_j=1$ , then by Lemma 3.4.1, $F_{n+1,j}$ followed by the projection to the weight space of $\chi'=(a_1,\ldots,0,\ldots,a_n,1)$ agrees on $P(K^n)_\chi$ with the map $P((n+1,j))$ , which of course we already knew is injective.

4.1 The main result

We recall that the tensor restriction theorem, its generalisation to polynomial representations, and Corollary 1.3.3 concern restriction-closed properties. We will simply use the term subset for such a property.

Theorem 4.1.2 (Noetherianity). Let P be a polynomial representation over the finite field K. Then any descending chain

\[ P \supseteq X_1 \supseteq X_2 \supseteq \cdots \]

of subsets stabilises.

Example 4.1.3. For polynomial functors of the form $P=S^{d_1} \oplus \cdots \oplus S^{d_k}$ with all $d_i<\operatorname{char} K$ , Theorem 4.1.2 can be derived from [Reference Kazhdan and ZieglerKZ20, Theorem 1.4 and Remark 1.8] as follows. Assume that $d_1 \leq \cdots \leq d_k$ . Let $X=X_1$ be a proper subset of P, and consider a descending chain $X_1 \supseteq X_2 \supseteq \cdots$ of subsets. Then by the results in [Reference Kazhdan and ZieglerKZ20], the Schmidt rank of tuples $(p_1,\ldots,p_k)$ in X is uniformly bounded. This means that, for some $i \in \{1,\ldots,k\}$ , some l, and some positive integers $e_1,\ldots,e_l<d_i$ , X is contained in the image of the map

\begin{align*}&\quad\qquad\!\! P':=S^{e_1} \oplus S^{d_i-e_1} \oplus \cdots \oplus S^{e_l}\oplus S^{d_i-e_l} \oplus \bigoplus_{j \neq i} S^{d_j} \to S^{d_1} \oplus S^{d_2} \oplus \cdots \oplus S^{d_k}, \\&(g_1,h_1,\ldots,g_l,h_l,p_1,\ldots,p_{i-1},p_{i+1},\ldots,p_k)\mapsto\Bigg(p_1,\ldots,p_{i-1},\sum_{j=1}^l g_j h_j,p_{i+1},\ldots,p_k\Bigg).\end{align*}

Let $X'_i \subseteq P'$ be the pre-image of $X_i$ . Since the tuple of degrees in P’ is lexicographically smaller than that in P, we may assume by induction that the chain $X'_1 \supseteq X'_2 \supseteq \cdots$ stabilises. And since the map $X'_i \to X_i$ is surjective, so does the chain $(X_i)_i$ .

The proof for general P uses a similar induction along a well-founded order introduced in § 2.7, but it is considerably more subtle. Moreover, rather than parameterising subsets of P by smaller P’, we embed subsets of P into smaller P’.

4.2 Irreducible decomposition of restriction-closed tensor properties

Before proceeding with the proof of Noetherianity, we deduce from it the fact that any subset of P admits a unique decomposition into irreducible subsets.

Theorem 4.2.2. For any subset X of a polynomial representation P over the finite field K, there is a unique decomposition

\[ X=X_1 \cup \cdots \cup X_k \]

where all $X_i$ are irreducible and none is contained in any other.

For an instructive example, we need the following lemma.

We note that the requirement that P(0) be zero is necessary for irreducibility; otherwise, one can take any partition $S_1 \sqcup S_2$ of the finite set P(0) into two nonempty parts, define $X_i(V)$ to be the set of elements in P(V) that map into the $S_i$ , and note that $P=X_1 \cup X_2$ .

Proof. Suppose that $X=X_1 \cup X_2$ where $X_i \subsetneq X$ for $i=1,2$ . Then $X_1$ has at least one forbidden restriction $T_1\in P(V_1)$ , and $X_2$ has at least one forbidden restriction $T_2 \in P(V_2)$ . Let $\iota_i: V_i \to V_1 \oplus V_2$ be the canonical inclusion, and write $T:=P(\iota_1)(T_1) + P(\iota_2)(T_2)$ . Let $\pi_i:V_1 \oplus V_2 \to V_i$ be the projection. Then $P(\pi_1 \circ \iota_2)=P(0_{V_2\to V_1})$ , which is the zero map since it factors via $P(0)=0$ ; and similarly $P(\pi_2 \circ \iota_1)=0$ . We conclude that

\[ T_1=P(\pi_1 \circ \iota_1)T_1 = P(\pi_1)(P(\iota_1)T_1 +P(\iota_2)T_2)=P(\pi_1)T, \]

so $T_1$ is a restriction of T. Similarly, $T_2$ is a restriction of T. It follows that T lies neither in $X_1(V_1 \oplus V_2)$ nor in $X_2(V_1 \oplus V_2)$ , a contradiction. Hence X is irreducible as claimed.

For $X_r$ (with r an integer) it is easy to write down a decomposition into irreducible subsets: for any of the $2^{d-1}-1$ unordered partitions $\{I,J\}$ of [d] into two nonempty sets, choose a number $r_{\{I,J\}}$ such that these numbers add up to r. This choice gives a natural map

\[ \prod_{\{I,J\}} (T^{|I|} \times T^{|J|})^{r_{\{I,J\}}} \to T^d\quad (f_{I,k},g_{J,k})_{\{I,J\},k} \mapsto\sum_{\{I,J\}} \sum_{k=1}^{r_{\{I,J\}}} f_{I,k} \otimes g_{J,k}\]

where $f_{I,k}$ and $g_{J,k}$ are tensors in the copies of V labelled by I and J, respectively. This parameterises the locus in $X_r$ of tensors with a partition rank at most r decomposition of a fixed type; this image is irreducible by virtue of Lemma 4.2.3 applied to the left-hand side above. The total number of components of $X_r$ that we find is the number of ways of partitioning r into $2^{d-1}-1$ nonnegative integers, which is polynomial in r.

Given the (almost) linear relation between partition rank and analytic rank [Reference Cohen and MoshkovitzCM23, Reference Moshkovitz and ZhuMZ22], it is natural to ask whether the number of components of $Y_r$ , too, is polynomial in r.

4.3 The vanishing ideal of a subset

We will prove Noetherianity by looking at functions that vanish identically on a subset.

We stress that conversely, since K is finite, X(V) is also the set of all common zeros of $I_X(V)$ in P(V).

4.4 Shifting and localising

Definition 2.6.1 can be extended to subsets of polynomial representations.

Definition 4.4.2. Given a subset X of P and a function $h \in K[P(0)]$ , we can think of h as a function on any P(V) via pullback along the linear map $P(V)\to P(0)$ , and hence also as a function on X(V). We define $X[1/h]$ as the functor

$V \mapsto X[1/h](V):=\{p \in X(V) \mid h(p) \neq 0\}.$

Clearly, $X[1/h]$ is a subset of P.

We will often combine a shift and a localisation: given a function $h\in K[P(U)]$ , we can think of h as a function on $K[(\mathrm{Sh}_U P)(0)]$ , and hence localise.

Example 4.4.3. Let $P:V \to V \otimes V$ and let X(V) be the set of tensors (matrices) of rank at most n in P(V). Let $h \in K[P(K^n)]$ be the $n \times n$ determinant and set $U:=K^n$ . Then $(\mathrm{Sh}_U X)[1/h]$ is isomorphic to the functor that sends V to $B \times V^{2n}$ , where $B:=X(U)[1/h]$ is the set of invertible $n \times n$ matrices, and the isomorphism $(\mathrm{Sh}_U X)[1/h](V)\to B \times V^{2n}$ comes from observing that

\[ (\mathrm{Sh}_U X)[1/h](V) \subseteq P(U \oplus V)=(K^n \otimes K^n) \times (K^n \otimes V) \times (V \otimes K^n) \times (V\otimes V), \]

and realising that the $V \otimes V$ component of a matrix of rank at most n is completely determined by its remaining three components, provided that the $K^n \otimes K^n$ component has nonzero determinant. This phenomenon, that X becomes an affine space up to shifting and localising, holds in greater generality, at least at a counting level (see Corollary 4.9.1).

4.5 Reduction to the prime field case

Now let P be a polynomial representation over K. Define a generic representation $P_F$ over F by setting, for a finite-dimensional F-vector space U, $P_F(U):=(P(K \otimes_F U))_F$ , and sending an F-linear map $\varphi:U \to V$ to the map $P_F(\varphi):=P(\operatorname{id}_K \otimes\varphi)$ , which is K-linear and therefore also F-linear. It is easy to see from the definitions that $P_F$ is polynomial of the same degree as P.

For a subset X of P, we define a subset of $P_F$ via $X_F(U):=X(K \otimes_F U)$ . If $X_1 \supseteq X_2 \supseteq \cdots$ is a chain of subsets in P, then $(X_1)_F \supseteq (X_2)_F \supseteq \cdots$ is a chain of subsets in $P_F$ . By assumption, the latter stabilises, say at $(X_{n_0})_F$ . Then it follows that, for any $n \geq n_0$ and any m,

\[ X_{n}(K^m)=X_{n}(K \otimes_F F^m)=(X_{n})_F(F^m)=(X_{n_0})_F(F^m)=X_{n_0}(K^m), \]

and this suffices to conclude that $X_n=X_{n_0}$ .

In view of Proposition 4.5.1, from now on we assume that K is a prime field. An important reason for this assumption is that we can then use Lemma 3.5.2. We believe that the proof below can be adapted to arbitrary finite fields, and this might actually give more general results. In particular, in the proof below we will act with the operators $F_{n+1,j}=F_{n+1,j}[1]$ ; and in the general case we would have to work with the operators $F_{n+1,j}[b]$ for $b \in \{1,\ldots,q-1\}$ . But the reasoning below is already rather subtle, and we prefer not to make it more opaque by the additional technicalities coming from nonprime fields.

4.6 The embedding theorem

We will prove Theorem 4.1.2 via an auxiliary result of independent interest. Let P be a polynomial representation of positive degree d and let R an irreducible subobject of $P_{>d-1}$ . Let $\pi:P\to P/R=:P'$ be the projection. Dually, this gives rise to an embedding $K[P/R] \subseteq K[P]$ . For a fixed $V \in \mathbf{Vec}$ , if we choose elements $y_1,\ldots,y_n \in P(V)^*$ that map to a basis of $R(V)^*$ , then we can write elements of K[P(V)] as reduced polynomials in $y_1,\ldots,y_n$ with coefficients that are elements of K[P’(V)]. We note, however, that R is typically not a direct summand of P. This implies, for instance, that when acting with $\operatorname{End}\nolimits(V)$ on $y_i$ , we typically do not stay within the linear span of the $y_1,\ldots,y_n$ but also get terms that are linear functions in K[P’(V)].

Let X be a subset of P, and let X’ be the image of X in $P/R$ , i.e., $X'(V):=\pi(X(V))$ (to simplify notation, we write $\pi$ instead of $\pi_V$ ).

Now there are two possibilities:

1. (1) $X=\pi^{-1}(X')$ , i.e., $X(V)=\pi^{-1}(X'(V))$ for all $V\!$ ; in this case, $I_X$ is generated by $I_{X'} \subseteq K[P'] \subseteq K[P]$ ;

2. (2) There exist a space V and an element $f \in I_X(V)$ such that f does not lie in $K[P] \cdot I_{X'}$ .

Here $(\mathrm{Sh}_U X)(V):=X(U \oplus V) \subseteq P(U \oplus V)=(\mathrm{Sh}_U P)(V)$ and $(\mathrm{Sh}_U X)[1/h]$ is the subset of $\mathrm{Sh}_U P$ consisting of points p where $h(p) \neq 0$ . A warning here is that h may actually vanish identically on $X(U) \subseteq \pi^{-1}(X'(U))$ , in which case the conclusion is trivial because $(\mathrm{Sh}_U X)[1/h]$ is empty. But in our application to the Noetherianity theorem, this will be irrelevant.

We will now first prove Theorem 4.1.2 using the embedding theorem. The proof of the embedding theorem itself is given in § 4.8.

4.7 Proof of Noetherianity from the embedding theorem

Proceeding by induction on P along the partial order from § 2.7, we may assume that Noetherianity holds for every representation $Q \prec P$ ; we call this the outer induction hypothesis.

Let d be the degree of P. If $d=0$ , then P(V) is a fixed finite set independent of V, and clearly any chain of subsets of this set stabilises. So we may assume that $d>0$ .

Let R be an irreducible subrepresentation in the subrepresentation $P_{>d-1}$ of P. Given a subset X of P, we write X’ for its projection in $P':=P/R$ .

We define $\delta_X \in \{1,2,\ldots,\infty\}$ as the minimal degree of a polynomial in $I_X \setminus K[P] \cdot I_{X'}$ ; this is $\infty$ if $I_X = K[P] \cdot I_{X'}$ .

For X,Y subsets of P we write $X > Y$ if we either have a strict inequality $X' \supsetneq Y'$ among the projections of X,Y in P’, or else $X'=Y'$ but $\delta_{X}>\delta_{Y}$ . Since, by the outer induction assumption, P’ is Noetherian, this is a well-founded partial order on subsets of P. To prove that a given subset $X \subseteq P$ is Noetherian, we may therefore assume that all subsets $Y \subseteq P$ with $Y<X$ are Noetherian; this is the inner induction hypothesis.

Now if $\delta_X=\infty$ , then any proper subset Y of X satisfies $Y<X$ , so we are done. We are therefore left with the case where $\delta_X\in {\mathbb Z}_{\geq 1}$ .

Then let $V \in \mathbf{Vec}$ and $f \in I_X(V) \setminus (K[P(V)] \cdot I_{X'}(V))$ be an element of degree $\delta_X$ . By the embedding theorem, there exists an element $h \in K[P(U)] \setminus I_{X}(U)$ of degree $<\delta_X$ such that $(\mathrm{Sh}_UX)[1/h] \to (\mathrm{Sh}_U P)/R$ is an injective map. Since $(\mathrm{Sh}_U P)/R \prec P$ by Lemma 2.7.3, $(\mathrm{Sh}_U X)[1/h]$ is Noetherian by the outer induction hypothesis.

Define Y as the subset of X defined by the vanishing of h. Explicitly,

\[ Y(V):=\{p \in X(V) \mid \forall \varphi \in \operatorname{Hom}\nolimits_{\mathbf{Vec}}(V,U):h(P(\varphi)p)=0\}. \]

Let $Y' \subseteq X'$ be the projection of Y in $P/R$ . If $Y' \subsetneq X'$ , then $Y<X$ and hence Y is Noetherian by the inner induction hypothesis. If $Y'=X'$ , then $h \in I_Y(U) \setminus (K[P(U)] \cdot I_{Y'}(U))$ and hence $\delta_Y \leq\deg(h) < \delta_X$ . So then, too, $Y<X$ , and Y is Noetherian by the inner induction hypothesis.

Now consider a chain

\[ X \supseteq X_1 \supseteq X_2 \supseteq \cdots \]

of subsets. By the above two paragraphs, from some point on both the chain $(X_i \cap Y)_i$ and the chain $((\mathrm{Sh}_UX_i)[1/h])_i$ have stabilised. We claim that then the chain $(X_i)_i$ has also stabilised.

Indeed, take $p \in X_i(W)$ . If $p \in X_i(W) \cap Y(W)$ , then also $p \in X_{i+1}(W) \cap Y(W)$ by the first chain, and we are done. If not, then let $\varphi:W \to U$ be a linear map such that $h(P(\varphi)p) \neq0$ . Let $\iota:W \to U \oplus W$ be the embedding $w \mapsto(\varphi(w),w)$ . Then we find that

\begin{align*} P(\iota)p \in X_i(U \oplus W)[1/h]&=(\mathrm{Sh}_U X_i)(W)[1/h]\\&=(\mathrm{Sh}_U X_{i+1})(W)[1/h]\subseteq X_{i+1}(U \oplus W).\end{align*}

Now if $\rho:U \oplus W \to W$ is the projection, then we find that $p=P(\rho)P(\iota)p \in P(\rho)X_{i+1}(U \oplus W)=X_{i+1}(W)$ , as desired.

4.8 Proof of the embedding theorem

Recall that P has degree $d>0$ , $X \subseteq P$ is a subset, R an irreducible subrepresentation of $P_{>d-1}$ , $\pi:P \to P/R$ is the projection, $X'=\pi(X)$ , $X \neq \pi^{-1}(X')$ , and $f \in I_X(V) \setminus (K[P(V)] \cdot I_{X'}(V))$ . Assume that f has degree $\delta$ . Recall from Lemma 2.8.2 that $V^* \mapsto K[P(V)]_{\leq \delta}$ is a polynomial representation. Furthermore, this has subrepresentations $V^* \mapsto I_X(V)_{\leq \delta}$ and $V^* \mapsto (K[P(V)] \cdot I_{X'}(V))_{\leq \delta}$ .

We may assume that $V=K^n$ and, without loss of generality, f is a weight vector. We will act on f with elements $g_{n+1,j}(s)^T$ (see Example 2.8.3 for an explanation of the transpose). The part of degree b in s is then captured by the operator $F_{n+1,j}[b]$ .

After acting repeatedly with operators $F_{n+1,j}[b]$ (for increasing values of n and possibly j and observing that this does not increase the degree of f), we may assume that the image of $f\in K[P(K^n)]$ in the quotient representation

\[ I_X(K^n)_{\leq \delta}/(K[P(K^n)] \cdot I_{X'}(K^n))_{\leq \delta} \]

is maximally spread out (see Proposition 3.4.3). After passing to a coordinate subspace, by Remark 3.5.4, this implies that the weight of f is $(1,\ldots,1)$ . Moreover, it implies that if we split, for any $j \in \{1,\ldots,n\}$ , $\tilde{f}:=F_{n+1,j} f$ as $\tilde{f}_0 +\tilde{f}_1$ where $\tilde{f}_0$ has weight $(1,\ldots,0,\ldots,1,1)$ and $\tilde{f}_1$ has weight $(1,\ldots,q-1,\ldots,1,1)$ , then $\tilde{f}_1$ vanishes identically on $X'(K^{n+1})$ ; indeed, otherwise $\tilde{f}_1$ would be a more spread-out polynomial that vanishes identically on X but not on X’.

Choose a basis $\mathbf{x}$ of $P'(K^n)^* \subseteq P(K^n)^*$ consisting of weight vectors, and extend this to a basis $\mathbf{x},\mathbf{y}$ of $P(K^n)^*$ of weight vectors. This means that $\mathbf{y}$ maps to a weight basis of $R(K^n)^*$ . Relative to these choices, we can write f as a reduced polynomial

(1) \begin{equation}f = \sum_\alpha f_\alpha(\mathbf{x}) \mathbf{y}^\alpha\end{equation}

for suitable exponent vectors $\alpha$ and nonzero functions $f_{\alpha} \in K[P'(K^n)]$ . We choose this expression minimal relative to $I_{X'}(K^n)$ in the following sense: no nonempty subset of the terms of any $f_{\alpha}$ add up to a polynomial in $I_{X'}(K^n)$ . This implies that no $f_{\alpha}$ is in the ideal of $I_{X'}(K^n)$ , but the requirement is a bit stronger than that.

Let $y_0$ be one of the elements in $\mathbf{y}$ that appears in f; we further choose $y_0$ such that the support in $\{1,\ldots,n\}$ of its weight is inclusionwise minimal. Consider the expression (coarser than (1))

\[ f=f_0(\mathbf{x},\mathbf{y}\setminus \{y_0\}) y_0^0 + \cdots +f_{c}(\mathbf{x},\mathbf{y} \setminus \{y_0\}) y_0^{c} \]

where every $f_e$ is a reduced polynomial in $\mathbf{x}$ and the variables in $\mathbf{y}$ except for $y_0$ , and where $f_{c} \neq 0$ and ${c} \in \{1,\ldots,q-1\}$ . Note that $f_0$ is a weight vector of the same weight as f. A priori, the coefficients $f_e$ with $e>0$ need not be weight vectors, since the weight monoid $(\{0,\ldots,q-1\}^n,\oplus)$ is not cancellative. However, all terms in $f_e$ have the same weight up to identifying 0 and $q-1$ , and upon adding e times the weight of $y_0$ to any of the weights of a term in $f_e$ (using the operation $\oplus$ ), one obtains the weight $(1,\ldots,1)$ of f.

We partition the variables $\mathbf{y}$ into three subsets: those whose weight has an entry 0 in position j are collected in the tuple $\mathbf{y}_0$ ; those with a 1 in position j in the tuple $\mathbf{y}_1$ ; and those with an entry greater than 1 there in the tuple $\mathbf{y}_{>1}$ .

We construct a weight basis of $P(K^{n+1})^*$ consisting of:

• – $\mathbf{x},\mathbf{y}_0,\mathbf{y}_1,$ and $\mathbf{y}_{>1}$ ;

• – the tuple $(n+1,j)\mathbf{y}_1$ obtained by applying $(n+1,j)$ to each variable in $\mathbf{y}_1$ (we suppress in this notation the polynomial generic representation in which these variables live);

• – the tuple $F_{n+1,j}\mathbf{y}_{>1}$ obtained by applying $F_{n+1,j}$ to each variable in the tuple $\mathbf{y}_{>1}$ ;

• – weight elements that together with $\mathbf{x}$ form a basis of $P'(K^{n+1})^*$ ; and

• – weight elements that along with $\mathbf{y}_0,\mathbf{y}_1,\mathbf{y}_{>1},(n+1,j)\mathbf{y}_1,F_{n+1,j}\mathbf{y}_{>1}$ project to a weight basis of $R(K^{n+1})^*$ .

The only nonobvious thing here is that the elements in $F_{n+1,j}\mathbf{y}_{>1}$ can be chosen as part of a set mapping to a basis of $R(K^{n+1})^*$ , and this follows from Lemma 3.5.2. Note that none of the variables in the tuples $(n+1,j)\mathbf{y}_1$ and $F_{n+1,j}\mathbf{y}_{>1}$ has a $q-1$ in position j of its weight.

Either $y_0$ belongs to $\mathbf{y}_1$ or to $\mathbf{y}_{>1}$ . In the first case we define $y_1:=(n+1,j)y_0$ , and in the second case we define $y_1:=F_{n+1,j}y_0$ . In both cases, $y_1$ is the (nonzero) weight- $(a_1,\ldots,a_j-1,\ldots,a_n,1)$ component of $F_{n+1,j}y_0$ and one of the chosen variables. (In the first case, this uses Lemma 3.4.1.)

Consider

(2) \begin{equation} g_{n+1,j}(s)^T f = \sum_{e=0}^{c} (f_e(g_{n+1,j}(s)^T\mathbf{x},g_{n+1,j}(s)^T (\mathbf{y}\setminus \{y_0\})))(g_{n+1,j}(s)^T y_0)^e,\end{equation}

where on both sides we have suppressed the polynomial generic representations where f and the coordinate tuples live (see Lemma 2.8.2 and Example 2.8.3). From the term with $e=c$ we get a contribution

(3) \begin{equation} c \cdot (f_{c} + F_{n+1,j}[q-1] f_{c}) \cdot s \cdot y_0^{{c}-1} \cdot y_1.\end{equation}

There is no cancellation in the $+$ in (3): the weights of monomials in $f_{c}$ are distinct from the weights of monomials in $F_{n+1,j}[q-1]f_{c}$ , because the latter have a positive entry on position $n+1$ . Furthermore, we have $0<c<q$ and q is prime, so the term $c \cdot f_{c} \cdot s \cdot y_0^{c-1} \cdot y_1$ is nonzero.

Rewriting (2) as a reduced polynomial in $s, y_0,y_1$ with coefficients that are reduced polynomials in the remaining chosen variables in $P(K^{n+1})^*$ , we claim that the coefficient of $s \cdot y_0^{c-1} \cdot y_1$ is precisely that in (3). Indeed, $y_0,y_1$ only appear in the terms $y_0=F_{n+1,j}[0] y_0$ and $F_{n+1,j} y_0$ from $g_{n+1,j}(s)^Ty_0$ and nowhere in $f_e(g_{n+1,j}(s)^T \mathbf{x}, g_{n+1,j}(s)^T(\mathbf{y}\setminus\{y_0\}))$ or in $F_{n+1,j}[b]y_0$ with $b>0$ because:

• – $g_{n+1,j}(s)^T$ maps the coordinates $\mathbf{x}$ into linear combinations of $\mathbf{x}$ and the further chosen variables in $P'(K^{n+1})^*$ ;

• – $y_0,y_1$ do not appear in $F_{n+1,j}[b] \mathbf{y}$ for $b>1$ for weight reasons; expressing the elements in the latter tuple on the basis of the chosen variables, all variables have weights with a $b>1$ at position $n+1$ , while $y_0,y_1$ have a 0 and 1 there, respectively;

• – $y_0$ does not appear in $F_{n+1,j} y$ for any variable y in $\mathbf{y}$ , again by comparing the weights in position $n+1$ ;

• – $y_1$ is different from all variables $F_{n+1,j}y$ where y ranges over the variables in $\mathbf{y}_{>1}$ (other than $y_0$ , if $y_0$ is in $\mathbf{y}_{>1}$ );

• – $y_1$ is different from all variables $(n+1,j)y$ where y ranges over the variables in $\mathbf{y}_1$ (other than $y_0$ , if $y_0$ is in $\mathbf{y}_1$ ); and indeed,

• – $y_1$ does not appear in the weight component $y'=F_{n+1,j}y - (n+1,j) y$ of any variable y in $\mathbf{y}_1$ . Indeed, if $(a_1',\ldots,1,\ldots,a_n')$ is the weight of y, then y’ has weight $(a_1',\ldots,q-1,\ldots,a_n',1)$ . But, as remarked earlier, the variable $y_1$ constructed from $y_0$ does not have a $q-1$ on position j in its weight.

This proves the claim. We conclude that, when writing $\tilde{f}=F_{n+1,j}f$ as a polynomial in all of the chosen variables, the terms divisible by $y_0^{{c}-1} y_1$ are precisely those in

\[ {c} \cdot (f_{c} + F_{n+1,j}[q-1]f_{c}) \cdot y_0^{{c}-1} y_1.\]

Now in $f_{c}$ , expanded as a reduced polynomial in $\mathbf{y} \setminus\{y_0\}$ with coefficients that are reduced polynomials in $\mathbf{x}$ , consider any nonzero term $\ell(\mathbf{x}) \cdot (\mathbf{y} \setminus \{y_0\})^\alpha$ .

Group the terms in $\ell(\mathbf{x})$ into two parts: $\ell(\mathbf{x})=\ell_0(\mathbf{x}) +\ell_1(\mathbf{x})$ , in such a manner that $\ell_0(\mathbf{x})\cdot (\mathbf{y} \setminus\{y_0\})^\alpha y_0^{{c}-1} y_1$ is the part of $\ell(\mathbf{x})\cdot (\mathbf{y}\setminus \{y_0\})^{\alpha} y_0^{{c}-1} y_1$ that has weight $(1,\ldots,0,\ldots,1,1)$ and hence is part of $({1}/{c})\tilde{f}_0$ ; and $\ell_1(\mathbf{x})\cdot (\mathbf{y}\setminus \{y_0\})^{\alpha}y_0^{{c}-1} y_1$ has weight $(1,\ldots,q-1,\ldots,1,1)$ and hence is part of $({1}/{c}) \tilde{f}_1$ .

For the same exponent vector $\alpha$ , $F_{n+1,j}[q-1]f_c$ may also contain a term $\tilde{\ell} \cdot (\mathbf{y} \setminus \{y_0\})^\alpha$ , where $\tilde{\ell}$ is a polynomial in $\mathbf{x}$ and the remaining chosen coordinates on $P'(K^{n+1})$ . We can similarly decompose $\tilde{\ell}=\tilde{\ell}_0+\tilde{\ell}_1$ , where $\tilde{\ell}_0 \cdot (\mathbf{y}\setminus \{y_0\})^\alpha y_0^{c-1} y_1$ is part of $({1}/{c}) \tilde{f}_0$ and $\tilde{\ell}_1 \cdot (\mathbf{y} \setminus\{0\})^\alpha y_0^{c-1} y_1$ is part of $({1}/{c}) \tilde{f}_1$ .

Since $\tilde{f}_1$ vanishes identically on $\pi^{-1}(X'(K^{n+1}))$ (this was the point of choosing f such that its image in a suitable representation is maximally spread out) we find that $\ell_1(\mathbf{x})+\tilde{\ell}_1$ vanishes identically on $X'(K^{n+1})$ . Furthermore, since the weight of every monomial in $F_{n+1,q}[q-1]f_c$ has a positive entry in position $n+1$ and the weight of $(\mathbf{y} \setminus \{y_0\})^\alpha$ has a zero in that position, the weight of every term in $\tilde{\ell}_1$ has a positive entry in position $n+1$ . This implies that $\tilde{\ell}_1$ vanishes identically on the linear subspace $P'(K^n) \subseteq P'(K^{n+1})$ . We conclude that $\ell_1(\mathbf{x})$ vanishes identically on $X'(K^n)$ . However, by minimality of the expression for f relative to $I_{X'}(K^n)$ , no nonempty subset of the terms of $\ell(\mathbf{x})$ add up to a polynomial that vanishes identically on $X'(K^n)$ . So we have $\ell_1(\mathbf{x})=0$ and $\ell(\mathbf{x})=\ell_0(\mathbf{x})$ . It follows that $\ell(\mathbf{x})(\mathbf{y}\setminus \{y_0\})^\alpha y_0^{{c}-1} y_1$ is a weight vector of weight $(1,\ldots,0,\ldots,1,1)$ . Since the term $\ell(\mathbf{x})(\mathbf{y} \setminus \{y_0\})^\alpha$ in $f_{c}(\mathbf{x},\mathbf{y} \setminus \{y_0\})$ was arbitrary, we find that $f_{c} y_0^{{c}-1} y_1$ is a weight vector of weight $(1,\ldots,0,\ldots,1,1)$ . As the weight of $y_0$ has a positive entry in position j, we find that ${c}-1=0$ and all weights appearing in $f_{c}$ have a 0 in position j.

Now j was arbitrary in the support of the weight of $y_0$ , so the weights appearing in $f_{c}$ all have disjoint support from that of $y_0$ . But the only way, in the weight monoid $\{0,1,\ldots,q-1\}^n$ , to obtain the weight $(1,\ldots,1)$ as a $\oplus$ -sum of two weights with disjoint supports is if, after a permutation, one weight is $(1^m,0^{n-m})$ and the other weight is $(0^m,1^{n-m})$ . Hence $f_{c}$ is a weight vector that, after that permutation, has the former weight, and then $y_0$ has the latter.