Separating invariants for certain representations of the elementary Abelian $p$-groups of rank two

1.   Introduction

Let $\rho:G\hookrightarrow \text{GL}(n, \mathbb{F})$ be a faithful representation of a finite group $G$ over a field $\mathbb{F}$ of arbitrary characteristic. Denote by $V = \mathbb{F}^n$ the $n$ dimensional representation space over $\mathbb{F}$. We write $\mathbb{F}[V]$ for the symmetric algebra $S(V^*)$ over the dual space $V^*$. The action of $G$ on $V$ induces an action on $\mathbb{F}[V]$: for $f\in \mathbb{F}[V]$ and $\textbf{v}\in V$, the action of $g\in G$ is given by $(g(f))(\textbf{v}) = f(g^{-1}(\textbf{v}))$. The ring of invariants $\mathbb{F}[V]^G$ is defined by

Here are some general methods to construct invariants of finite groups. Let $f\in \mathbb{F}[V]$, then the transfer of $f$ is defined by

Let $H\leq G$ be a subgroup. Then the relative transfer is defined as

where $G/H$ denotes a set of left coset representatives of $H$ in $G$. $\text{Tr}^G_H$ is independent of the choice of the coset representatives. The norm of $f$ is defined by

Note that the transfer, relative transfer and norm are invariant polynomials. If the characteristic of $\mathbb{F}$ divides the group order $|G|$, we speak of the modular case. Otherwise, we are in the nonmodular case, which includes $\text{char}(\mathbb{F}) = 0$.

Separating orbits of a group action on some geometric or algebraic space is likely to have been one of the original motivations of invariant theory. It has regained particular attention following the influential textbook of Derksen and Kemper ^[1]. Since then, separating invariants have been extensively studied within the last decade.

Definition 1. A subset $S\subseteq \mathbb{F}[V]^G$ is said to be separating if for any two points $\textbf{v}, \textbf{v}^{\prime}\in V$, we have: If there exists an invariant $f\in \mathbb{F}[V]^G$ with $f(\textbf{v})\neq f(\textbf{v}^{\prime})$, then there exists an element $h\in S$ with $h(\textbf{v})\neq h(\textbf{v}^{\prime})$.

If $G$ is finite, then for $\textbf{v}, \textbf{v}^{\prime}\in V$ with distinct $G$-orbits, there exists $f\in \mathbb{F}[V]^G$ such that $f(\textbf{v})\neq f(\textbf{v}^{\prime})$. It follows that a subset $S\subseteq \mathbb{F}[V]^G$ is separating if any two $G$-orbits can be separated by invariants from $S$ for any finite group ^[2]. While the ring of invariants forms a separating set, computing the ring of invariants for a modular representation is typically a difficult problem. Moreover, separating invariants are better behaved than generating ones. For instance, the Noether degree bound and Weyl theorem hold for separating invariants without any hypothesis on $\text{char}(\mathbb{F})$, see ^[2,3]. In ^[4], Dufresne introduced a geometric notion of separating algebra and gave two geometric formulations of this notion. Geometric separating sets and separating invariants over finite fields were considered in ^[5,6]. For more background on separating invariants we direct readers to ^{[7,8,9,10,11,12,13]}.

In the study of explicit separating invariants, it is natural to take $p$-groups as a starting point. The work of Sezer ^[14] gives us a good understanding in the case of the cyclic group of order $p$. Since then, explicit separating invariants have also been calculated for various groups such as cyclic $p$-groups, the Klein four group, etc ^{[15,16,17,18,19]}. The next step is to look at elementary abelian $p$-groups. With a few notable exceptions, the modular representation theory of an elementary abelian $p$-group is wild, see for example [20, Theorem 4.4.4]. In the modular case, the degrees of the generators can become arbitrarily big. Therefore, computing the invariants of elementary abelian $p$-groups in the modular case is particularly difficult and explicit generating sets are available only for a handful of cases. The ring of invariants for all two dimensional representations of $(\mathbb{Z}/p)^r$ and the ring of invariants for all three dimensional representations of $(\mathbb{Z}/p)^2$ have been worked out in ^[21]. See also ^[22] for further research. Four families of finite dimensional representations of $(\mathbb{Z}/p)^2$ over an algebraically closed field $\mathbb{F}$ of characteristic $p$, where $p$ is an odd prime, is given in ^[23] and their invariant rings have not been computed. In this paper, we give explicit separating sets, including transfers and norms for each representation. Transfers and norms are basic invariants that are easier to obtain. These invariants usually do not suffice to generate the entire ring of invariants $\mathbb{F}[V]^G$ in the modular case. Since the dual of a subrepresentation sits in the dual of higher dimensional representation of $(\mathbb{Z}/p)^2$, this allows us to reduce the problem to separating two points whose coordinates are all the same except a few coordinates. Consequently, we show that the separating set for a representation of $(\mathbb{Z}/p)^2$ can be obtained by adding some transfers and norms to any separating set for the subrepresentation. It is worth pointing out that the size of the separating set depends only on the dimension of the representation. Our work can be viewed as the generalization of the Klein four group (the elementary abelian $2$-groups of rank two) ^[17] to the elementary abelian $p$-groups of rank two for arbitrary odd prime $p$. However, the latter case needs more complicated computation and additional separating invariants.

2.   Constructing separating invariants

Let $G = \langle\sigma_1, \sigma_2\rangle\cong (\mathbb{Z}/p)^2$ be the elementary abelian $p$-group of rank two of order $p^2$, where $p$ is an odd prime. Let $\sigma_3 = \sigma_1\sigma_2$ and $H_i$ denote the subgroup of $G$ which is generated by $\sigma_i$ for $1\leq i\leq3$. The complete list of indecomposable representations of the Klein four group is described in [20, Theorem 4.3.3]. However, the modular representation theory of an elementary abelian $p$-group of rank two is wild. Here, we study the natural generalization of the irreducible representations of the Klein four group. There are four families of finite dimensional representations of $G$ over an algebraically closed field $\mathbb{F}$ of characteristic odd prime $p$, which are given in ^[23]. For each representation in each family, we construct a finite separating set recursively. In the following, $I_n$ denotes the $n\times n$ identity matrix and for any element $\lambda$ of the field $\mathbb{F}$, $J_\lambda$ denotes the $n\times n$ Jordan block (lower triangular) with eigenvalues $\lambda$.

Type $(\text{I})$ For every even dimension $2n$ there are representations $V_{2n, \lambda}$,

Type $(\text{II})$ For every even dimension $2n$ there are representations $V_{2n, \infty}$,

Type $(\text{III})$ For every odd dimension $2n-1$ there are representations $V_{-(2n-1)}$,

Type $(\text{IV})$ For every odd dimension $2n-1$ there are representations $V_{2n-1}$,

Notice that the matrix group associated with $V_{2n, \infty}$ in type $(\text{II})$ is the same as the matrix group associated with $V_{2n, 0}$ in type $(\text{I})$. Therefore, their invariant rings are equal, and a separating set for $V_{2n, 0}$ is also a separating set for $V_{2n, \infty}$. Each representation $V_{-(2n-1)}$ in type $(\text{III})$ is isomorphic to a subrepresentation of $V_{2n, p-1}$ in type $(\text{I})$, which we will explain in detail later. So we study the separating sets for types $(\text{I})-(\text{III})$ in Subsection $2.1$ and for type $(\text{IV})$ in Subsection $2.2$.

2.1.   Separating invariants for types $(\text{I})$–$(\text{III})$

We start with the action of $G$ on the representation space $V_{2n, \lambda}$. Let $\boldsymbol{\varepsilon_1}$, $\boldsymbol{\varepsilon_2}$, $\cdots$, $\boldsymbol{\varepsilon_n}$, $\boldsymbol{\xi_1}$, $\boldsymbol{\xi_2}$, $\cdots$, $\boldsymbol{\xi_n}$ be the basis for $V_{2n, \lambda}$ with $\sigma_{1}(\boldsymbol{\varepsilon_i}) = \boldsymbol{\varepsilon_i}+\boldsymbol{\xi_i}$, $\sigma_{1}(\boldsymbol{\xi_i}) = \boldsymbol{\xi_i}$, $\sigma_{2}(\boldsymbol{\xi_i}) = \boldsymbol{\xi_i}$ for $1\leq i\leq n$, $\sigma_{2}(\boldsymbol{\varepsilon_n}) = \boldsymbol{\varepsilon_n}+\lambda\boldsymbol{\xi_n}$ and $\sigma_{2}(\boldsymbol{\varepsilon_i}) = \boldsymbol{\varepsilon_i}+\lambda\boldsymbol{\xi_i}+\boldsymbol{\xi_{i+1}}$ for $1\leq i\leq n-1$. We identify each $\boldsymbol{\varepsilon_i}$ with the column vector with $1$ on the $i$-th coordinate and zero elsewhere, and each $\boldsymbol{\xi_i}$ with the column vector with $1$ on the $(n+i)$-th coordinate and zero elsewhere. Let $x_1, x_2, \cdots, x_n, y_1, y_2, \cdots, y_n$ denote the corresponding elements in the dual space $V^*_{2n, \lambda}$. In fact, $x_1, x_2, \cdots, x_n, y_1, y_2, \cdots, y_n$ form the basis for $V^*_{2n, \lambda}$ in the reverse order: we have $\sigma^{-1}_1(x_i) = x_i$, $\sigma^{-1}_1(y_i) = x_i+y_i$, $\sigma^{-1}_2(x_i) = x_i$ for $1\leq i\leq n$, $\sigma^{-1}_2(y_1) = \lambda x_1+y_1$ and $\sigma^{-1}_2(y_i) = x_{i-1}+\lambda x_i+y_i$ for $2\leq i\leq n$. For simplicity we will use the generators $\sigma^{-1}_i$ instead of $\sigma_i$ for the rest of the paper and change the notation by writing $\sigma_{i}$ for the new generators for $1\leq i\leq 3$. Note also that $\mathbb{F}[V_{2n, \lambda}] = \mathbb{F}[x_1, x_2, \cdots, x_n, y_1, y_2, \cdots, y_n]$. Pick a point $\textbf{v} = (v_1, v_2, \cdots, v_n, w_1, w_2, \cdots, w_n)$ in $V_{2n, \lambda}$. The surjection $\varphi:V_{2n, \lambda}\rightarrow V_{2n-2, \lambda}$ given by $(v_1, v_2, \cdots, v_n, w_1, w_2, \cdots, w_n)\mapsto (v_1, v_2, \cdots, v_{n-1}, w_1, w_2, \cdots, w_{n-1})$ is $G$-equivariant, as for $g\in G$, $\textbf{v}\in V_{2n, \lambda}$, $g(\varphi(\textbf{v})) = \varphi(g(\textbf{v}))$. Dual to this surjection, the subspace in $V^*_{2n, \lambda}$ generated by $x_1, x_2, \cdots, x_{n-1}, y_1, y_2, \cdots, y_{n-1}$ is closed under the $G$-action and isomorphic to $V^*_{2n-2, \lambda}$. Hence $\mathbb{F}[V_{2n-2, \lambda}] = \mathbb{F}[x_1, x_2, \cdots, x_{n-1}, y_1, y_2, \cdots, y_{n-1}]$ is a subalgebra in $\mathbb{F}[V_{2n, \lambda}]$.

The following three lemmas are very useful in studying the image of the transfer for modular groups. We will use these formulas repeatedly in the proofs of Lemmas 4–6.

Lemma 1. Let $k$ be a positive integer. Then $\sum_{0\leq l\leq p-1}l^k\equiv -1\; \text{mod}\; p$ if $p-1$ divides $k$ and $\sum_{0\leq l\leq p-1}l^k\equiv 0\; \text{mod}\; p$, otherwise.

Proof. See [24, Lemma 9.0.2] for a proof for this statement. □

Lemma 2. Let $k$ and $l$ be positive integers such that $0\leq k\leq p-1$, $k\leq l\leq p-1$. There holds

Proof. This statement can be proved by induction on $k$ and $l$ and we omit the detailed proof. □

Lemma 3. (1) Let $k$ be a positive integer such that $1\leq k\leq p$. Then

(2) Let $l$ and $k$ be a positive integer such that $2\leq l\leq p$ and $1\leq k\leq l-1$. Then

Proof. It is a simple matter to prove the two identities above by the definition of binomial coefficient. □

From now on all congruences are modulo $\mathbb{F}[x_1, x_2, \cdots, x_n, y_1, y_2, \cdots, y_{n-1}]$ in Subsection $2.1$. The congruences of separating invariants in the following two lemmas will play an important part in the proof of Theorem 1.

Lemma 4. (1) ${\rm Tr}^G(y^{p-1}_iy^{p-1}_jy_n)\equiv (x_{i-1}x_j-x_ix_{j-1})^{p-1}y_n$ for $2\leq i, j\leq n-1$.

(2) ${\rm Tr}^G(y^{p-1}_1y^{p-1}_2y_n)\equiv x^{2(p-1)}_1y_n$.

Proof. Here we only prove for $(1)$. It is easy to verify that $\text{Tr}^G = \text{Tr}^G_{H}\circ \text{Tr}^{H}$ for any subgroup $H$ of $G$. This suggest that we may compute $\text{Tr}^G$ by first computing $\text{Tr}^{H}$ and then computing $\text{Tr}^G_{H}$. Thus we may work with the two smaller groups $H$ and $G/H$.

By the definition of transfer we have

By Lemma 1, we see that

The last congruence follows since ${p-1\choose k}^2\equiv 1\; \text{mod}\; p$. Similarly, for each $k$ with $0\leq k\leq p-1$ we have

Thus

It follows from Lemma 2 that the coefficient of $x^{p-1-l}_{i-1}x^l_ix^{m-1}_{j-1}x^{p-m}_jy_n$ is

Moreover, ${p\choose l-(m-1)}\equiv 1\; \text{mod}\; p$ if $l = m-1$ and ${p\choose l-(m-1)}\equiv 0\; \text{mod}\; p$, otherwise. From the above it follows that

□

Lemma 5. (1) ${\rm Tr}^G(y^{(p+1)(p-1)}_{n-1}y_n)\!\equiv\!\sum_{1\leq k\leq p}(x_{n-2}\!+\!\lambda x_{n-1})^{(p+1-k)(p-1)}x^{k(p-1)}_{n-1}y_n$.

(2) ${\rm N}^{H_3}(x_{n-1}y_n-x_ny_{n-1})\equiv x^p_{n-1}y^p_n-x_{n-1}(x^2_{n-1}-x_{n-2}x_n)^{p-1}y_n.$

Proof. $(1)$ By the definition of transfer, we have

By Lemma 1 and Lemma 3(1) we see that

For each $k$ with $1\leq k\leq p+1$,

By Lemma 1 and Lemma 3(1) we have

For $1\leq k\leq p$, we see that ${(p+1-k)(p-1)\choose s(p-1)}\equiv 0\; \text{mod}\; p$ unless $s = p+1-k$ by Lemma 3(2) in which case we have

For $k = p+1$, $\text{Tr}^{H_2}(x^{k(p-1)}_{n-1}y^{(p+1-k)(p-1)}_{n-1}y_n)\equiv \sum_{0\leq l\leq p-1}x^{(p+1-k)(p-1)}_{n-1}y_n\equiv 0$. Thus,

$(2)$ Note that $x_{n-1}y_n-x_ny_{n-1}$ is $\sigma_1$-invariant, so the $H_3$-orbit product of this polynomial is $G$-invariant. Thus, we have

□

Theorem 1. Let $\mathbb{F}[V_{2n, \lambda}] = \mathbb{F}[x_1, x_2, \cdots, x_n, y_1, y_2, \cdots, y_n]$. Then

is a separating set for $V_{2, \lambda}$. And $S_2 = S_1\bigcup T_2$ is a separating set for $V_{4, \lambda}$, where

Let $n\geq3$ and $S_{n-1}\subseteq \mathbb{F}[V_{2n-2, \lambda}]^G$ be a separating set for $V_{2n-2, \lambda}$. Then $S_n = S_{n-1}\bigcup T_n$ is a separating set for $V_{2n, \lambda}$, where

Moreover, a separating set for $V_{2n, 0}$ is a separating set for $V_{2n, \infty}$.

Proof. The cases $V_{2, \lambda}$ and $V_{4, \lambda}$ are easy to check so we only prove the case of $n$. Consider any two points $\textbf{v} = (v_1, \cdots, v_n, w_1, \cdots, w_n), \; \textbf{v}^{\prime} = (v^{\prime}_1, \cdots, v^{\prime}_n, w^{\prime}_1, \cdots, w^{\prime}_n)\in V_{2n, \lambda}$ that do not lie in the same $G$-orbit, and suppose that $f(\textbf{v}) = f(\textbf{v}^{\prime})$ for all $f\in S_n$. We will show that there exists $g\in G$ such that $\textbf{v}^{\prime} = g(\textbf{v})$. This contradicts our assumption that $\textbf{v}$ and $\textbf{v}^{\prime}$ do not lie in the same $G$-orbit and this contradiction shows that $S_n$ is a separating set for $V_{2n, \lambda}$. We assume that every invariant in $S_n$ takes the same value on $\textbf{v}$ and $\textbf{v}^{\prime}$ from now on.

If $(v_1, \cdots, v_{n-1}, w_1, \cdots, w_{n-1}), \; (v^{\prime}_1, \cdots, v^{\prime}_{n-1}, w^{\prime}_1, \cdots, w^{\prime}_{n-1})\in V_{2n-2, \lambda}$ do not lie in the same $G$-orbit, then there exists a polynomial in $S_{n-1}$ that separates the two points because $S_{n-1}\subseteq \mathbb{F}[V_{2n-2, \lambda}]^G$ is separating. Therefore this polynomial separates $\textbf{v}$ and $\textbf{v}^{\prime}$ as well. Hence by replacing $\textbf{v}^{\prime}$ with a suitable element in its $G$-orbit we may assume that $v^{\prime}_i = v_i$ and $w^{\prime}_i = w_i$ for $1\leq i\leq n-1$. Since $x_n\in T_n$, we may assume that $v^{\prime}_n = v_n$. Note that with this assumption we must have $w^{\prime}_n\neq w_n$.

First, by Lemma 4(2), $\text{Tr}^G(y^{p-1}_1y^{p-1}_2y_n)(\textbf{v}) = \text{Tr}^G(y^{p-1}_1y^{p-1}_2y_n)(\textbf{v}^{\prime})$ implies

As $w^{\prime}_n\neq w_n$, we obtain

Similarly, $\text{Tr}^G(y^{p-1}_iy^{p-1}_{i+1}y_n)(\textbf{v}) = \text{Tr}^G(y^{p-1}_iy^{p-1}_{i+1}y_n)(\textbf{v}^{\prime})$ implies

for $2\leq i\leq n-2$ by setting $j = i+1$ in Lemma 4 (1). Since $v_1 = 0$ and $w^{\prime}_n\neq w_n$, we get

successively. Since $v_{n-2} = 0$, $\text{Tr}^G(y^{(p+1)(p-1)}_{n-1}y_n)(\textbf{v}) = \text{Tr}^G(y^{(p+1)(p-1)}_{n-1}y_n)(\textbf{v}^{\prime})$ implies

by Lemma 5(1). Notice that $\lambda^{p\!-\!1}(\lambda^{p-1}-1)^{p-1} = 0$ if and only if $\lambda\in \mathbb{F}_p$, so the following proof falls into two parts depending on whether $\lambda$ is in $\mathbb{F}_p$ or not.

If $\lambda\notin \mathbb{F}_p$, then we have

by (2.1). As $\text{N}^G(y_n) = \prod_{0\leq k, l\leq p-1}(lx_{n-1}+(l\lambda+k)x_n+y_n)$, we have $\text{N}^G(y_n)(\textbf{v}) = \prod_{0\leq k, l\leq p-1}((l\lambda+k)v_n+w_n)$. We define a polynomial

in $\mathbb{F}[X]$. Notice that $\text{N}^G(y_n)(\textbf{v}) = P(w_n)$ and that $P(w_n) = P(w_n+(l\lambda+k)v_n)$ for all $0\leq k, l\leq p-1$. Since $P(X)$ is a polynomial of degree $p^2$, it follows that $w_n+(l\lambda+k)v_n$ for $0\leq k, l\leq p-1$ are the only solutions of $P(X)-P(w_n) = 0$. Therefore the equality of $\text{N}^G(y_n)(\textbf{v}^{\prime}) = P(w^{\prime}_n)$ and $\text{N}^G(y_n)(\textbf{v}) = P(w_n)$ implies $w_n$ must be equal to $w^{\prime}_n+(l\lambda+k)v_n$ for some $0\leq k, l\leq p-1$. Hence $\textbf{v}^{\prime} = \sigma^k_1\sigma^l_2(\textbf{v})$. This is a contradiction because $\textbf{v}$ and $\textbf{v}^{\prime}$ lie in the same $G$-orbit.

Next we turn to the case $\lambda\in \mathbb{F}_p$. Since $v_{n-2} = 0$, then $\text{N}^{H_3}(x_{n-1}y_n-x_ny_{n-1})$ taking the same value on $\textbf{v}, \textbf{v}^{\prime}$ implies

by Lemma 5(2). If $v_{n-1}\neq0$, then we have $(w_n-w^{\prime}_n)^{p-1}-v^{p-1}_{n-1} = 0$, i.e. $(w_n-w^{\prime}_n)^{p-1} = v^{p-1}_{n-1}$, i.e. $w_n-w^{\prime}_n = lv_{n-1}$ for some $1\leq l\leq p-1$. There must exist $k$ with $0\leq k\leq p-1$ such that $l\lambda+k = 0$. Hence $\textbf{v}^{\prime} = \sigma^k_1\sigma^l_2(\textbf{v})$. This is a contradiction. So now assume $v_{n-1} = 0$. Then $\text{N}^G(y_n)(\textbf{v}) = \text{N}^G(y_n)(\textbf{v}^{\prime})$ implies $(\prod_{0\leq l\leq p-1}(l v_n+w_n))^p = (\prod_{0\leq l\leq p-1}(l v_n+w^{\prime}_n))^p$. Thus $0 = (\prod_{0\leq l\leq p-1}(l v_n+w_n))^p-(\prod_{0\leq l\leq p-1}(l v_n+w^{\prime}_n))^p = (\prod_{0\leq l\leq p-1}(l v_n+w_n)-\prod_{0\leq l\leq p-1}(l v_n+w^{\prime}_n))^p$ and therefore $w_n = w^{\prime}_n+kv_n$ for some $1\leq k\leq p-1$. Hence $\textbf{v}^{\prime} = \sigma^k_1(\textbf{v})$. This is also a contradiction.

The final statement follows because the matrix group associated with $V_{2n, \infty}$ is the same as the matrix group associated with $V_{2n, 0}$, so their invariant rings are equal, and a separating set for $V_{2n, 0}$ is also a separating set for $V_{2n, \infty}$. □

Remark 1. From the proof of Theorem 1, we see that the separating set for each representation $V_{2n, \lambda}$ we obtained is minimal. Moreover, the size of separating set for $V_{2n, \lambda}$ is $\frac{n(n+3)}{2}$, which only depends on the dimension of the representation. Nevertheless, the maximal degree of an invariant in this set is the group order $p^2$.

Since $V_{-(2n-1)}$ is isomorphic to the submodule of $V_{2n, p-1}$ spanned by $\boldsymbol{\varepsilon_1}$, $\cdots$, $\boldsymbol{\varepsilon_{n-1}}$, $\boldsymbol{\xi_1}$, $\cdots$, $\boldsymbol{\xi_n}$, where $\boldsymbol{\varepsilon_1}$, $\cdots$, $\boldsymbol{\varepsilon_n}$, $\boldsymbol{\xi_1}$, $\cdots$, $\boldsymbol{\xi_n}$ is the basis for $V_{2n, p-1}$. Dual to this inclusion, there is a restriction map $\mathbb{F}[V_{2n, p-1}]^G\rightarrow \mathbb{F}[V_{-(2n-1)}]^G, \; f\mapsto f|_{V_{-(2n-1)}}$ which sends separating sets to separating sets by [1, Theorem 2.4.9]. Therefore, in view of Theorem 2.1, we have the following statement.

Corollary 1. Let

and

Then

is a separating set for $V_{-1}$. Additionally,

is a separating set for $V_{-3}$. Let $n\geq3$ and $S_{n-1}\subseteq \mathbb{F}[V_{2n-2, p-1}]^G$ be a separating set for $V_{2n-2, p-1}$. Then the polynomials in $S_n = S_{n-1}\bigcup T_n$ restricted to $V_{-(2n-1)}$ form a separating set for $V_{-(2n-1)}$, where

2.2.   Separating sets for type $(\text{IV})$

We consider type $(\text{I})$ representations $V_{2n, p-1}$. In view of $\langle \boldsymbol{\xi_1}\rangle\subset V_{2n, p-1}$ is a $G$-submodule, we have $V_{2n-1}\cong V_{2n, p-1}/\langle \boldsymbol{\xi_1}\rangle$ with basis $\boldsymbol{\tilde{\varepsilon}_i}: = \boldsymbol{\varepsilon_i}+\langle \boldsymbol{\xi_1}\rangle$ for $1\leq i\leq n$, $\boldsymbol{\tilde{\xi}_i}: = \boldsymbol{\xi_i}+\langle \boldsymbol{\xi_1}\rangle$ for $2\leq i\leq n$, and a $G$-algebra inclusion $\mathbb{F}[V_{2n-1}] = \mathbb{F}[x_1, \cdots, x_n, y_2, \cdots, y_n]\subset \mathbb{F}[V_{2n, p-1}]$. The action of $\sigma_1$ and $\sigma_3$ on the variables are given by

and

Pick a point $(v_1, \cdots, v_n, w_2, \cdots, w_n)$ in $V_{2n-1}$. There is a $G$-equivariant surjection $V_{2n-1}\rightarrow V_{2n-3}$ given by

Hence

is a subalgebra in $\mathbb{F}[V_{2n-1}]$.

Note that all congruences are modulo $\mathbb{F}[x_1, \cdots, x_n, y_2, \cdots, y_{n-1}]$ in subsection $2.2$.

Lemma 6. (1) ${\rm Tr}^G(y^{p-1}_iy^{p-1}_jy_n)\equiv (x_{i-1}x_j-x_ix_{j-1})^{p-1}y_n$ for $2\leq i, j\leq n-1$.

(2) ${\rm Tr}^G(y^{(p+1)(p-1)}_iy_n)\equiv x^{p-1}_{i-1}x^{p-1}_i(x^{p-1}_{i-1}-x^{p-1}_i)^{p-1}y_n$ for $2\leq i\leq n-1$.

(3) ${\rm N}^{H_3}(x_{n-1}y_n-x_ny_{n-1})\equiv x^p_{n-1}y^p_n-x_{n-1}(x^2_{n-1}-x_{n-2}x_n)^{p-1}y_n.$

(4) ${\rm Tr}^G((y_i+\alpha y_{n-1}\!)^{(p+1)(p-1)}y_n)\!\equiv\! (\!x_{i-1}+\alpha x_{n-2}\!)^{p-1}(x_i+\alpha x_{n-1})^{p-1}((x_{i-1}+\alpha x_{n-2})^{p-1}-(x_i+\alpha x_{n-1})^{p-1})^{p-1}y_n$ for every $\alpha\in \mathbb{F}\backslash \mathbb{F}_p$ and $2\leq i\leq n-2$.

Proof. The above congruences follow by the same methods as Lemmas 4 and 5. □

Theorem 2. Let $\mathbb{F}[V_{2n-1}] = \mathbb{F}[x_1, \cdots, x_n, y_2, \cdots, y_n]$. Then $S_1 = \{x_1\}$, $S_2 = \big\{x_1, x_2, {\rm N}^G(y_2)\big\}$ and $S_3 = S_2\bigcup T_3$ are separating sets for $V_1, V_3$ and $V_5$ respectively, where

Let $n\geq4$ and $S_{n-1}\subseteq \mathbb{F}[V_{2n-3}]^G$ be a separating set for $V_{2n-3}$. Choose an element $\alpha\in \mathbb{F}$ with $\alpha$ not in the prime field $\mathbb{F}_p$. Then $S_n = S_{n-1}\bigcup T_n$ is a separating set for $V_{2n-1}$, where

Proof. We first prove the cases $n\geq4$. Consider any two points $\textbf{v} = (v_1, \cdots, v_n, w_2, \cdots, w_n), \; \textbf{v}^{\prime} = (v^{\prime}_1, \cdots, v^{\prime}_n, w^{\prime}_2, \cdots, w^{\prime}_n)\in V_{2n-1}$ that do not lie in the same $G$-orbit, and suppose that $f(\textbf{v}) = f(\textbf{v}^{\prime})$ for all $f\in S_n$. We show that there exists $g\in G$ such that $\textbf{v}^{\prime} = g(\textbf{v})$. This contradicts our assumption that $\textbf{v}$ and $\textbf{v}^{\prime}$ do not lie in the same $G$-orbit and this contradiction shows that $S_n$ is a separating set for $V_{2n-1}$. We assume that every invariant in $S_n$ takes the same value on $\textbf{v}$ and $\textbf{v}^{\prime}$ from now on. We may assume that $v^{\prime}_i = v_i$ for $1\leq i\leq n$, $w^{\prime}_i = w_i$ for $2\leq i\leq n-1$ and $w^{\prime}_n\neq w_n$ as the proof of Theorem 2.1.

$Case\; 1$. We assume that there exists $v_i = 0$ for $1\leq i\leq n-1$ and let $j$ be maximal with this property.

First, $\text{Tr}^G(y^{p-1}_iy^{p-1}_{i+1}y_n)(\textbf{v}) = \text{Tr}^G(y^{p-1}_iy^{p-1}_{i+1}y_n)(\textbf{v}^{\prime})$ implies

for $1\leq i\leq n-3$ by setting $j = i+1$ in Lemma 6(1). As $w^{\prime}_n\neq w_n$, this suggests that: If $v_{i-1} = 0$, then $v_i = 0$ for $2\leq i\leq n-2$. Therefore $j\geq n-2$.

If $j = n-2$, then $v_{n-1}\neq0$. Again, $\text{Tr}^G(y^{p-1}_iy^{p-1}_{i+1}y_n)(\textbf{v}) = \text{Tr}^G(y^{p-1}_iy^{p-1}_{i+1}y_n)(\textbf{v}^{\prime})$ implies $(v_{i-1}v_{i+1}-v^2_i)^{p-1}(w_n-w^{\prime}_n) = 0$ for $1\leq i\leq n-3$. As $w^{\prime}_n\neq w_n$, so the last equation suggests that: If $v_{i+1} = 0$, then $v_i = 0$ for $1\leq i\leq n-3$. Since $v_{n-2} = 0$, we get

successively. However, $\text{N}^{H_3}(x_{n-1}y_n-x_ny_{n-1})$ taking the same value on $\textbf{v}, \textbf{v}^{\prime}$ implies

by Lemma 6(3). As $v_{n-1}\neq0$ and $w^{\prime}_n\neq w_n$, then we have $w_n = w^{\prime}_n+lv_{n-1}$ for some $1\leq l\leq p-1$. Hence $\textbf{v}^{\prime} = \sigma^l_1\sigma^l_2(\textbf{v})$ which is a contradiction.

If $j = n-1$, namely $v_{n-1} = 0$. $\text{Tr}^G(y^{p-1}_2y^{p-1}_{n-1}y_n)$ taking the same value on $\textbf{v}$, $\textbf{v}^{\prime}$ implies

by setting $i = 2$, $j = n-2$ in Lemma 6(1). We have that $v_2 = 0$ or $v_{n-2} = 0$. Whether $v_2 = 0$ or $v_{n-2} = 0$, we have

successively by $\text{Tr}^G(y^{p-1}_iy^{p-1}_{i+1}y_n)(\textbf{v}) = \text{Tr}^G(y^{p-1}_iy^{p-1}_{i+1}y_n)(\textbf{v}^{\prime})$ for $2\leq i\leq n-2$. Furthermore, $\text{N}^G(y_n)(\textbf{v}) = \text{N}^G(y_n)(\textbf{v}^{\prime})$ implies

Thus $w_n = w^{\prime}_n+(-l+k)v_n$ for some $0\leq k, l\leq p-1$. Hence $\textbf{v}^{\prime} = \sigma^k_1\sigma^l_2(\textbf{v})$ which is also a contradiction.

$Case\; 2$. We assume that $v_i\neq0$ for $1\leq i\leq n-1$. Then $\text{Tr}^G(y^{(p+1)(p-1)}_iy_n)(\textbf{v}) = \text{Tr}^G(y^{(p+1)(p-1)}_iy_n)(\textbf{v}^{\prime})$ implies

for $2\leq i\leq n-1$ by Lemma 6(2). So we have

We claim that

for $1\leq i\leq n-3$. Given this, $\text{N}^{H_3}(x_{n-1}y_n-x_ny_{n-1})$ taking the same value on $\textbf{v}, \textbf{v}^{\prime}$ implies

by Lemma 6(3). Thus, there exists some $1\leq l\leq p-1$ such that $w_n-w^{\prime}_n = lv_{n-1}-l\gamma v_n$. There must exist $k$ with $0\leq k\leq p-1$ such that $-l+k+l\gamma = 0$. Then $\textbf{v}^{\prime} = \sigma^k_1\sigma^l_2(\textbf{v})$. This is a contradiction.

Now we prove for the claim. For $1\leq i\leq n-2$, We define

It is obvious that $\gamma_i\in \mathbb{F}^*_p$. Because of $\text{Tr}^G((y_i+\alpha y_{n-1})^{(p+1)(p-1)}y_n)(\textbf{v}) = \text{Tr}^G((y_i+\alpha y_{n-1})^{(p+1)(p-1)}y_n)(\textbf{v}^{\prime})$, we have that

by Lemma 6(4). Since $v^{p-1}_1 = v^{p-1}_2 = \cdots = v^{p-1}_{n-1}\neq0$ and $\alpha\in \mathbb{F}\backslash \mathbb{F}_p$, we have that $v_{i-1}+\alpha v_{n-2}\neq0$ and $v_i+\alpha v_{n-1}\neq0$. Since $w_n\neq w^{\prime}_n$, we obtain that

Substituting $\gamma_i = \frac{v_i}{v_{n-1}}$ into (2.2), and because of $v^{p-1}_{n-1}\neq0$ and $\gamma^{p-1}_{n-2} = 1$ we get

Consider the following polynomial

in $\mathbb{F}_p[X]$. It is obvious that the degree of $Q(X)$ is strictly less than $p-1$. We next show that there are at least $p-1$ different roots of $Q(X)$ and consequently $Q(X) = 0$.

Since $Q(\alpha) = (\alpha+\frac{\gamma_{i-1}}{\gamma_{n-2}})^{p-1}-(\alpha+\gamma_i)^{p-1} = 0$, $\alpha+\frac{\gamma_{i-1}}{\gamma_{n-2}}\neq0$ and $\alpha+\gamma_i\neq0$, then we have

Set

for each $a\in \mathbb{F}_p\backslash \{\pm1\}$. It is easy to see that $M_a\in \mathbb{F}_p$ and $M_a\neq-1$. Indeed, if $M_a = -1$, then $\alpha = -(a\frac{\gamma_{i-1}}{\gamma_{n-2}}+\gamma_i)/(a+1)\in \mathbb{F}_p$, which contradicts $\alpha\in \mathbb{F}\backslash \mathbb{F}_p$.

Now consider

Equation (2.4) suggests that $(a\alpha+\alpha+a\frac{\gamma_{i-1}}{\gamma_{n-2}}+\frac{\gamma_{i-1}}{\gamma_{n-2}})^{p-1} = (M_a\alpha+\alpha+M_a\gamma_i+\gamma_i)^{p-1}$ and we see that $(M_a\alpha+\alpha+M_a\gamma_i+\gamma_i)^{p-1} = (a\alpha+\alpha+a\frac{\gamma_{i-1}}{\gamma_{n-2}}+\gamma_i)^{p-1}$ by (2.3). Thus $(a\alpha+\alpha+a\frac{\gamma_{i-1}}{\gamma_{n-2}}+\frac{\gamma_{i-1}}{\gamma_{n-2}})^{p-1} = (a\alpha+\alpha+a\frac{\gamma_{i-1}}{\gamma_{n-2}}+\gamma_i)^{p-1}$, i.e.

for each $a\in \mathbb{F}_p\backslash \{\pm1\}$. For any $a_1\neq a_2\in \mathbb{F}_p\backslash \{\pm1\}$, $a_1\alpha+\alpha+a_1\frac{\gamma_{i-1}}{\gamma_{n-2}}\neq a_2\alpha+\alpha+a_2\frac{\gamma_{i-1}}{\gamma_{n-2}}$. Moreover, $Q(0) = 0$. Therefore there are at least $p-1$ different roots of $Q(X)$. We have proved $Q(X) = 0$.

Substituting $-\gamma_i$ into $Q(X) = 0$ we obtain

i.e.

for $2\leq i\leq n-2$. This establish the claim.

Now, we prove the cases $1\leq n\leq 3$. Obviously, $\mathbb{F}[V_1]^G = \mathbb{F}[x_1]$. Since $\big\{x_1, x_2, \text{N}^G(y_2)\big\}$ forms a homogeneous system of parameters for $\mathbb{F}[V_3]^G$ and the product of their degrees is equal to the order of $G$, it follows from [1, Theorem 3.9.4] that $\mathbb{F}[V_3]^G = \mathbb{F}[x_1, x_2, \text{N}^G(y_2)]$. Naturally $S_2 = \big\{x_1, x_2, \text{N}^G(y_2)\big\}$ is a separating set for $V_3$. Then, $S_3 = S_2\bigcup T_3$ is a separating set for $V_5$, where

The proof is analogous to the proof for $n\geq4$. □

Remark 2. Theorem 2 yields a minimal separating set for each representation $V_{2n-1}$. Moreover, the size of separating set for $V_{2n-1}$ is $\frac{n(3n-2)}{2}$, which depends only on the dimension of the representation. Incidently, the maximal degree of an invariant in this set is the group order $p^2$.

3.   Conclusions

In this paper, we determine explicit separating sets for four families of finite dimensional representations of the elementary abelian $p$-groups of rank two $(\mathbb{Z}/p)^2$ over an algebraically closed field of characteristic $p$, where $p$ is an odd prime. The size of every separating set depends only on the dimension of the representation.

Author contributions

Panpan Jia: Conceptualization, Methodology, Writing-original draft preparation, Writing-review and editing; Jizhu Nan: Methodology, Writing-review and editing, Funding acquisition; Yongsheng Ma: Methodology, Writing-review and editing. All authors have read and approved the final version of the manuscript for publication.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

This work was supported by the National Natural Science Foundation of China (No.12171194).

Conflict of interest

The authors declare no conflicts of interest in this paper.