Research article

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

  • Received: 11 July 2024 Revised: 24 August 2024 Accepted: 28 August 2024 Published: 03 September 2024
  • MSC : 13A50

  • For a finite group acting linearly on a vector space, a separating set is a subset of the invariant ring that separates the orbits. In this paper, we determined explicit separating sets in the corresponding rings of invariants for four families of finite dimensional representations of the elementary abelian p-groups (Z/p)2 of rank two over an algebraically closed field of characteristic p, where p is an odd prime. Our construction was recursive. The separating sets consisted only of transfers and norms, and the size of every separating set depended only on the dimension of the representation.

    Citation: Panpan Jia, Jizhu Nan, Yongsheng Ma. Separating invariants for certain representations of the elementary Abelian p-groups of rank two[J]. AIMS Mathematics, 2024, 9(9): 25603-25618. doi: 10.3934/math.20241250

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  • For a finite group acting linearly on a vector space, a separating set is a subset of the invariant ring that separates the orbits. In this paper, we determined explicit separating sets in the corresponding rings of invariants for four families of finite dimensional representations of the elementary abelian p-groups (Z/p)2 of rank two over an algebraically closed field of characteristic p, where p is an odd prime. Our construction was recursive. The separating sets consisted only of transfers and norms, and the size of every separating set depended only on the dimension of the representation.



    Let ρ:GGL(n,F) be a faithful representation of a finite group G over a field F of arbitrary characteristic. Denote by V=Fn the n dimensional representation space over F. We write F[V] for the symmetric algebra S(V) over the dual space V. The action of G on V induces an action on F[V]: for fF[V] and vV, the action of gG is given by (g(f))(v)=f(g1(v)). The ring of invariants F[V]G is defined by

    F[V]G:={fF[V]|g(f)=fforallgG}.

    Here are some general methods to construct invariants of finite groups. Let fF[V], then the transfer of f is defined by

    TrG(f):=gGg(f).

    Let HG be a subgroup. Then the relative transfer is defined as

    TrGH:F[V]HF[V]G,fgHG/Hg(f),

    where G/H denotes a set of left coset representatives of H in G. TrGH is independent of the choice of the coset representatives. The norm of f is defined by

    NG(f):=gGg(f).

    Note that the transfer, relative transfer and norm are invariant polynomials. If the characteristic of F divides the group order |G|, we speak of the modular case. Otherwise, we are in the nonmodular case, which includes char(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 SF[V]G is said to be separating if for any two points v,vV, we have: If there exists an invariant fF[V]G with f(v)f(v), then there exists an element hS with h(v)h(v).

    If G is finite, then for v,vV with distinct G-orbits, there exists fF[V]G such that f(v)f(v). It follows that a subset SF[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 char(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 (Z/p)r and the ring of invariants for all three dimensional representations of (Z/p)2 have been worked out in [21]. See also [22] for further research. Four families of finite dimensional representations of (Z/p)2 over an algebraically closed field 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 F[V]G in the modular case. Since the dual of a subrepresentation sits in the dual of higher dimensional representation of (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 (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.

    Let G=σ1,σ2(Z/p)2 be the elementary abelian p-group of rank two of order p2, where p is an odd prime. Let σ3=σ1σ2 and Hi denote the subgroup of G which is generated by σi for 1i3. 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 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, In denotes the n×n identity matrix and for any element λ of the field F, Jλ denotes the n×n Jordan block (lower triangular) with eigenvalues λ.

    Type (I) For every even dimension 2n there are representations V2n,λ,

    σ1(In0InIn),
    σ2(In0JλIn).

    Type (II) For every even dimension 2n there are representations V2n,,

    σ1(In0J0In),
    σ2(In0InIn).

    Type (III) For every odd dimension 2n1 there are representations V(2n1),

    σ1(In10In101×(n1)In),
    σ3(In1001×(n1)In1In).

    Type (IV) For every odd dimension 2n1 there are representations V2n1,

    σ1(In00(n1)×1In1In1),
    σ3(In0In10(n1)×1In1).

    Notice that the matrix group associated with V2n, in type (II) is the same as the matrix group associated with V2n,0 in type (I). Therefore, their invariant rings are equal, and a separating set for V2n,0 is also a separating set for V2n,. Each representation V(2n1) in type (III) is isomorphic to a subrepresentation of V2n,p1 in type (I), which we will explain in detail later. So we study the separating sets for types (I)(III) in Subsection 2.1 and for type (IV) in Subsection 2.2.

    We start with the action of G on the representation space V2n,λ. Let ε1, ε2, , εn, ξ1, ξ2, , ξn be the basis for V2n,λ with σ1(εi)=εi+ξi, σ1(ξi)=ξi, σ2(ξi)=ξi for 1in, σ2(εn)=εn+λξn and σ2(εi)=εi+λξi+ξi+1 for 1in1. We identify each εi with the column vector with 1 on the i-th coordinate and zero elsewhere, and each ξi with the column vector with 1 on the (n+i)-th coordinate and zero elsewhere. Let x1,x2,,xn,y1,y2,,yn denote the corresponding elements in the dual space V2n,λ. In fact, x1,x2,,xn,y1,y2,,yn form the basis for V2n,λ in the reverse order: we have σ11(xi)=xi, σ11(yi)=xi+yi, σ12(xi)=xi for 1in, σ12(y1)=λx1+y1 and σ12(yi)=xi1+λxi+yi for 2in. For simplicity we will use the generators σ1i instead of σi for the rest of the paper and change the notation by writing σi for the new generators for 1i3. Note also that F[V2n,λ]=F[x1,x2,,xn,y1,y2,,yn]. Pick a point v=(v1,v2,,vn,w1,w2,,wn) in V2n,λ. The surjection φ:V2n,λV2n2,λ given by (v1,v2,,vn,w1,w2,,wn)(v1,v2,,vn1,w1,w2,,wn1) is G-equivariant, as for gG, vV2n,λ, g(φ(v))=φ(g(v)). Dual to this surjection, the subspace in V2n,λ generated by x1,x2,,xn1,y1,y2,,yn1 is closed under the G-action and isomorphic to V2n2,λ. Hence F[V2n2,λ]=F[x1,x2,,xn1,y1,y2,,yn1] is a subalgebra in F[V2n,λ].

    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 0lp1lk1modp if p1 divides k and 0lp1lk0modp, 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 0kp1, klp1. There holds

    (k0)(p(k+1)lk)+(k+11)(p(k+2)l(k+1))++(llk)(p(l+1)0)=(plk).

    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 1kp. Then

    ((p+1)(p1)k(p1))1modp.

    (2) Let l and k be a positive integer such that 2lp and 1kl1. Then

    (l(p1)k(p1))0modp.

    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 F[x1,x2,,xn,y1,y2,,yn1] 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) TrG(yp1iyp1jyn)(xi1xjxixj1)p1yn for 2i,jn1.

    (2) TrG(yp11yp12yn)x2(p1)1yn.

    Proof. Here we only prove for (1). It is easy to verify that TrG=TrGHTrH for any subgroup H of G. This suggest that we may compute TrG by first computing TrH and then computing TrGH. Thus we may work with the two smaller groups H and G/H.

    By the definition of transfer we have

    TrH1(yp1iyp1jyn)=0lp1(lxi+yi)p1(lxj+yj)p1(lxn+yn)0lp1(lxi+yi)p1(lxj+yj)p1yn0lp10s,tp1(p1s)(p1t)(lxi)syp1si(lxj)p1tytjyn0lp10s,tp1(p1s)(p1t)lp1+stxsixp1tjyp1siytjyn.

    By Lemma 1, we see that

    TrH1(yp1iyp1jyn)0lp10kp1(p1k)2lp1xkixp1kjyp1kiykjyn0lp1lp1(0kp1(p1k)2xkixp1kjyp1kiykjyn)0kp1xkixp1kjyp1kiykjyn.

    The last congruence follows since (p1k)21modp. Similarly, for each k with 0kp1 we have

    TrH2(xkixp1kjyp1kiykjyn)=0lp1xkixp1kj(lxi1+lλxi+yi)p1k(lxj1+lλxj+yj)k(lxn1+lλxn+yn)0lp1xkixp1kj(lxi1+lλxi+yi)p1k(lxj1+lλxj+yj)kynxkixp1kj(xi1+λxi)p1k(xj1+λxj)kyn.

    Thus

    TrG(yp1iyp1jyn)=TrGH1(TrH1(yp1iyp1jyn))TrGH1(0kp1xkixp1kjyp1kiykjyn)0kp1xkixp1kj(xi1+λxi)p1k(xj1+λxj)kyn=0kp10sp1k0tk(p1ks)(kt)λs+txp1ksi1xk+sixktj1xp1k+tjyn.

    It follows from Lemma 2 that the coefficient of xp1li1xlixm1j1xpmjyn is

    λlm+1((m10)(pml(m1))+(m1)(p(m+1)lm)++(ll(m1))(p(l+1)0))=λlm+1(pl(m1)).

    Moreover, (pl(m1))1modp if l=m1 and (pl(m1))0modp, otherwise. From the above it follows that

    TrG(yp1iyp1jyn)0kp10sp1k0tk(p1ks)(kt)λs+txp1ksi1xk+sixktj1xp1k+tjyn=(xp1i1xp1j+xp2i1xixj1xp2j++xp1ixp1j1)yn=(xi1xjxixj1)p1yn.

    Lemma 5. (1) TrG(y(p+1)(p1)n1yn)1kp(xn2+λxn1)(p+1k)(p1)xk(p1)n1yn.

    (2) NH3(xn1ynxnyn1)xpn1ypnxn1(x2n1xn2xn)p1yn.

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

    TrH1(y(p+1)(p1)n1yn)=0lp1(lxn1+yn1)(p+1)(p1)(lxn+yn)0lp1(lxn1+yn1)(p+1)(p1)yn.

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

    TrH1(y(p+1)(p1)n1yn)0lp10kp+1((p+1)(p1)k(p1))lk(p1)xk(p1)n1y(p+1k)(p1)n1yn1kp+1xk(p1)n1y(p+1k)(p1)n1yn.

    For each k with 1kp+1,

    TrH2(xk(p1)n1y(p+1k)(p1)n1yn)=0lp1xk(p1)n1(lxn2+lλxn1+yn1)(p+1k)(p1)(lxn1+lλxn+yn)0lp1xk(p1)n1(lxn2+lλxn1+yn1)(p+1k)(p1)yn.

    By Lemma 1 and Lemma 3(1) we have

    TrH2(xk(p1)n1y(p+1k)(p1)n1yn)0lp10sp+1k((p+1k)(p1)s(p1))ls(p1)(xn2+λxn1)s(p1)xk(p1)n1y(p+1ks)(p1)n1yn.

    For 1kp, we see that ((p+1k)(p1)s(p1))0modp unless s=p+1k by Lemma 3(2) in which case we have

    TrH2(xk(p1)n1y(p+1k)(p1)n1yn)0lp1l(p+1k)(p1)(xn2+λxn1)(p+1k)(p1)xk(p1)n1yn(xn2+λxn1)(p+1k)(p1)xk(p1)n1yn.

    For k=p+1, TrH2(xk(p1)n1y(p+1k)(p1)n1yn)0lp1x(p+1k)(p1)n1yn0. Thus,

    TrG(y(p+1)(p1)n1yn)=TrGH1(TrH1(y(p+1)(p1)n1yn))TrGH1(1kp+1xk(p1)n1y(p+1k)(p1)n1yn)1kp(xn2+λxn1)(p+1k)(p1)xk(p1)n1yn.

    (2) Note that xn1ynxnyn1 is σ1-invariant, so the H3-orbit product of this polynomial is G-invariant. Thus, we have

    NH3(xn1ynxnyn1)=Π0lp1(xn1(lxn1+(lλ+l)xn+yn)xn(lxn2+(lλ+l)xn1+yn1))=Π0lp1(l(x2n1xn2xn)+(xn1ynxnyn1))=(xn1ynxnyn1)p(xn1ynxnyn1)(x2n1xn2xn)p1xpn1ypnxn1(x2n1xn2xn)p1yn.

    Theorem 1. Let F[V2n,λ]=F[x1,x2,,xn,y1,y2,,yn]. Then

    S1={x1,fλ={NG(y1)forλFp,NH1(y1)forλFp}

    is a separating set for V2,λ. And S2=S1T2 is a separating set for V4,λ, where

    T2={x2,NG(y2),fλ={TrG(y(p+1)(p1)1y2)forλFp,NH3(x1y2x2y1)forλFp}.

    Let n3 and Sn1F[V2n2,λ]G be a separating set for V2n2,λ. Then Sn=Sn1Tn is a separating set for V2n,λ, where

    Tn={xn,NG(yn),TrG(yp1iyp1i+1yn)for2in1,
    fλ={TrG(y(p+1)(p1)n1yn)forλFp,NH3(xn1ynxnyn1)forλFp}.

    Moreover, a separating set for V2n,0 is a separating set for V2n,.

    Proof. The cases V2,λ and V4,λ are easy to check so we only prove the case of n. Consider any two points v=(v1,,vn,w1,,wn),v=(v1,,vn,w1,,wn)V2n,λ that do not lie in the same G-orbit, and suppose that f(v)=f(v) for all fSn. We will show that there exists gG such that v=g(v). This contradicts our assumption that v and v do not lie in the same G-orbit and this contradiction shows that Sn is a separating set for V2n,λ. We assume that every invariant in Sn takes the same value on v and v from now on.

    If (v1,,vn1,w1,,wn1),(v1,,vn1,w1,,wn1)V2n2,λ do not lie in the same G-orbit, then there exists a polynomial in Sn1 that separates the two points because Sn1F[V2n2,λ]G is separating. Therefore this polynomial separates v and v as well. Hence by replacing v with a suitable element in its G-orbit we may assume that vi=vi and wi=wi for 1in1. Since xnTn, we may assume that vn=vn. Note that with this assumption we must have wnwn.

    First, by Lemma 4(2), TrG(yp11yp12yn)(v)=TrG(yp11yp12yn)(v) implies

    0=TrG(yp11yp12yn)(v)TrG(yp11yp12yn)(v)=v2(p1)1wnv2(p1)1wn=v2(p1)1(wnwn).

    As wnwn, we obtain

    v1=0.

    Similarly, TrG(yp1iyp1i+1yn)(v)=TrG(yp1iyp1i+1yn)(v) implies

    0=TrG(yp1iyp1i+1yn)(v)TrG(yp1iyp1i+1yn)(v)=(vi1vi+1v2i)p1(wnwn)

    for 2in2 by setting j=i+1 in Lemma 4 (1). Since v1=0 and wnwn, we get

    v2=v3==vn2=0

    successively. Since vn2=0, TrG(y(p+1)(p1)n1yn)(v)=TrG(y(p+1)(p1)n1yn)(v) implies

    0=TrG(y(p+1)(p1)n1yn)(v)TrG(y(p+1)(p1)n1yn)(v)=1kp(vn2+λvn1)(p+1k)(p1)vk(p1)n1(wnwn)=1kpλ(p+1k)(p1)v(p+1)(p1)n1(wnwn)=(λp1+λ2(p1)++λp(p1))v(p+1)(p1)n1(wnwn)=λp1(λp11)p1v(p+1)(p1)n1(wnwn) (2.1)

    by Lemma 5(1). Notice that λp1(λp11)p1=0 if and only if λFp, so the following proof falls into two parts depending on whether λ is in Fp or not.

    If λFp, then we have

    vn1=0

    by (2.1). As NG(yn)=0k,lp1(lxn1+(lλ+k)xn+yn), we have NG(yn)(v)=0k,lp1((lλ+k)vn+wn). We define a polynomial

    P(X):=0k,lp1(X+(lλ+k)vn)

    in F[X]. Notice that NG(yn)(v)=P(wn) and that P(wn)=P(wn+(lλ+k)vn) for all 0k,lp1. Since P(X) is a polynomial of degree p2, it follows that wn+(lλ+k)vn for 0k,lp1 are the only solutions of P(X)P(wn)=0. Therefore the equality of NG(yn)(v)=P(wn) and NG(yn)(v)=P(wn) implies wn must be equal to wn+(lλ+k)vn for some 0k,lp1. Hence v=σk1σl2(v). This is a contradiction because v and v lie in the same G-orbit.

    Next we turn to the case λFp. Since vn2=0, then NH3(xn1ynxnyn1) taking the same value on v,v implies

    0=NH3(xn1ynxnyn1)(v)NH3(xn1ynxnyn1)(v)=(vpn1wpnv2p1n1wn)(vpn1wpnv2p1n1wn)=vpn1(wnwn)((wnwn)p1vp1n1)

    by Lemma 5(2). If vn10, then we have (wnwn)p1vp1n1=0, i.e. (wnwn)p1=vp1n1, i.e. wnwn=lvn1 for some 1lp1. There must exist k with 0kp1 such that lλ+k=0. Hence v=σk1σl2(v). This is a contradiction. So now assume vn1=0. Then NG(yn)(v)=NG(yn)(v) implies (0lp1(lvn+wn))p=(0lp1(lvn+wn))p. Thus 0=(0lp1(lvn+wn))p(0lp1(lvn+wn))p=(0lp1(lvn+wn)0lp1(lvn+wn))p and therefore wn=wn+kvn for some 1kp1. Hence v=σk1(v). This is also a contradiction.

    The final statement follows because the matrix group associated with V2n, is the same as the matrix group associated with V2n,0, so their invariant rings are equal, and a separating set for V2n,0 is also a separating set for V2n,.

    Remark 1. From the proof of Theorem 1, we see that the separating set for each representation V2n,λ we obtained is minimal. Moreover, the size of separating set for V2n,λ is 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 p2.

    Since V(2n1) is isomorphic to the submodule of V2n,p1 spanned by ε1, , εn1, ξ1, , ξn, where ε1, , εn, ξ1, , ξn is the basis for V2n,p1. Dual to this inclusion, there is a restriction map F[V2n,p1]GF[V(2n1)]G,ff|V(2n1) 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

    F[V2n,p1]=F[x1,x2,,xn,y1,y2,,yn]

    and

    F[V(2n1)]=F[x1,x2,,xn1,y1,y2,,yn].

    Then

    T1={y1}

    is a separating set for V1. Additionally,

    T2={x1,NH1(y1),NH3(y2)}

    is a separating set for V3. Let n3 and Sn1F[V2n2,p1]G be a separating set for V2n2,p1. Then the polynomials in Sn=Sn1Tn restricted to V(2n1) form a separating set for V(2n1), where

    Tn={NG(yn),NH3(xn1ynxnyn1),TrG(yp1iyp1i+1yn)for2in1}.

    We consider type (I) representations V2n,p1. In view of ξ1V2n,p1 is a G-submodule, we have V2n1V2n,p1/ξ1 with basis ˜εi:=εi+ξ1 for 1in, ˜ξi:=ξi+ξ1 for 2in, and a G-algebra inclusion F[V2n1]=F[x1,,xn,y2,,yn]F[V2n,p1]. The action of σ1 and σ3 on the variables are given by

    {σ1(xi)=xifor1in1,σ1(yi)=xi+yifor2in

    and

    {σ3(xi)=xifor1in1,σ3(yi)=xi1+yifor2in.

    Pick a point (v1,,vn,w2,,wn) in V2n1. There is a G-equivariant surjection V2n1V2n3 given by

    (v1,,vn,w2,,wn)(v1,,vn1,w2,,wn1).

    Hence

    F[V2n3]=F[x1,,xn1,y2,,yn1]

    is a subalgebra in F[V2n1].

    Note that all congruences are modulo F[x1,,xn,y2,,yn1] in subsection 2.2.

    Lemma 6. (1) TrG(yp1iyp1jyn)(xi1xjxixj1)p1yn for 2i,jn1.

    (2) TrG(y(p+1)(p1)iyn)xp1i1xp1i(xp1i1xp1i)p1yn for 2in1.

    (3) NH3(xn1ynxnyn1)xpn1ypnxn1(x2n1xn2xn)p1yn.

    (4) TrG((yi+αyn1)(p+1)(p1)yn)(xi1+αxn2)p1(xi+αxn1)p1((xi1+αxn2)p1(xi+αxn1)p1)p1yn for every αFFp and 2in2.

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

    Theorem 2. Let F[V2n1]=F[x1,,xn,y2,,yn]. Then S1={x1}, S2={x1,x2,NG(y2)} and S3=S2T3 are separating sets for V1,V3 and V5 respectively, where

    T3={x3,NG(y3),TrG(y(p+1)(p1)2y3),NH3(x2y3x3y2)}.

    Let n4 and Sn1F[V2n3]G be a separating set for V2n3. Choose an element αF with α not in the prime field Fp. Then Sn=Sn1Tn is a separating set for V2n1, where

    Tn={xn,NG(yn),NH3(xn1ynxnyn1),TrG(yp12yp1n1yn),
    TrG(yp1iyp1i+1yn)for2in2,TrG(y(p+1)(p1)iyn)for2in1,TrG((yi+αyn1)(p+1)(p1)yn)for2in2}.

    Proof. We first prove the cases n4. Consider any two points v=(v1,,vn,w2,,wn),v=(v1,,vn,w2,,wn)V2n1 that do not lie in the same G-orbit, and suppose that f(v)=f(v) for all fSn. We show that there exists gG such that v=g(v). This contradicts our assumption that v and v do not lie in the same G-orbit and this contradiction shows that Sn is a separating set for V2n1. We assume that every invariant in Sn takes the same value on v and v from now on. We may assume that vi=vi for 1in, wi=wi for 2in1 and wnwn as the proof of Theorem 2.1.

    Case1. We assume that there exists vi=0 for 1in1 and let j be maximal with this property.

    First, TrG(yp1iyp1i+1yn)(v)=TrG(yp1iyp1i+1yn)(v) implies

    0=TrG(yp1iyp1i+1yn)(v)TrG(yp1iyp1i+1yn)(v)=(vi1vi+1v2i)p1(wnwn)

    for 1in3 by setting j=i+1 in Lemma 6(1). As wnwn, this suggests that: If vi1=0, then vi=0 for 2in2. Therefore jn2.

    If j=n2, then vn10. Again, TrG(yp1iyp1i+1yn)(v)=TrG(yp1iyp1i+1yn)(v) implies (vi1vi+1v2i)p1(wnwn)=0 for 1in3. As wnwn, so the last equation suggests that: If vi+1=0, then vi=0 for 1in3. Since vn2=0, we get

    v1=v2==vn2=0

    successively. However, NH3(xn1ynxnyn1) taking the same value on v,v implies

    0=NH3(xn1ynxnyn1)(v)NH3(xn1ynxnyn1)(v)=(vpn1wpnv2p1n1wn)(vpn1wpnv2p1n1wn)=vpn1(wnwn)((wnwn)p1vp1n1)

    by Lemma 6(3). As vn10 and wnwn, then we have wn=wn+lvn1 for some 1lp1. Hence v=σl1σl2(v) which is a contradiction.

    If j=n1, namely vn1=0. TrG(yp12yp1n1yn) taking the same value on v, v implies

    0=TrG(yp12yp1n1yn)(v)TrG(yp12yp1n1yn)(v)=vp12vp1n2(wnwn)

    by setting i=2, j=n2 in Lemma 6(1). We have that v2=0 or vn2=0. Whether v2=0 or vn2=0, we have

    v1=v2==vn2=0

    successively by TrG(yp1iyp1i+1yn)(v)=TrG(yp1iyp1i+1yn)(v) for 2in2. Furthermore, NG(yn)(v)=NG(yn)(v) implies

    0=NG(yn)(v)NG(yn)(v)=(0l,kp1(lvn1+(l+k)vn+wn))(0l,kp1(lvn1+(l+k)vn+wn))=(0l,kp1((l+k)vn+wn))(0l,kp1((l+k)vn+wn)).

    Thus wn=wn+(l+k)vn for some 0k,lp1. Hence v=σk1σl2(v) which is also a contradiction.

    Case2. We assume that vi0 for 1in1. Then TrG(y(p+1)(p1)iyn)(v)=TrG(y(p+1)(p1)iyn)(v) implies

    0=TrG(y(p+1)(p1)iyn)(v)TrG(y(p+1)(p1)iyn)(v)=vp1i1vp1i(vp1i1vp1i)p1wnvp1i1vp1i(vp1i1vp1i)p1wn=vp1i1vp1i(vp1i1vp1i)p1(wnwn)

    for 2in1 by Lemma 6(2). So we have

    vp11=vp12==vp1n10.

    We claim that

    vivi+1=vn2vn1=γFp

    for 1in3. Given this, NH3(xn1ynxnyn1) taking the same value on v,v implies

    0=NH3(xn1ynxnyn1)(v)NH3(xn1ynxnyn1)(v)=(vpn1wpnvn1(v2n1vn2vn)p1wn)(vpn1wpnvn1(v2n1vn2vn)p1wn)=(vpn1wpnvn1(v2n1γvn1vn)p1wn)(vpn1wpnvn1(v2n1γvn1vn)p1wn)=vpn1(wnwn)((wnwn)p1(vn1γvn)p1)

    by Lemma 6(3). Thus, there exists some 1lp1 such that wnwn=lvn1lγvn. There must exist k with 0kp1 such that l+k+lγ=0. Then v=σk1σl2(v). This is a contradiction.

    Now we prove for the claim. For 1in2, We define

    γi:=vivn1.

    It is obvious that γiFp. Because of TrG((yi+αyn1)(p+1)(p1)yn)(v)=TrG((yi+αyn1)(p+1)(p1)yn)(v), we have that

    0=TrG((yi+αyn1)(p+1)(p1)yn)(v)TrG((yi+αyn1)(p+1)(p1)yn)(v)=(vi1+αvn2)p1(vi+αvn1)p1((vi1+αvn2)p1(vi+αvn1)p1)p1wn(vi1+αvn2)p1(vi+αvn1)p1((vi1+αvn2)p1(vi+αvn1)p1)p1wn=(vi1+αvn2)p1(vi+αvn1)p1((vi1+αvn2)p1(vi+αvn1)p1)p1(wnwn)

    by Lemma 6(4). Since vp11=vp12==vp1n10 and αFFp, we have that vi1+αvn20 and vi+αvn10. Since wnwn, we obtain that

    (vi1+αvn2)p1(vi+αvn1)p1=0. (2.2)

    Substituting γi=vivn1 into (2.2), and because of vp1n10 and γp1n2=1 we get

    (α+γi1γn2)p1(α+γi)p1=0.

    Consider the following polynomial

    Q(X):=(X+γi1γn2)p1(X+γi)p1

    in Fp[X]. It is obvious that the degree of Q(X) is strictly less than p1. We next show that there are at least p1 different roots of Q(X) and consequently Q(X)=0.

    Since Q(α)=(α+γi1γn2)p1(α+γi)p1=0, α+γi1γn20 and α+γi0, then we have

    ((α+γi1γn2)/(α+γi))p1=1.

    Set

    Ma=a(α+γi1γn2)/(α+γi) (2.3)

    for each aFp{±1}. It is easy to see that MaFp and Ma1. Indeed, if Ma=1, then α=(aγi1γn2+γi)/(a+1)Fp, which contradicts αFFp.

    Now consider

    (α+γi1γn2)(a+1)(α+γi)(Ma+1)=aα+α+aγi1γn2+γi1γn2Maα+α+Maγi+γiFp. (2.4)

    Equation (2.4) suggests that (aα+α+aγi1γn2+γi1γn2)p1=(Maα+α+Maγi+γi)p1 and we see that (Maα+α+Maγi+γi)p1=(aα+α+aγi1γn2+γi)p1 by (2.3). Thus (aα+α+aγi1γn2+γi1γn2)p1=(aα+α+aγi1γn2+γi)p1, i.e.

    0=(aα+α+aγi1γn2+γi1γn2)p1(aα+α+aγi1γn2+γi)p1=Q(aα+α+aγi1γn2)

    for each aFp{±1}. For any a1a2Fp{±1}, a1α+α+a1γi1γn2a2α+α+a2γi1γn2. Moreover, Q(0)=0. Therefore there are at least p1 different roots of Q(X). We have proved Q(X)=0.

    Substituting γi into Q(X)=0 we obtain

    γi=γi1γn2,

    i.e.

    vi1vi=vn2vn1

    for 2in2. This establish the claim.

    Now, we prove the cases 1n3. Obviously, F[V1]G=F[x1]. Since {x1,x2,NG(y2)} forms a homogeneous system of parameters for F[V3]G and the product of their degrees is equal to the order of G, it follows from [1, Theorem 3.9.4] that F[V3]G=F[x1,x2,NG(y2)]. Naturally S2={x1,x2,NG(y2)} is a separating set for V3. Then, S3=S2T3 is a separating set for V5, where

    T3={x3,NG(y3),TrG(y(p+1)(p1)2y3),NH3(x2y3x3y2)}.

    The proof is analogous to the proof for n4.

    Remark 2. Theorem 2 yields a minimal separating set for each representation V2n1. Moreover, the size of separating set for V2n1 is n(3n2)2, which depends only on the dimension of the representation. Incidently, the maximal degree of an invariant in this set is the group order p2.

    In this paper, we determine explicit separating sets for four families of finite dimensional representations of the elementary abelian p-groups of rank two (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.

    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.

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

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

    The authors declare no conflicts of interest in this paper.



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