Counterexamples to the local–global divisibility over elliptic curves

Abstract

Let $p \ge 5$ be a prime number. We find all the possible subgroups G of $\mathrm{GL}_2 ({\mathbb Z}/ p {\mathbb Z})$ such that there exist a number field k and an elliptic curve ${\mathcal {E}}$ defined over k such that the $\mathrm{Gal}(k ({\mathcal {E}}[p])/k)$-module ${\mathcal {E}}[p]$ is isomorphic to the G-module $({\mathbb Z}/ p {\mathbb Z})^2$ and there exists $n \in {\mathbb N}$ such that the local–global divisibility by $p^n$ does not hold over ${\mathcal {E}}(k)$.

1 Introduction

Let k be a number field, and let ${\mathcal {A}}$ be a commutative algebraic group defined over k. Several papers have been written on the following classical question, known as the Local–Global Divisibility Problem.

PROBLEM: Let $P \in {\mathcal {A}}(k)$. Assume that for all but finitely many valuations v of k, there exists $D_v \in {\mathcal {A}}(k_v )$ such that $P = qD_v$, where q is a positive integer. Is it possible to conclude that there exists $D\in {\mathcal {A}}(k)$ such that $P=qD$?

By Bézout’s identity, to get answers for a general integer it is sufficient to solve it for powers $p^n$ of a prime. In the classical case of ${\mathcal {A}}={\mathbb {G}}_m$, the answer is positive for p odd, and negative for instance for $q=8$ (and $P=16$) (see for example [1, 19]).

For general commutative algebraic groups, Dvornicich and Zannier gave a cohomological interpretation of the problem (see [5] and [7]) that we shall explain. Let $\Gamma$ be a group and let M be a $\Gamma$-module. We say that a cocycle $Z :\Gamma \rightarrow M$ satisfies the local conditions if for every $\gamma \in \Gamma$, there exists $m_\gamma \in M$ such that $Z_\gamma = \gamma (m_\gamma ) - m_\gamma$. The set of the classes of cocycles in $H^1 (\Gamma , M)$ that satisfy the local conditions is a subgroup of $H^1 (\Gamma , M)$. We call it the first local cohomology group $H^1_{\mathrm{loc}} (\Gamma , M)$. Dvornicich and Zannier [5, Proposition 2.1] proved the following result.

Proposition 1

Let p be a prime number, let n be a positive integer, let k be a number field and let ${\mathcal {A}}$ be a commutative algebraic group defined over k. If $H^1_{\mathrm{loc}} (\mathrm{Gal}(k ({\mathcal {A}}[p^n]) / k) , {\mathcal {A}}[p^n]) = 0$, then the local–global divisibility by $p^n$ over ${\mathcal {A}}(k)$ holds.

The converse of Proposition 1 is not true, but if the group $H^1_{\mathrm{loc}} (\mathrm{Gal}(k ({\mathcal {A}}[p^n]) / k) , {\mathcal {A}}[p^n])$ is not trivial, we can find an extension L of k such that $L \cap k ( {\mathcal {A}}[p^n]) = k$, and the local–global divisibility by $p^n$ over ${\mathcal {A}}(L)$ does not hold (see [7, Theorem 3] for the details).

Several mathematicians got criterions for the validity of the local–global divisibility principle for various commutative algebraic groups, as algebraic tori [5] and [12], elliptic curves [3,4,5,6,7,8, 14,15,16,17], and very recently polarized abelian surfaces [9] and $\mathrm{GL}_2$-type varieties [10].

In this paper, we focus on elliptic curves. Let p be a prime number, let k be a number field, and let ${\mathcal {E}}$ be an elliptic curve defined over k. Dvornicich and Zannier [7, Theorem 1] found a very interesting criterion for the validity of the local–global divisibility by a power of p over ${\mathcal {E}}(k)$, in the case when $k \cap {\mathbb Q}(\zeta _p) = {\mathbb Q}$.

In a joint work with Paladino and Viada (see [16], and Sect. 2), we refined this criterion, by proving that if k does not contain ${\mathbb Q}(\zeta _p + \overline{\zeta _p})$ and ${\mathcal {E}}(k)$ does not admit a point of order p, then for every positive integer n, the local–global divisibility by $p^n$ holds over ${\mathcal {E}}( k)$. In another joint work with Paladino and Viada [17], we improved our previous criterion and the new criterion allowed us to show that if $k = {\mathbb Q}$ and $p \ge 5$, for every positive integer n the local–global divisibility by $p^n$ holds for ${\mathcal {E}}({\mathbb Q})$.

Very recently, Lawson and Wutrich [13] found a very strong criterion for the triviality of $H^1 (\mathrm{Gal}(k ({\mathcal {E}}[p^n]) / k) , {\mathcal {E}}[p^n])$ (then for the validity of the local–global principle by $p^n$ over ${\mathcal {E}}(k)$, see Proposition 1), but still in the case when $k \cap {\mathbb Q}(\zeta _p) = {\mathbb Q}$.

Finally, Dvornicich and Zannier [6] and Paladino [14] studied the case when $p = 2$ and Paladino [15] and Creutz [3] studied the case when $p = 3$.

Thus we have a fairly good understanding of the local–global divisibility by a power of p over ${\mathcal {E}}(k)$ either when $p \in \{ 2, 3 \}$ or k does not contain ${\mathbb Q}(\zeta _p + \overline{\zeta _p} )$ and ${\mathcal {E}}(k)$ does not admit a point of order p. In this paper we prove the following result:

Theorem 2

Let $p \ge 5$ be a prime number, let k be a number field and let ${\mathcal {E}}$ be an elliptic curve defined over k. Suppose that there exists a positive integer n such that the local–global divisibility by $p^n$ does not hold over ${\mathcal {E}}(k)$. Let $G_1$ be $\mathrm{Gal}(k ({\mathcal {E}}[p])/k)$. Then one of the following holds:

• 1. $p \equiv 2 \mod (3)$ and $G_1$ is isomorphic to a subgroup of $S_3$ of order divisible by 3;

• 2. $G_1$ is cyclic of order dividing $p-1$, and it is generated by an element that has an eigenvalue equal to 1;

• 3. $G_1$ is contained in a Borel subgroup, and it is generated by an element $\sigma$ of order p and an element g of order dividing 2 such that $\sigma$ and g have one common eigenvector for the eigenvalue 1.

Moreover, for every case $i \in \{1, 2, 3 \}$ there exist a number field $L_i$ and an elliptic curve ${\mathcal {E}}_i$ defined over $L_i$, such that the $\mathrm{Gal}(L_i ({\mathcal {E}}_i[p]) / L_i)$-module ${\mathcal {E}}_i[p]$ is isomorphic to the $G_1$-module ${\mathcal {E}}[p]$ of the case i and the local–global divisibility by $p^2$ does not hold over ${\mathcal {E}}(L_i)$.

Proof

By Proposition 4 and Lemma 15, we are in one of the three cases of the statement. The elliptic curves exist in case 1 by Remark 6 and Corollary 9, in case 2 by Remark 6 and Lemma 10, in case 3 by Remark 6 and Lemma 11. $\square$

Clearly the case 2 of Theorem 2 corresponds to the case when ${\mathcal {E}}(k)$ has a point of order p defined over k. The cases 1 and 3 of Theorem 2 correspond to the case when ${\mathbb Q}( \zeta _p + \overline{\zeta _p}) \subseteq k$.

By the main result of [16] and Theorem 2, we have the following corollary:

Corollary 3

Let $p \ge 5$ be a prime number, let k be a number field and let ${\mathcal {E}}$ be an elliptic curve defined over k. If $p \equiv 1 \mod (3)$ and ${\mathcal {E}}$ does not admit any point of order p over k, then for every positive integer n, the local–global divisibility by $p^n$ holds over ${\mathcal {E}}(k)$. If $p \equiv 2 \mod (3)$, ${\mathcal {E}}$ does not admit any point of order p over k and $[k ({\mathcal {E}}[p]) : k]$ is not 3 or 6, then for every positive integer n the local–global divisibility by $p^n$ holds over ${\mathcal {E}}(k)$.

Proof

If k does not contain ${\mathbb Q}(\zeta _p + \overline{\zeta _p})$, just apply the main result of [16]. If $p \equiv 1 \mod (3 )$ and k contains ${\mathbb Q}(\zeta _p + \overline{\zeta _p})$, and if there exists $n \in {\mathbb N}$ such that the local–global divisibility by $p^n$ does not hold over ${\mathcal {E}}(k)$, then either case 2 or case 3 of Theorem 2 applies. Thus ${\mathcal {E}}$ admits a point of order p defined over k.

If $p \equiv 2 \mod (3)$, ${\mathcal {E}}$ does not admit any point of order p over k, and there exists a positive integer n such that the local–global divisibility by $p^n$ does not hold over ${\mathcal {E}}(k)$, then case 1 of Theorem 2 applies. Hence $k ({\mathcal {E}}[p]) / k$ is either an extension of degree 3 or an extension of degree 6. $\square$

2 Known results

In the following proposition, we combine the main results of [16] and [17] with results of [9].

Proposition 4

Let k be a number field and let ${\mathcal {E}}$ be an elliptic curve defined over k. Let p be a prime number and, for every $m \in {\mathbb N}$, let $G_m$ be $\mathrm{Gal}(k ({\mathcal {E}}[p^m]) / k)$. Suppose that there exists $n \in {\mathbb N}$ such that $H^1_{\mathrm{loc}} (G_n , {\mathcal {E}}[p^n]) \ne 0$. Then one of the following cases holds:

• 1. If p does not divide $\vert G_1 \vert$, then either $G_1$ is cyclic of order dividing $p-1$, generated by an element fixing a point of order p of ${\mathcal {E}}$, or $p \equiv 2 \mod (3)$ and $G_1$ is a group isomorphic either to $S_3$ or to a cyclic group of order 3;

• 2. If p divides $\vert G_1 \vert$ then $G_1$ is contained in a Borel subgroup, and it is either cyclic of order p, or it is generated by an element of order p and an element of order 2 distinct from $-Id$.

Proof

Suppose first that p does not divide $\vert G_1 \vert$. By the argument in [7, p. 29], we have that $G_1$ is isomorphic to its projective image. By [18, Proposition 16], then $G_1$ is either cyclic, or dihedral or isomorphic to one of the following groups: $A_4$, $S_4$, $A_5$.

Suppose that the last case holds. Then $G_1$ should contain a subgroup isomorphic to ${\mathbb Z}/ 2 {\mathbb Z}\times {\mathbb Z}/ 2 {\mathbb Z}$, and so it contains $-Id$. This contradicts the fact that $G_1$ is isomorphic to its projective image.

Suppose that $G_1$ is dihedral. Then $G_1$ is generated by $\tau$ and $\sigma$ with $\sigma$ of order 2 and $\sigma \tau = \tau ^{-1} \sigma$. In particular all the elements of $G_1$ have determinant either 1 or $-1$. Suppose that there exists $i \in {\mathbb N}$ such that $\tau ^i$ has order dividing $p-1$, and distinct from 1. Observe that since p does not divide $\vert G_1 \vert$, we have $H^1 (G_1 , {\mathcal {E}}[p]) = 0$. Then, by [9, Theorem 2], we get that $\tau ^i$ has at least an eigenvalue equal to 1. Thus, since $\tau ^i$ is not the identity, it has determinant $-1$. Then $\tau ^i$ has order 2. Since $\sigma \tau = \tau ^{-1} \sigma$, we get $\sigma \tau ^i = \tau ^{-i} \sigma = \tau ^i \sigma$, because $\tau ^i$ has order 2. Then, since $G_1$ is not cyclic, $\tau ^i$ and $\sigma$ are two distinct elements of order 2 which commute. Thus, like in the previous case, $G_1$ contains a subgroup isomorphic to ${\mathbb Z}/ 2 {\mathbb Z}\times {\mathbb Z}/ 2 {\mathbb Z}$, and so it contains $-Id$. This contradicts the fact that $G_1$ is isomorphic to its projective image. Then $\tau$ has odd order dividing $p+1$. In particular it has two eigenvalues over ${\mathbb F}_{p^2}$: $\lambda$ and $\lambda ^p$. By [9, Proposition 17, Lemma 18] (or see [2, Sect. 3]), if there exists $n \in {\mathbb N}$ such that $H^1 (G_n , {\mathcal {E}}[p^n]) \ne 0$, then the intersection between the sets $\{ 1 , \lambda ^{p-1} , \lambda ^{1-p} \}$ and $\{ \lambda , \lambda ^p \}$ is not trivial. It follows that $\tau$ has order 3. Then 3 divides $p+1$ and $G_1$ is isomorphic to $S_3$.

Finally suppose that $G_1$ is cyclic. If $G_1$ is generated by an element of order dividing $p-1$, by [9, Theorem 2] we have that such an element has an eigenvalue equal to 1. On the other hand if the generator of $G_1$ has order not dividing $p-1$, again by [9, Proposition 17, Lemma 18] (see the dihedral case) we get that such an element has order 3 and 3 divides $p+1$.

Suppose now that p divides $\vert G_1 \vert$. Since p divides the order of $G_1$, by [18, Proposition 15] and the fact that $G_1$ is isomorphic to its projective image, we have that $G_1$ is contained in a Borel subgroup. In particular the p-Sylow subgroup N of $G_1$ is normal. Suppose that $G_1/ N$ is not cyclic. Then $G_1$ is not isomorphic to its projective image. Thus $G_1$ is generated by an element $\sigma$ of order p, which generates N, and an element g of order dividing $p-1$. Suppose that 1 is not an eigenvalue for g. Then by [9, Theorem 2] (in particular notice that, by [9, Remark 16], the hypothesis $H^1 (G_1 , {\mathcal {E}}[p]) = 0$ is not necessary), we have $H^1_{\mathrm{loc}} (G_m , {\mathcal {E}}[p^m]) = 0$ for every $m \in {\mathbb N}$ and so we get a contradiction. Then g has an eigenvalue equal to 1. Suppose that g has order $\ge 3$. Then its determinant has order $\ge 3$ and so, since the determinant is the pth cyclotomic character, k does not contain ${\mathbb Q}(\zeta _p + \overline{\zeta _p})$. Then if g and $\sigma$ do not fix the same point of order p, by [16, Theorem 1] we get a contradiction. On the other hand, since p divides the order of $G_1$, we have $k ({\mathcal {E}}[p]) \ne k (\zeta _p)$. Then by [17, Theorem 3], we get a contradiction.

We conclude that $G_1$ is either cyclic of order p, or it is generated by an element g of order 2 distinct from $-Id$ and an element of order p (which generates a normal subgroup of $G_1$). $\square$

We now recall some properties of the Galois action over the torsion points on an elliptic curve over a number field. In [8] we proved the following Lemma, which is a direct consequence of very interesting results of Greicius [11] and Zywina [20].

Lemma 5

Given a prime number p, a positive integer n and a subgroup G of $\mathrm{GL}_2 ({\mathbb Z}/p^n {\mathbb Z})$, there exists a number field k and an elliptic curve ${\mathcal {E}}$ defined over k such that there are an isomorphism $\phi :\mathrm{Gal}(k ({\mathcal {E}}[p^n]) / k) \rightarrow G$ and a ${\mathbb Z}/ p^n {\mathbb Z}$-linear homomorphism $\tau :{\mathcal {E}}[p^n] \rightarrow ({\mathbb Z}/ p^n {\mathbb Z})^2$ such that, for all $\sigma \in \mathrm{Gal}( k ({\mathcal {E}}[p^n]) / k)$ and $v \in {\mathcal {E}}[p^n]$, we have $\phi (\sigma ) \tau (v) = \tau (\sigma (v))$.

Proof

See [8, Lemma 11]. $\square$

Remark 6

Given a prime number p, a positive integer n and a subgroup G of $\mathrm{GL}_2 ({\mathbb Z}/p^n {\mathbb Z})$, if we suppose $H^1_{\mathrm{loc}} (G , ({\mathbb Z}/ p^n {\mathbb Z})^2) \ne 0$, then by Lemma 5, there exist a number field k and an elliptic curve ${\mathcal {E}}$ defined over k such that $H^1_{\mathrm{loc}} (G_n , {\mathcal {E}}[p^n]) \ne 0$. Hence, by [7, Theorem 3], there exists a finite extension L of k such that $L \cap k ({\mathcal {E}}[p^n]) = k$ and the local–global divisibility by $p^n$ does not hold over ${\mathcal {E}}(L)$.

3 Auxiliary results in the prime to p case

Let $p \equiv 2 \mod (3)$ be a prime number. In [9, Sect. 5] we already found a subgroup G of $\mathrm{GL}_2 ({\mathbb Z}/ p^2 {\mathbb Z})$ such that $H^1_{\mathrm{loc}} (G, ({\mathbb Z}/ p^2 {\mathbb Z})^2) \ne 0$ and the quotient of G by the subgroup H of the elements congruent to the identity modulo p is a cyclic group of order 3. We use the following remark and the following proposition to extend the example to a group $G^\prime$ containing G such that $G^\prime / H$ is isomorphic to $S_3$.

Remark 7

Let p be a prime number, let m be a positive integer, let V be $({\mathbb Z}/ p^2 {\mathbb Z})^{2m}$, let G be a subgroup of $\mathrm{GL}_{2m} ({\mathbb Z}/ p^2 {\mathbb Z})$ and let H be the subgroup of G of the elements congruent to the identity modulo p. Then we have the following inflation–restriction exact sequence:

$$\begin{aligned} 0 \rightarrow H^1 (G/H , V[p]) \rightarrow H^1 (G , V[p]) \rightarrow H^1 (H, V[p])^{G/ H} \rightarrow H^2 (G/H , V[p]). \end{aligned}$$

(3.1)

Moreover, the exact sequence

$$\begin{aligned} 0 \rightarrow V[p] \rightarrow V \rightarrow V[p] \rightarrow 0 \end{aligned}$$

(the first map is the inclusion and the second map the multiplication by p) induces the following exact sequence:

$$\begin{aligned} H^0 (G , V[p]) \rightarrow H^1 (G , V[p]) \rightarrow H^1 (G , V) \rightarrow H^1 (G, V[p]). \end{aligned}$$

(3.2)

Proposition 8

Let p be a prime number, let m be a positive integer, let V be $({\mathbb Z}/ p^2 {\mathbb Z})^{2m}$, let G be a subgroup of $\mathrm{GL}_{2m} ({\mathbb Z}/ p^2 {\mathbb Z})$, and let H be the subgroup of G of the elements congruent to the identity modulo p. Suppose that:

• 1. G has an element $\delta$ not fixing any element of V;

• 2. H is isomorphic, as an G / H-module, to a non-trivial G / H-submodule of V[p];

• 3. For every $h \in H$ distinct from the identity, the endomorphism $h - Id :V / V[p] \rightarrow V / V[p]$ is an isomorphism;

• 4. G / H has order not divisible by p.

Then $H^1_{\mathrm{loc}} (G, V) \ne 0$.

Proof

By Hypothesis 4, we know that the groups $H^1 (G/H , {\mathcal {A}}[p])$ and $H^2 (G/H , {\mathcal {A}}[p])$ in (3.1) are trivial, and hence the restriction map is an isomorphism. Since the action of H over V[p] is trivial and H is an abelian group of exponent p, we have that $H^1 (H , V[p])^{G/H}$ is isomorphic to $\mathrm{Hom}_{{\mathbb Z}/ p {\mathbb Z}[G/ H]} (H, V[p])$. By Hypothesis 2, there exists $\phi :H \rightarrow V[p]$ an injective homomorphism of ${\mathbb Z}/ p {\mathbb Z}[G/ H]$-modules. Let [Z] be in $H^1 (G , V[p])$ such that its image in $H^1 (H, V[p])^{G/ H}$ is the class of $\phi$. In particular, we have $[Z] \ne 0$ because $\phi$ is injective and the restriction map is an isomorphism.

Now observe that $H^0 (G , V[p]) = 0$ by Hypothesis 1. Then, by Remark 7, we have the following exact sequence of G-modules

$$\begin{aligned} 0 \rightarrow H^1 (G , V[p]) \rightarrow H^1 (G , V) \rightarrow H^1 (G, V[p]). \end{aligned}$$

Let us call $[W] \in H^1 (G , V)$ the image of $[Z] \in H^1 (G , V[p])$ defined above by the injective map $H^1 (G , V[p]) \rightarrow H^1 (G , V)$. Since $[Z] \ne 0$, the same holds for [W]. Moreover, since G / H is not divisible by p, the restriction $H^1 (G , V) \rightarrow H^1 (H , V)$ is injective. We conclude because by Hypothesis 3, the image of [W] under this map is in $H^1_{\mathrm{loc}} (H , V)$. $\square$

Corollary 9

Let p be an odd prime such that $p \equiv 2 \mod (3)$. Let G be the subgroup of $\mathrm{GL}_2 ({\mathbb Z}/ p^2 {\mathbb Z})$ generated by

$$\begin{aligned} \tau = \left( \begin{array}{c@{\quad }c} 1 &{} -3 \\ 1 &{} -2 \\ \end{array} \right) \end{aligned}$$

(which has order 3), by an element $\sigma$ of order 2 such that $\sigma \tau \sigma ^{-1} = \tau ^2$ and by

$$\begin{aligned} H = \bigg \{ \left( \begin{array}{c@{\quad }c} 1 + p (a - 2b) &{} 3p (b-a) \\ -pb &{} 1 - p (a - 2b) \\ \end{array} \right) , \ a, b \in {\mathbb Z}/ p^2 {\mathbb Z}\bigg \}. \end{aligned}$$

Then $H^1_{\mathrm{loc}} (G , ({\mathbb Z}/ p^2 {\mathbb Z})^2) \ne 0$.

Proof

It suffices to show that the conditions of Proposition 8 hold for G. Conditions 1 and 4 are clear and condition 3 holds by [9, Sect. 5]. Observe that G / H is isomorphic to $S_3$ and recall that $S_3$ has a unique irreducible representation of dimension 2 over ${\mathbb F}_p$. To prove condition 2 we equivalently prove that H is stable by the conjugation by $\tau$ and $\sigma$. In [9, Sect. 5] we proved that the conjugation by $\tau$ sends H to H.

Let us show that $\sigma H \sigma ^{-1} = H$. A straightforward computation shows that if $\overline{\sigma }$ has order 2 in G / H and $\overline{\sigma } \overline{\tau } \overline{\sigma }^{-1} = \overline{\tau }^2$, then there exists $\alpha , \beta \in {\mathbb F}_p$ such that

$$\begin{aligned} \overline{\sigma } = \left( \begin{array}{c@{\quad }c} \alpha - 2\beta &{} 3 (\beta -\alpha ) \\ \beta &{} 2 \beta - \alpha \\ \end{array} \right) . \end{aligned}$$

Let

$$\begin{aligned} W = \bigg \{ \left( \begin{array}{c@{\quad }c} 1 + p c &{} pd \\ pe &{} 1 - p c \\ \end{array} \right) , \ c, d, e \in {\mathbb Z}/ p^2 {\mathbb Z}\bigg \}. \end{aligned}$$

It is a subgroup of $\mathrm{GL}_2 ({\mathbb Z}/ p^2 {\mathbb Z})$ and a ${\mathbb {F}}_p$-vector space of dimension 3. Observe that W is the subgroup of the group of the matrices congruent to the identity modulo p and having trace 2. Since the trace is invariant under conjugation, we have that $\sigma W \sigma ^{-1} = W$. Let $\phi _\sigma$ be the automorphism of W such that, for every $w \in W$, $\phi _\sigma (w) = \sigma w \sigma ^{-1}$. Observe that since $\sigma$ has order 2, and it is distinct from Id and $-Id$, $\phi _\sigma$ has an eigenspace $W_1$ of dimension 1 for the eigenvalue 1, which is generated by the element $h_1 \in H$ with $a = \alpha$, $b = \beta$, and an eigenspace $W_2$ of dimension 2 for the eigenvalue $-1$. Let h be in H and $h \not \in W_1$. Then $h \in W$ and, since $W = W_1 \bigoplus W_2$, there exist $r \in {\mathbb Z}$ and $h_2 \in W_2$ distinct from the identity such that $h = h_1^r h_2$. Thus $h_2 = h h_1^{-r} \in H$. Since $h_1$ and $h_2$ are linearly indipendent, they generate H. Moreover, $\phi _\sigma ( h_2) = h_2^{-1} \in H$. Then $\phi _\sigma (H) = \sigma H \sigma ^{-1} = H$. $\square$

Lemma 10

Let p be a prime number and let V be $({\mathbb Z}/ p^2 {\mathbb Z})^2$. Let $\lambda \in ({\mathbb Z}/ p^2 {\mathbb Z})^*$ be of order dividing $p-1$ and let G be the following subgroup of $\mathrm{GL}_2 (V)$:

$$\begin{aligned} G = \bigg \langle g = \left( \begin{array}{c@{\quad }c} \lambda &{} 0 \\ 0 &{} 1 \\ \end{array} \right) , h (1, 0) = \left( \begin{array}{c@{\quad }c} 1 + p &{} 0 \\ 0 &{} 1 - p \\ \end{array} \right) , h (0, 1) = \left( \begin{array}{c@{\quad }c} 1 &{} p \\ 0 &{} 1 \\ \end{array} \right) \bigg \rangle . \end{aligned}$$

Then $H^1_{\mathrm{loc}} (G, V) \ne 0$.

Proof

Observe that the subgroup H of G of the elements congruent to the identity modulo p is the group generated by h(1, 0) and h(0, 1). Since G / H has order not divisible by p, $H^1 ( G/ H , V[p]) = 0$ and $H^2 (G/ H , V[p]) = 0$. Then, from the exact sequence (3.1) in Remark 7, we get an isomorphism from $H^1 (G/ H , V[p])$ to $H^1 (H , V[p])^{G/ H}$. Since H acts like the identity over V[p] and since the groups V[p] and H are abelian with exponent p, we have $H^1 (H , V[p])^{G/ H} = \mathrm{Hom}_{{\mathbb Z}/ p {\mathbb Z}[G/H]} (H , V[p])$. Observe that $g h (0, 1) g^{-1} = h (0, 1)^\lambda$ and $g (p , 0) = \lambda (p, 0)$. Then we can define a non-trivial ${\mathbb Z}/ p {\mathbb Z}[G/H]$ homomorphism $\phi$ from H to V[p] by sending h(0, 1) to (p, 0) and h(1, 0) to (0, 0) and extending it by linearity. Let Z be a cocycle representing the class [Z] in $H^1 (G , V[p])$ corresponding to $\phi$. By (3.2) of Remark 7, we have an homomorphism from $H^1 (G, V[p])$ to $H^1 (G , V)$. Let [W] be the image of [Z] for such homomorphism. Let us show that $[W] \in H^1_{\mathrm{loc}} (G, V)$ and $[W] \ne 0$. Since G / H has order not divisible by p, it is sufficient to prove that the image of [W] under the restriction to $H^1 (H , V)$ is in $H^1_{\mathrm{loc}} (H , V)$. For all integers a, b define $h (a , b) := a h (1, 0) + b h (0, 1 )$. Then, by the definition of [Z], we have that h(a, b) is sent to (bp, 0). An easy calculation shows that for every a, b, there exist x, y in ${\mathbb Z}/ p^2 {\mathbb Z}$ such that $(h - Id) (x, y) = (bp, 0)$. This proves that $[W] \in H^1_{\mathrm{loc}} (G, V )$.

Finally observe that for every x, y in ${\mathbb Z}/ p^2 {\mathbb Z}$ such that $(h (1, 0) - Id) (x, y) = (0, 0)$, we have $x \equiv 0 \mod (p)$ and $y \equiv 0 \mod (p)$. On the other hand, for every x, y in ${\mathbb Z}/ p^2 {\mathbb Z}$ such that $(h (1, 0) - Id) (x, y) = (p, 0)$, we have $y \equiv 1 \mod (p)$. Thus $[W] \ne 0$. $\square$

4 Auxiliary results in the p-dividing case

In this section we first prove the following result.

Lemma 11

Let V be $({\mathbb Z}/ p^2 {\mathbb Z})^2$ and let G be the following subgroup of $\mathrm{GL}_2 ({\mathbb Z}/ p^2 {\mathbb Z})$:

$$\begin{aligned} G = \bigg \langle g = \left( \begin{array}{c@{\quad }c} 1 &{} 0 \\ 0 &{} -1 \\ \end{array} \right) , \sigma = \left( \begin{array}{c@{\quad }c} 1 + p &{} 1 \\ 2 p &{} 1 + p \\ \end{array} \right) , h = \left( \begin{array}{c@{\quad }c} 1 + p &{} 0 \\ 0 &{} 1 - p \\ \end{array} \right) \bigg \rangle . \end{aligned}$$

Then $H^1_{\mathrm{loc}} (G, V) \ne 0$.

Proof

Let H be the subgroup of G of the elements congruent to 1 modulo p. Let $\overline{g}$ and $\overline{\sigma }$ be the classes of g and $\sigma$ modulo H. We have that $H^1 (G /H , V[p]) \ne 0$. In fact we can define a cocycle $Z :G/ H \rightarrow V[p]$, which is not a coboundary, by sending, for every integer $i_1 , i_2$, $Z_{\overline{g}^{i_1} \overline{\sigma }^{i_2}}$ to $(p i_2 (i_2 - 1) / 2 , (-1)^{i_1} p i_2)$. Since H is normal, we have an injective homomorphism (the inflation) from $H^1 (G/ H , V[p])$ to $H^1 (G , V[p])$. By abuse of notation we still call Z a cocycle representing the image of the class of Z in $H^1 (G , V[p])$. Moreover, see Remark 7 and in particular the sequence (3.2), we have a homomorphism from $H^1 (G , V[p])$ to $H^1 (G , V)$. It maps the class of Z in $H^1 (G, V[p])$ to some class $[W] \in H^1 (G , V)$. We shall prove that $[W] \in H^1_{\mathrm{loc}} (G, V)$ and $[W] \ne 0$.

First of all let us observe that for every $a, b, c, d \in {\mathbb Z}/ p^2 {\mathbb Z}$, we have

$$\begin{aligned} \left( \begin{array}{c@{\quad }c} 1 + ap &{} 1 + bp \\ cp &{} 1 + dp \\ \end{array} \right) ^p = \left( \begin{array}{c@{\quad }c} 1 &{} p \\ 0 &{} 1 \\ \end{array} \right) . \end{aligned}$$

To verify this write

$$\begin{aligned} \left( \begin{array}{c@{\quad }c} 1 + ap &{} 1 + bp \\ cp &{} 1 + dp \\ \end{array} \right) = \left( \begin{array}{c@{\quad }c} 1 &{} 0 \\ 0 &{} 1 \\ \end{array} \right) + \left( \begin{array}{c@{\quad }c} ap &{} 1 + bp \\ cp &{} dp \\ \end{array} \right) \end{aligned}$$

and observe that

$$\begin{aligned} \left( \begin{array}{c@{\quad }c} ap &{} 1 + bp \\ cp &{} dp \\ \end{array} \right) ^2 \equiv \left( \begin{array}{c@{\quad }c} 0 &{} 0 \\ 0 &{} 0 \\ \end{array} \right) \mod (p), \ \left( \begin{array}{c@{\quad }c} ap &{} 1 + bp \\ cp &{} dp \\ \end{array} \right) ^4 = \left( \begin{array}{c@{\quad }c} 0 &{} 0 \\ 0 &{} 0 \\ \end{array} \right) . \end{aligned}$$

Thus the subgroup H of G of the elements congruent to the identity modulo p is

$$\begin{aligned} H = \bigg \langle \left( \begin{array}{c@{\quad }c} 1 &{} p \\ 0 &{} 1 \\ \end{array} \right) , \left( \begin{array}{c@{\quad }c} 1 + p &{} 0 \\ 0 &{} 1 - p \\ \end{array} \right) \bigg \rangle . \end{aligned}$$

Now observe that, since H and $\langle \sigma , H \rangle$ are normal in G, for every $\tau \in G$ there exist integers $i_1 , i_2 , i_3$ and $h \in H$ such that $\tau = g^{i_1} \sigma ^{i_2} h^{i_3}$. If W is a representant for [W], we have $W_\tau = (p (i_2 - 1) , (-1)^{i_1} p i_2)$. If $i_2 \equiv 0 \mod (p)$, then clearly $W_\tau = (0, 0)$ and so $W_\tau = (\tau - Id) (( 0, 0))$. Then we can suppose $i_2 \not \equiv 0 \mod (p)$. It is simple to prove by induction on $i_2$ that

$$\begin{aligned} \sigma ^{i_2} = \left( \begin{array}{c@{\quad }c} 1 + ap &{} i_2 + b p \\ 2 i_2 p &{} 1 + c p \\ \end{array} \right) \end{aligned}$$

holds for some $a, b, c \in {\mathbb Z}/ p^2 {\mathbb Z}$. Moreover $\sigma ^{i_2} h^{i_3}$ has again the top right entry congruent to $i_2$ modulo p and the bottom left entry equal to $2 i_2 p$. From these remarks is an easy exercise to prove that there exist $\alpha$ and $\beta \in {\mathbb Z}/ p^2 {\mathbb Z}$ such that $W_\tau = (\tau - Id) ((\alpha , p \beta ))$. Then [W] is in $H^1_{\mathrm{loc}} (G, V)$.

Finally let us observe that W is not a coboundary. Let $\alpha$, $\beta \in {\mathbb Z}/ p^2 {\mathbb Z}$ be such that $W_\sigma = (0, p) = (\sigma - Id) ((\alpha , \beta ))$. Then $\alpha \not \equiv 0 \mod (p )$. On the other hand, let $h \in H$ be such that

$$\begin{aligned} h = \left( \begin{array}{c@{\quad }c} 1 + p &{} 0 \\ 0 &{} 1 - p \\ \end{array} \right) . \end{aligned}$$

Then $W_h = (0, 0)$ and so for every $\alpha , \beta \in {\mathbb Z}/ p^2 {\mathbb Z}$ such that $(h - id) ((\alpha , \beta )) = (0, 0)$, we have $\alpha \equiv 0 \mod (p)$. Hence W is not a coboundary. $\square$

Remark 12

For every $a, b, c, d \in {\mathbb Z}/ p^2 {\mathbb Z}$, we have

$$\begin{aligned} \left( \begin{array}{c@{\quad }c} 1 + ap &{} 1 + bp \\ cp &{} 1 + dp \\ \end{array} \right) ^p = \left( \begin{array}{c@{\quad }c} 1 &{} p \\ 0 &{} 1 \\ \end{array} \right) . \end{aligned}$$

In a similar way, for every integer $m \ge 2$, and every $a_m, b_m, c_m, d_m \in {\mathbb Z}/ p^m {\mathbb Z}$, we have

$$\begin{aligned} \left( \begin{array}{c@{\quad }c} 1 + a_mp &{} 1 + b_mp \\ c_mp &{} 1 + d_mp \\ \end{array} \right) ^{p^{m-1}} = \left( \begin{array}{c@{\quad }c} 1 &{} p^m \\ 0 &{} 1 \\ \end{array} \right) . \end{aligned}$$

Corollary 13

Let V be $({\mathbb Z}/ p^2 {\mathbb Z})^2$ and let G be the following subgroup of $\mathrm{GL}_2 ({\mathbb Z}/ p^2 {\mathbb Z})$:

$$\begin{aligned} \widetilde{G} = \bigg \langle \sigma = \left( \begin{array}{c@{\quad }c} 1 + p &{} 1 \\ 2 p &{} 1 + p \\ \end{array} \right) , h = \left( \begin{array}{c@{\quad }c} 1 + p &{} 0 \\ 0 &{} 1 - p \\ \end{array} \right) \bigg \rangle . \end{aligned}$$

Then $H^1_{\mathrm{loc}} (\widetilde{G}, V) \ne 0$

Proof

Observe that $\widetilde{G}$ is a subgroup of index 2 of the group G of Lemma 11. Since $p \ne 2$, the restriction $H^1_{\mathrm{loc}} (G, V) \rightarrow H^1_{\mathrm{loc}} ( \widetilde{G}, V)$ is injective and the result follows. $\square$

Before proving the last result of this section, we need a result of linear algebra.

Lemma 14

Let $n \in {\mathbb N}$ and let G be a subgroup of $\mathrm{GL}_2 ({\mathbb Z}/ p^n {\mathbb Z})$. Let H be the subgroup of G of the elements congruent to the identity modulo p. Suppose that G / H is contained in a Borel subgroup, and it is generated by an element g of order 2 and an element $\sigma$ of order p such that $\sigma$ and g do not fix the same element of order p. Let $\tau$ be in H and let $\sigma _n \in G$ be such that $\sigma _n$ is sent to $\sigma$ by the projection of G over G / H. Then there exist $\tau _d, \tau _l \in H$, $\lambda \in {\mathbb N}$, such that $\tau _d$ is diagonal, $\tau _l$ is lower unitriangular and $\tau = \tau _d \tau _l \sigma _n^{p \lambda }$. In other words H is generated by its subgroups of the diagonal matrices, its subgroup of the lower unitriangular matrices and $\sigma _n^p$.

Proof

Fix a basis of $({\mathbb Z}/ p^n {\mathbb Z})^2$ such that

$$\begin{aligned} \sigma _n \equiv \left( \begin{array}{c@{\quad }c} 1 &{} 1 \\ 0 &{} 1 \\ \end{array} \right) \mod (p). \end{aligned}$$

Then, since g has order 2 and p is odd, there exists an element $g_n$ of G such that

$$\begin{aligned} g_n = \left( \begin{array}{c@{\quad }c} -1 &{} 0 \\ 0 &{} 1 \\ \end{array} \right) . \end{aligned}$$

We remark that $\sigma _n^p \in H$. In fact $\sigma _n^p \equiv Id \mod (p)$.

We first show that every $\tau \in H$ can be written as a product of a lower triangular matrix $\tau _L \in H$ and a power of $\sigma _n^p$. Since $\tau \in H$, $\tau \equiv Id \mod (p)$ and so there exist $e, g, m, r \in {\mathbb Z}/ p^n {\mathbb Z}$ such that

$$\begin{aligned} \tau = \left( \begin{array}{c@{\quad }c} 1 + pe &{} pg \\ pm &{} 1 + pr\\ \end{array} \right) . \end{aligned}$$

We prove by induction that for every integer $i \ge 1$, there exists $\lambda _i \in {\mathbb Z}/ p^n {\mathbb Z}$ such that

$$\begin{aligned} \tau \sigma _n^{p \lambda _i} = \left( \begin{array}{c@{\quad }c} 1 + pe_i &{} p^i g_i \\ pm_i &{} 1 + pr_i\\ \end{array} \right) , \end{aligned}$$

(4.1)

for some $e_i, g_i, m_i, r_i \in {\mathbb Z}/p^n {\mathbb Z}$. If $i = 1$ then for $\lambda _1 = 0$ the relation (4.1) is satisfied. Suppose that (4.1) is satisfied for an integer $i \ge 1$. Then there exists $\lambda _i \in {\mathbb Z}/ p^n {\mathbb Z}$ such that

$$\begin{aligned} \tau \sigma _n^{p \lambda _i} = \left( \begin{array}{c@{\quad }c} 1 + pe_i &{} p^i g_i \\ pm_i &{} 1 + pr_i\\ \end{array} \right) , \end{aligned}$$

for some $e_i, g_i, m_i, r_i \in {\mathbb Z}/p^n {\mathbb Z}$. Choose an element $\lambda _{i+1}$ of ${\mathbb Z}/p^n {\mathbb Z}$ such that $p \lambda _{i+1} = p \lambda _i - p^i g_i$. Observe that this element exists because $i \ge 1$. By Remark 12 we have

$$\begin{aligned} \sigma _n^{-p^i g_i}&= \left( \begin{array}{c@{\quad }c} 1 + p^{i+1}a_{i+1} &{} p^i + p^{i+1}b_{i+1} \\ p^{i+1}c_{i+1} &{} 1 + p^{i+1}d_{i+1}\\ \end{array} \right) ^{-g_i} \\&= \left( \begin{array}{c@{\quad }c} 1 + p^{i+1}a^\prime _{i+1} &{} -p^i g_i + p^{i+1}b^\prime _{i+1} \\ p^{i+1}c^\prime _{i+1} &{} 1 + p^{i+1}d^\prime _{i+1}\\ \end{array} \right) , \end{aligned}$$

for some $a^\prime _{i+1}, b^\prime _{i+1}, c^\prime _{i+1}, d^\prime _{i+1} \in {\mathbb Z}/p^n {\mathbb Z}$. By a short computation

$$\begin{aligned} \tau \sigma _n^{p \lambda _{i+1}}&= \tau \sigma _n^{p\lambda _i}\sigma _n^{-p^i g_i} \\&= \left( \begin{array}{c@{\quad }c} 1 + pe_i &{} p^i + p^ig_i \\ pm_i &{} 1 + pr_i\\ \end{array} \right) \left( \begin{array}{c@{\quad }c} 1 + p^{i+1}a^\prime _{i+1} &{} -p^i g_i + p^{i+1}b^\prime _{i+1} \\ p^{i+1}c^\prime _{i+1} &{} 1 + p^{i+1}d^\prime _{i+1}\\ \end{array} \right) \\&= \left( \begin{array}{c@{\quad }c} 1 + pe_{i+1} &{} + p^{i+1}g_{i+1} \\ pm_{i+1} &{} 1 + pr_{i+1}\\ \end{array} \right) , \end{aligned}$$

for some $e_{i+1}, g_{i+1}, m_{i+1}, r_{i+1} \in {\mathbb Z}/p^n {\mathbb Z}$. Then (4.1) is verified for $\lambda _{i+1}$ that satisfies $p\lambda _{i+1} = p\lambda _i -p^i g_i$. In particular for $i = n$ we have

$$\begin{aligned} \tau \sigma _n^{p \lambda _n} = \left( \begin{array}{c@{\quad }c} 1 + pe_n &{} 0 \\ pm_n &{} 1 + pr_n\\ \end{array} \right) . \end{aligned}$$

Then, setting $\tau _L = \tau \sigma _n^{p \lambda _n}$ and $\lambda = -\lambda _n$, we have shown that $\tau$ can be written as a product of a lower triangular matrix $\tau _L \in H$ and the power $\sigma _n^{p \lambda }$ of $\sigma _n^p$.

Observe that, to conclude the proof, it is sufficient to show that $\tau _L$ can be written as the product of a diagonal matrix $\tau _d \in H$ and a lower unitriangular matrix $\tau _l \in H$. Since H is normal in G, $g_n \tau _L g_n^{-1} \in H$. Then $g_n \tau _L g_n^{-1} \tau _L^{-1} \in H$. Moreover, by a simple computation, we have

$$\begin{aligned} g_n \tau _L g_n^{-1} \tau _L^{-1} = \left( \begin{array}{c@{\quad }c} 1 &{} 0 \\ -2pm_n / (pe_n + 1) &{} 1 \\ \end{array} \right) . \end{aligned}$$

Thus

$$\begin{aligned} (g_n \tau _L g_n^{-1} \tau _L^{-1})^{- (pe_n + 1) / 2 (pr_n + 1 )} = \left( \begin{array}{c@{\quad }c} 1 &{} 0 \\ pm_n / (pr_n + 1) &{} 1 \\ \end{array} \right) \in H. \end{aligned}$$

Call such a matrix $\tau _l$ and observe that

$$\begin{aligned} \tau _L = \left( \begin{array}{c@{\quad }c} 1 + pe_n &{} 0 \\ 0 &{} 1 + pr_n\\ \end{array} \right) \left( \begin{array}{c@{\quad }c} 1 &{} 0 \\ pm_n / (pr_n + 1) &{} 1 \\ \end{array} \right) . \end{aligned}$$

Call the diagonal matrix $\tau _d$. Since $\tau _L, \tau _l \in H$, also $\tau _d \in H$, proving the statement. $\square$

Lemma 15

Under the assumptions and with the notation of Lemma 14, let $V_n$ be $({\mathbb Z}/ p^n {\mathbb Z})^2$. Then $H^1_{\mathrm{loc}} (G, V_n) = 0$.

Proof

By replacing V with $V_n$, by observing that $H^0 (G , V_n[p^{n-1}]) = 0$ because the group generated by g and $\sigma$ do not fix any element of $V_n[p^{n-1}]$, and by using the Remark 7, we get the following exact sequence

$$\begin{aligned} 0 \rightarrow H^1 (G, V_n[p]) \rightarrow H^1 (G, V_n) \rightarrow H^1 (G, V_n[p^{n-1}]). \end{aligned}$$

(4.2)

Suppose that $H^1_{\mathrm{loc}} (G, V_n) \ne 0$. Then $H^1_{\mathrm{loc}} (G, V_n)[p] \ne 0$ and so let Z be a cocycle representing a non-trivial class $[Z] \in H^1_{\mathrm{loc}} (G, V_n)[p]$. Let us observe that [Z] is in the kernel of the map $H^1 (G, V_n) \rightarrow H^1 (G, V_n[p^{n-1}])$ (here we generalize the proof of [9, Lemma 13]). Since [Z] has order p, then pZ is a coboundary and so there exists $v \in V_n$ such that, for every $\tau \in G$, $pZ_\tau = \tau (v) - v$. Let us observe that $v \in V_n[p^{n-1}]$. Since for every $\tau$ we have $\tau (v) - v \in V_n[p^{n-1}]$, and we get that $v \in \cap _{\tau \in G} \ker (p^{n-1} ( \tau - Id))$. Since G does not fix any element of order p, the unique possibility is that $v \in V_n[p^{n-1}]$. Then (see the sequence (4.2)) [Z] is in the image of $H^1 (G, V_n[p]) \rightarrow H^1 (G, V_n)$. By abuse of notation we call [Z] the class in $H^1 (G, V_n[p])$ sent to [Z].

Consider now the inflation–restriction sequence

$$\begin{aligned} 0 \rightarrow H^1 (G / H , V_n[p]) \rightarrow H^1 (G, V_n[p]) \rightarrow H^1 (H , V_n[p])^{G/ H}. \end{aligned}$$

(4.3)

Let us observe that $H^1 (G/ H , V_n[p]) = 0$. Let $W :G/ H \rightarrow V_n[p]$ be a cocycle. Since $\sigma$ and g are contained in a Borel subgroup, g has order 2, and g and $\sigma$ do not fix any nonzero element of $V_n[p^{n-1}]$, we can choose a basis of $V_n$ such that $(p^{n-1} , 0)$ is fixed by $\sigma$, $g ((p^{n-1} , 0)) = (- p^{n-1} , 0)$ and $(0 , p^{n-1})$ is sent to $(p^{n-1} , p^{n-1})$ by $\sigma$ and fixed by g. Observe that, since summing a coboundary to W does not change its class, we can suppose that $W_\sigma = (0 , p^{n-1})$. Then, for every integer i, we have $W_{\sigma ^i} = (p^{n-1} i ( i-1) / 2 , p^{n-1} i)$. Observe that since g has order 2, we have $W_{g^2} = W_g + g W_g = (0, 0)$. In particular there exists $a \in {\mathbb Z}/ p^n {\mathbb Z}$ such that $W_g = (p^{n-1} a , 0)$, and which is fixed by $\sigma$. Thus $W_{g \sigma g^{-1}} = g W_\sigma = ( p^{n-1} , -p^{n-1})$. On the other hand, $g \sigma g^{-1} = \sigma ^{-1}$ and so $W_{\sigma ^{-1}} = (- p^{n-1} , -p^{n-1})$. We then get a contradiction. Thus, by the sequence (4.3), to every class of $H^1 (G, V_n[p])$ we can associate a class in $H^1 (H , V_n[p])^{G/ H}$. Since H acts as the identity over $V_n[p]$, we have that $H^1 (H , V_n[p])^{G/ H}$ is a subgroup of $\mathrm{Hom} (H , V_n[p])$. In particular, we can associate with $[Z] \in H^1 (G , V[p])$ defined above a homomorphism from H to $V_n[p]$. By Lemma 14, for every $\tau \in H$ there exist $\tau _l \in H$ a lower unitriangular matrix, $\tau _D \in H$ a diagonal matrix and $\lambda \in {\mathbb Z}$ such that $\tau = \tau _l \tau _D \sigma _n^{\lambda p}$. Consider the homorphism associated with $[Z] \in H^1 (G, V_n[p])$. Since the cocycle Z has values in $V_n[p]$, in particular $Z_{\sigma _n} \in V_n[p]$ and, by the cocycle property, $Z_{\sigma _n^p} = (0, 0)$. On the other hand, since $g_n \tau _D g_n^{-1} = \tau _D$, there exists $b \in {\mathbb Z}/ p^n {\mathbb Z}$ such that $Z_{\tau _D} = (0, p^{n-1} b)$. If $p^{n-1} b$ is distinct from 0, then $(0, p^{n-1} b)$ generates V[p] as an G / H-module. Since $g_n \tau _l g_n^{-1} = \tau _l^{-1}$, there exists $a \in {\mathbb Z}/ p^n {\mathbb Z}$ such that $Z_{\tau _l} = (p^{n-1} a , 0 )$. Observe that for every $(\alpha , \beta ) \in V_n$, we have that $(\tau _l - Id) (\alpha , \beta ) = (p^{n-1} a , 0)$ only if $p^{n-1} a = 0$. Then if the image of Z satisfies the local conditions over $V_n$, the homomorphism associated with Z is trivial, and so Z is a coboundary. $\square$