Ovals in \({\mathbb {Z}}^2_{2p}\)

Abstract

By an oval in \({\mathbb {Z}}^2_{2p},\)p odd prime, we mean a set of \(2p+2\) points, such that no three of them are on a line. It is shown that ovals in \({\mathbb {Z}}^2_{2p}\) only exist for \(p=3,5\) and they are unique up to an isomorphism.

Introduction

A k-arc is a set of k points, such that no three of them are collinear. Arcs are well studied in projective geometry, see, e.g., [1, 9] and [2] for further references. Recall that any non-singular conic of \(\mathrm {PG}(2,q)\) is a \((q + 1)\)-arc. If K is any k-arc of \(\mathrm {PG}(2,q)\) with q odd, then \(k \le q + 1.\) A \((q+1)\)-arc is called an oval. A famous theorem of Segre [9] tells us that for q odd, every oval of \(\mathrm {PG}(2,p)\) is a non-singular conic.

Many authors continue to study the classical problem in the context of Hjelmslev geometry, see, e.g., [3, 4, 8] and [6] for the definition of the abstract Hjelmslev plane. In this article, we consider similar questions in \({\mathbb {Z}}_n^2.\) This modified problem is still interesting and it was investigated in [5, 7] and [10]. If \(n=p\) is prime, the resulting space is just \(\mathrm {AG}(2,p).\) The correspondence between projective and affine planes leads to the following facts: The maximum size of an arc in \({\mathbb {Z}}^2_{p}\) is \(p+1;\) every \((p+1)\)-arc in \({\mathbb {Z}}^2_{p}\) is a non-singular conic.

The maximum size of a cap in \({\mathbb {Z}}^2_{2p},\)p odd prime, is \(2p+2\) (see Lemma 4.1 in [7] and Theorem 3.1 in [10]). We will call a \((2p+2)\)-arc in \({\mathbb {Z}}^2_{2p}\) an oval in \({\mathbb {Z}}^2_{2p}.\) In this article, we completely solve the problem of the existence and the uniqueness of ovals in \({\mathbb {Z}}^2_{2p}.\)

We define a line in \({\mathbb {Z}}_n^2\) to be a subset of \({\mathbb {Z}}_n^2\) of the form:

$$\begin{aligned} \left\{ (x; y) : ax + by+c = 0\right\} , \end{aligned}$$

where \(\gcd (a; b; n) = 1.\) This line is denoted by \(\ell _{\left[ a;b;c\right] }.\)

By an automorphism of \({\mathbb {Z}}_{n}^{2}\), we mean a mapping from \({\mathbb {Z}}_{n}^{2}\) to \({\mathbb {Z}}_{n}^{2}\) which preserves arcs. Let XY be arcs in \({\mathbb {Z}}_{n}^2.\) We say that X and Y are isomorphic if there exists an automorphism of \({\mathbb {Z}}_{n}^{2}\) mapping X to Y.

The General Case

Due to the Chinese remainder theorem, we have the following.

Lemma 2.1

Let p be an odd prime, \(\phi _2:{\mathbb {Z}}^2_{2p}\rightarrow {\mathbb {Z}}^2_{2},\)\(\phi _p:{\mathbb {Z}}^2_{2p}\rightarrow {\mathbb {Z}}^2_{p}\) be reduction maps. Then, any three points \(x,y,z\in {\mathbb {Z}}^{2}_{2p}\) are collinear if and only if \(\phi _2(x),\phi _2(y),\phi _2(z) \in {\mathbb {Z}}^{2}_{2}\) and \(\phi _p(x),\phi _p(y),\phi _p(z) \in {\mathbb {Z}}^{2}_{p}\) are collinear.

From now on, we will use the notation of the previous Lemma.

Lemma 2.2

Let p be an odd prime. Let \(\sigma \) be an arbitrary permutation of \({\mathbb {Z}}^{2}_{2}\). Then, there exists an automorphism F of \({\mathbb {Z}}^{2}_{2p}\), such that \(\phi _{2}\circ F =\sigma \circ \phi _{2}\) and \(\phi _{p}\circ F =Id \circ \phi _{p}.\)

Proof

Let \(F_{A}\) and \(F_{B}\) be linear transformations determined by matrices \(A=\left( {\begin{matrix} 1 &{} 0\\ p &{} 1\\ \end{matrix}}\right) \) and \(B=\left( {\begin{matrix} p+1 &{} p\\ p &{} p+1\\ \end{matrix}}\right) \), respectively. Denote by \(t_{[0,p]}\) the translation by a vector [0, p]. Note that the group of permutations of the set \({\mathbb {Z}}^{2}_{2}\) is generated by transpositions \(\sigma _{1}=\left( {\scriptstyle (0,0), (0,1)}\right) ,\)\(\sigma _{2}=\left( {\scriptstyle (0,1), (1,0)}\right) ,\)\(\sigma _{3}=\left( {\scriptstyle (1,0), (1,1)}\right) .\)

One can verify by a straightforward calculation that \(\phi _{2}\circ F_{A}\circ t_{\left[ 0,p\right] }=\sigma _{1}\circ \phi _{2},\)\(\phi _{2}\circ F_{B}=\sigma _{2}\circ \phi _{2} \) and \(\phi _{2}\circ F_{A}=\sigma _{3}\circ \phi _{2}.\)\(\square \)

Lemma 2.3

Let X be an oval in \({\mathbb {Z}}_{2p}.\) Then, the following holds.

  1. (1)

    \(|\phi _2(X)|=4.\)

  2. (2)

    \(|\phi ^{-1}_2(a)\cap X|=\frac{p+1}{2}\) for every \(a\in {\mathbb {Z}}_2^2.\)

  3. (3)

    \(|\phi _p(X)|=2p+2.\)

Proof

To prove (1), suppose that \(|\phi _2(X)|<4.\) Note that in this case, there exists a line l in \( {\mathbb {Z}}^{2}_{2}\), such that \(|\phi ^{-1}_2(l)\cap X|>p+1.\) As a consequence of Lemma 2.1 together with the fact that the maximum size of an arc in \({\mathbb {Z}}_p\) is \(p + 1,\) we obtain a contradiction with the assumption that X is an arc. (2) follows immediately from (1) and Lemma 2.2. To prove (3), suppose, to the contrary, that \(\phi _p(a)=\phi _p(b)\) for some \(a,b \in X\). By (2), there exists a \(c\in X\), such that \(\phi _2(a)=\phi _2(c),\) so there exists a c, such that \(\phi _2(a),\)\(\phi _2(b)\) and \(\phi _2(c)\) are collinear. Moreover, it is clear that \(\phi _p(a),\)\(\phi _p(b),\) and \(\phi _p(c)\) are collinear. Therefore, abc is a collinear triple by Lemma 2.1, a contradiction. \(\square \)

Theorem 2.4

There is no oval in \({\mathbb {Z}}_{2p}\) for \(p\ge 7.\)

Proof

It is shown in [10] by computer calculations that the maximum size of an arc in \({\mathbb {Z}}_{14}\) is 12. Assume, to the contrary, that there is an oval X in \({\mathbb {Z}}_{2p},\) where \(p>7\) is an odd prime. By Lemma 2.3 (1), we have \(\phi _2(X)={\mathbb {Z}}_2^{2}.\) Let \(a\in {\mathbb {Z}}_2^2.\) By Lemma 2.3 (2),(3), \(|\phi _p\left( \phi ^{-1}_2(a)\cap X \right) |>4.\) Denote by \(\ell \) a line through a in \({\mathbb {Z}}_2^2.\) By Lemma 2.3 (2) and (3) and Lemma 2.1, \(\phi _p(\phi _2^{-1}(\ell )\cap X)\) is a \((p+1)-\)arc in \({\mathbb {Z}}_p^2,\) i.e., \(\phi _p(\phi _2^{-1}(\ell )\cap X)\) is a non-singular conic. Note that there are exactly three distinct lines through a in \({\mathbb {Z}}_2^2,\) Therefore, we have three distinct non-singular conics through \(\phi _p \left( \phi ^{-1}_2(a)\cap X\right) \) in \(Z_{p}.\) However, this is a contradiction to the fact that there exists a unique non-singular conic through five points, no three collinear. \(\square \)

The Cases \(p=3,5\)

The existence of ovals in \({\mathbb {Z}}_{6}^2\) and \({\mathbb {Z}}_{10}^2\) was proved in [10], where appropriate examples were given, see also Figs. 1 and 2. Therefore, it remains to show the uniqueness.

First, we introduce some notation. Let \(X\subset {\mathbb {Z}}^{2}_{2p}\) be an oval. For \((i,j)\in {\mathbb {Z}}_2^2,\) let \(X_{ij}=\phi _p(\phi _2^{-1}(i,j)\cap X).\) A d-secant of K is a line l, such that \(|l\cap K|=d.\) Denote by \({\mathcal {S}}^{d}_K\) the set of d-secants of K.

Fig. 1
figure1

An oval in \({\mathbb {Z}}^2_6\)

Fig. 2
figure2

An oval in \({\mathbb {Z}}^2_{10}\)

The following Lemma is an immediate consequence of Lemma 2.1.

Lemma 3.1

Let X be an oval in \({\mathbb {Z}}^{2}_{2p},\)\(a,b\in X\). If \(\phi _2(a)=\phi _2(b)\), then the line through \(\phi _p(a)\) and \(\phi _p(b)\) is a \(2-\)secant of \(\phi _p(X).\)

Theorem 3.2

An oval in \({\mathbb {Z}}_6\) is unique up to an isomorphism.

Proof

Let Y be an oval in \({\mathbb {Z}}_6\) given in Fig. 1. This means that:

$$\begin{aligned} Y=\left\{ (0,0),(0,1),(1,0),(1,2),(2,1),(2,2),(3,5),(5,3)\right\} . \end{aligned}$$

Then, \(|\phi _3(Y)|=8,\)\(Y_{00}=\left\{ (0,0),(2,2)\right\} ,\)\(Y_{01}=\left\{ (0,1),(2,1)\right\} \), \(Y_{10}=\left\{ (1,0),(1,2)\right\} \), \(Y_{11}=\left\{ (0,2),(2,0)\right\} .\) Let X be another oval in \({\mathbb {Z}}_6^2.\) We will show that X is isomorphic to Y. By Lemma 2.3 (3), \(|\phi _3(X)|=8.\) Let \((i,j)\in {\mathbb {Z}}^2_2.\) Using Lemma 2.3(2) and (3), we obtain \(|X_{ij}|=2.\) From Lemma 3.1, it follows that \(X_{ij}\) is a 2-secant of \(\phi _3(X).\) Since the translations are automorphisms of \({\mathbb {Z}}^2_{6}\), we can assume that \((1,1)\notin \phi _3(X).\) This means that the points of \(X_{ij}\) lie on one of the 4 lines through (1, 1). Thus, there exists \((i',j')\in {\mathbb {Z}}^2_2\), such that \(X_{ij}=Y_{i'j'}.\) Consequently, we obtain the permutation \(\sigma \) of \({\mathbb {Z}}_2^2\) given by \(\sigma (i,j)=(i',j').\) The proof is completed by Lemma  2.2.

\(\square \)

The proof of Theorem 3.6 is straightforward but tedious. Therefore, we collect some useful Lemmas.

Lemma 3.3

Let XY be ovals in \({\mathbb {Z}}_{10}^2\). If \(\phi _5(X)\) and \(\phi _5(Y)\) are isomorphic, then X and Y are isomorphic.

Proof

First, we will show that for every automorphism f of \(Z_{5}^2\), there exists an automorphism F of \(Z_{10}^2\), such that \(\phi _5\circ F=f\circ \phi _5.\) If f is a translation by a vector, then we may take \(F=f.\) Obviously, F satisfies the required condition. Let f be an automorphism arising from multiplication by \(A\in GL(2,{\mathbb {Z}}_5).\) It can be verified that at least one of A :  \(A+ \left( {\begin{matrix} 5 &{} 0\\ 0&{} 0\\ \end{matrix}}\right) ,\)\(A+ \left( {\begin{matrix} 0 &{} 5\\ 0&{} 0\\ \end{matrix}}\right) ,\)\(A+ \left( {\begin{matrix} 0 &{} 0\\ 5&{} 0\\ \end{matrix}}\right) ,\)\(A+ \left( {\begin{matrix} 0 &{} 0\\ 0&{} 5\\ \end{matrix}}\right) ,\)\(A+ \left( {\begin{matrix} 5 &{} 0\\ 0&{} 5\\ \end{matrix}}\right) \) is irreducible in \({\mathbb {Z}}_{10},\) call it \({\tilde{A}}\). Then, F given by multiplication by \({\tilde{A}}\) satisfies the required condition.

Next, we claim that \(f(X_{ij})=Y_{i'j'}\). Suppose, to the contrary, that \(f(X_{ij})\) has a nonempty intersection with at least two of \(Y_{00},\)\(Y_{01},\)\(Y_{10},\)\(Y_{11}.\) Recall that by Lemma 2.1, there exist six conics \(C_i,\)\(i=1,\ldots ,6,\) in Y corresponding to six lines in \({\mathbb {Z}}_2^2.\) Consider a conic C containing \(X_{ij}.\) Then, f(C) is the conic different from \(C_i,\)\(i=1,\ldots ,6.\) Moreover, there exists \(C_i\) for some i, such that \(|C\cup C_i|\ge 5.\) However, this contradicts the fact that there exists a unique non-singular conic through five points, no three collinear.

Finally, consider the permutation \(\sigma \) of \({\mathbb {Z}}_2^2\) given by \(\sigma (i,j)=(i',j').\) Denote by \(F_1\) an automorphism described in Lemma 2.2. Then, the composition of F and \(F_1\) maps X onto Y. \(\square \)

Lemma 3.4

Let \(X\subset {\mathbb {Z}}^{2}_{10}\) be an oval, \(F(x,y)=(-x,y).\) If \(F(X_{00})=X_{01}\) and \(F(X_{10})=X_{00}\), then \(\ell _{[1,0,0]}\notin {\mathcal {S}}^2_{X_{10}}\cup {\mathcal {S}}^2_{X_{11}}.\)

Proof

Since \(X_{10}\) is an arc, we have \(\ell _{[1,0,0]}\in {\mathcal {S}}^k_{X_{10}}\) for \(k=0,1,2.\) Suppose, to the contrary, that \(\ell _{[1,0,0]}\in {\mathcal {S}}^2_{X_{10}}\), i.e., \(|X_{10}\cap \ell _{[1,0,0]}|=2.\) Obviously \(|X_{10}\cap F(X_{10})|=2\) and \(X_{10}\ne F(X_{10})\). We obtain ovals, i.e., non-singular conics, \(C_1=X_{00}\cup X_{10}\) and \(C_2=F(X_{01}\cup X_{10})=X_{00}\cup F(X_{10})\) with \(|C_1\cap C_2|=5.\) However, this contradicts the fact that there exists a unique non-singular conic through five points, no three collinear. Similarly, \(\ell _{[1,0,0]}\notin {\mathcal {S}}^k_{X_{11}}.\)\(\square \)

Lemma 3.5

Let \(X\subset {\mathbb {Z}}^{2}_{10}\) be an oval. Then, the following holds:

  1. (1)

    \({\mathcal {S}}^2_{X_{10}}\cup {\mathcal {S}}^2_{X_{11}}\subset {\mathcal {S}}^0_{X_{00}\cup X_{01}};\)

  2. (2)

    \(X_{ij}=\left\{ l_1\cap l_2, l_1\cap l_3, l_2\cap l_3\right\} ,\) where \({\mathcal {S}}^2_{X_{ij}}=\left\{ l_1,l_2,l_3\right\} .\)

Proof

(1) is a direct consequence of Lemma 2.1. (2) follows from the definition of \({\mathcal {S}}^2_{X_{ij}}\) together with the fact that both \(X_{ij}\) and \({\mathcal {S}}^2_{X_{ij}}\) have size 3.

\(\square \)

We are now ready for the proof of Theorem 3.6.

Theorem 3.6

An oval in \({\mathbb {Z}}_{10}\) is unique up to an isomorphism.

Proof

Let Y be an oval in \({\mathbb {Z}}_{10}\) given in Fig. 2. This means that:

$$\begin{aligned} Y=\left\{ (0,0),(1,0),(0,1),(1,1),(4,2),(7,2),(5,4),(6,4),(5,7),(6,7),(4,9),(7,9)\right\} . \end{aligned}$$

Let X be another arc of size 12 in \({\mathbb {Z}}_{10}^2.\) We will show that X is isomorphic to Y. Recall that \(X_{00}\cup X_{01}\) is a non-singular conic C in \({\mathbb {Z}}_5^2.\) Thus, C can be transformed by some automorphism of \({\mathbb {Z}}_{5}^{2}\) to \(C_1=\left\{ (x,y):x^2+2y^2=1\right\} .\) By Lemma 3.3, we can write \(X_{00}\cup X_{01}=\left\{ (1,0),(2,1),(2,4),(4,0),(3,1),(3,4) \right\} .\)

It is straightforward to check that in this case, we have:

$$\begin{aligned} {\mathcal {S}}^0_{X_{00}\cup X_{01}}= \left\{ \ell _{[1;0;0]}, \ell _{[1;1;0]}, \ell _{[4;1;0]}, \ell _{[0;1;2]}, \ell _{[0;1;3]}, \ell _{[3;1;1]}, \ell _{[3;1;4]}, \ell _{[2;1;1]}, \ell _{[2;1;4]}\right\} .\nonumber \\ \end{aligned}$$
(3.0.1)

We need to split \(X_{00}\cup X_{01}\) into two subsets \(X_{00},\)\(X_{01}.\) By Lemma 2.2, we may assume that \((1,0)\in X_{00}.\) There are ten cases for \(X_{00}.\) To reduce the number of cases, we use Lemma 3.3 and the automorphism G of \({\mathbb {Z}}_5^2\) given by \(G(x,y)=(3x+3y,x+3y).\) A straightforward computation shows that it remains to consider the following three cases: (1) \(X_{00}=\left\{ (1,0),(3,1),(3,4) \right\} ,\) (2) \(X_{00}=\left\{ (1,0),(2,1),(2,4) \right\} ,\) (3) \(X_{00}=\left\{ (1,0),(2,1),(3,4) \right\} .\)

We shall consider the cases (1) and (2) separately, and then, we will show that cases 1 and 3 give isomorphic ovals.

Case 1\(X_{00}=\left\{ (1,0),(3,1),(3,4) \right\} .\) Then, \(X_{01}=\left\{ (4,0),(2,1),(2,4) \right\} .\) It is straightforward to check that \({\mathcal {S}}^2_{X_{00}}=\left\{ \ell _{[2;1;3]},\ell _{[3;1;2]},\ell _{[1;0;2]}\right\} \) and \({\mathcal {S}}^2_{X_{01}}=\left\{ \ell _{[3;1;3]},\ell _{[2;1;2]},\ell _{[1;0;3]} \right\} .\) Let \(\langle {\mathcal {S}}^2_{X_{ij}}\rangle =\left\{ p\in \ell | \ell \in {\mathcal {S}}^2_{X_{ij}} \right\} .\) Denote by \(\overline{\langle {\mathcal {S}}^2_{X_{ij}}\rangle }\) the complement of \(\langle {\mathcal {S}}^2_{X_{ij}}\rangle .\) Next, one can verify that \(\overline{\langle {\mathcal {S}}^2_{X_{00}}\rangle }\cap \overline{\langle {\mathcal {S}}^2_{X_{01}}\rangle }= \left\{ (0,0),(0,1),(0,4),(1,2),(1,3),(4,2),(4,3) \right\} .\) It follows from Lemma 3.1 that \( X_{10}\cup X_{11}\subset \overline{\langle {\mathcal {S}}^2_{X_{00}}\rangle }\cap \overline{\langle {\mathcal {S}}^2_{X_{01}}\rangle }.\) Furthermore, note that (0, 0) is only one point of \(\ell _{[4;1;0]}\) in \(\overline{\langle {\mathcal {S}}^2_{X_{00}}\rangle }\cap \overline{\langle {\mathcal {S}}^2_{X_{01}}\rangle }.\) Thus, \(\ell _{[4;1;0]}\notin {\mathcal {S}}^2_{X_{10}}\cup {\mathcal {S}}^2_{X_{11}}.\) By the same argument, \(\ell _{[1;1;0]}\notin {\mathcal {S}}^2_{X_{10}}\cup {\mathcal {S}}^2_{X_{11}}.\) Lemma 3.4 implies that \(\ell _{[1;0;0]}\notin {\mathcal {S}}^2_{X_{10}}\cup {\mathcal {S}}^2_{X_{11}}.\) The above considerations together with Lemma 3.5 (1) and Eq. (3.0.1) imply that \({\mathcal {S}}^2_{X_{10}}\cup {\mathcal {S}}^2_{X_{11}}=\left\{ \ell _{[0;1;2]}, \ell _{[0;1;3]},\ell _{[3;1;1]}, \ell _{[3;1;4]},\ell _{[2;1;1]}, \ell _{[2;1;4]}\right\} .\) (We use the fact that \(|{\mathcal {S}}^2_{X_{10}}\cup {\mathcal {S}}^2_{X_{11}}|=6\).) Lemma 3.5 (2) and simple calculations show that \((0,0)\notin X_{10}\cup X_{11}.\) Since \(|X_{10}\cup X_{11}|=6\), \(X_{10}\cup X_{11}=\left\{ (0,1),(0,4),(1,2),(1,3),(4,2),(4,3) \right\} .\) Now, we need to split \(X_{10}\cup X_{11}\) into two subsets \(X_{10},\)\(X_{11}.\) By Lemma 2.2, we may assume that \((0,1)\in X_{10}.\) It can be verified that \(\left\{ (0,1)\right\} =\ell _{[3;1;4]}\cap \ell _{[2;1;4]}.\) By Lemma 3.5 (2), \(\ell _{[3;1;4]},\ell _{[2;1;4]}\in {\mathcal {S}}^2_{X_{10}}.\) One can verify that \(\ell _{[3;1;4]}\cap (X_{10}\cup X_{11})=\left\{ (0,1),(1,3) \right\} \) and \(\ell _{[2;1;4]}\cap (X_{10}\cup X_{11})=\left\{ (0,1),(4,3) \right\} .\) Hence, \(X_{10}=\left\{ (0,1),(1,3),(4,3)\right\} \), and so, \(X_{11}=\left\{ (0,4),(1,2),(4,2)\right\} \)

Thus:

$$\begin{aligned} X=\left\{ (6,0),(8,6),(8,4),(4,5),(2,1),(2,9),(5,6),(1,8),(9,8), (5,9),(1,7),(9,7)\right\} . \end{aligned}$$

Finally, one can check that \(H(X)=Y,\) where \(H(x,y)=(3x+7y+8,4x+9y+3).\)

Case 2\(X_{00}=\left\{ (1,0),(2,1),(2,4) \right\} .\) Then, \(X_{01}=\left\{ (4,0),(3,1),(3,4) \right\} .\) In this case, \({\mathcal {S}}^2_{X_{00}}=\left\{ \ell _{[4;1;1] },\ell _{[1;0;3]},\ell _{[1;1;4] }\right\} \) and \({\mathcal {S}}^2_{X_{01}}=\left\{ \ell _{[4;1;4] },\ell _{[1;0;2]},\ell _{[1;1;1] } \right\} .\) With the notation of the above case, it can be verified that:

$$\begin{aligned} \overline{\langle {\mathcal {S}}^2_{X_{00}}\rangle }\cap \overline{\langle {\mathcal {S}}^2_{X_{01}}\rangle }= \left\{ (0,0),(0,2),(0,3),(1,1),(1,4),(4,1),(4,4)\right\} . \end{aligned}$$

Hence, \( X_{10}\cup X_{11}\subset \left\{ (0,0),(0,2),(0,3),(1,1),(1,4),(4,1),(4,4)\right\} ,\) see Lemma 3.1. By the same arguments as in the previous case, \(\ell _{[1;0;0]},\ell _{[0;1;3]},\ell _{[0;1;2]}\notin {\mathcal {S}}^2_{X_{10}}\cup {\mathcal {S}}^2_{X_{11}}\) and, therefore, \({\mathcal {S}}^2_{X_{10}}\cup {\mathcal {S}}^2_{X_{11}}=\big \{ \ell _{[1;1;0]}, \ell _{[4;1;0]}, \ell _{[3;1;1]}, \ell _{[3;1;4]}, \ell _{[2;1;1]}, \ell _{[2;1;4]} \big \}.\) It can be verified that \(\ell _{[3;1;4]}\cap \ell _{[2;1;4]}=\left\{ (0,1)\right\} .\) However, \((0,1)\notin X_{10}\cup X_{11}.\) This contradicts Lemma 3.5 (2). Hence, this case does not occur.

Finally, we will show that cases 1 and 3 give isomorphic ovals. For \((i,j)\in {\mathbb {Z}}_2^2\) and \(k=1,3\), denote by \(X^{k}_{ij}\) the set \(X_{ij}\) from Case k. With this notation, \(X^3_{00}=\left\{ (1,0),(2,1),(3,4) \right\} \) and \(X^3_{01}=\left\{ (4,0),(2,4),(3,1) \right\} .\) Consider f given by \(f(x,y)=(3x+2y,3x+3y+2).\) One easily verifies that \(f(X_{11}^1)=X_{00}^3\) and \(f(X_{01}^1)=X_{01}^3.\) The result now follows from Lemma 3.3. The proof is complete.

\(\square \)

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Stȩpień, Z. Ovals in \({\mathbb {Z}}^2_{2p}\). Ann. Comb. (2020). https://doi.org/10.1007/s00026-020-00503-6

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Keywords

  • Arc
  • Collinearity

Mathematics Subject Classification

  • 05B99