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.

1 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 X,  Y 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.

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

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

• (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, a, b, c 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. $\square$

3 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

An oval in ${\mathbb {Z}}^2_6$

Fig. 2

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 X,  Y 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) ${\mathcal {S}}^2_{X_{10}}\cup {\mathcal {S}}^2_{X_{11}}\subset {\mathcal {S}}^0_{X_{00}\cup X_{01}};$

• (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.

$\square$