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 karc 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 nonsingular conic of \(\mathrm {PG}(2,q)\) is a \((q + 1)\)arc. If K is any karc 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 nonsingular 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 nonsingular 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:
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.
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 nonsingular conic. Note that there are exactly three distinct lines through a in \({\mathbb {Z}}_2^2,\) Therefore, we have three distinct nonsingular 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 nonsingular 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 dsecant of K is a line l, such that \(l\cap K=d.\) Denote by \({\mathcal {S}}^{d}_K\) the set of dsecants of K.
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:
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 2secant 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 nonsingular 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., nonsingular 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 nonsingular 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:
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 nonsingular 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:
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:
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:
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/s00026020005036
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Keywords
 Arc
 Collinearity
Mathematics Subject Classification
 05B99