Infinite families of 2-designs from linear codes

Abstract

Interplay between coding theory and combinatorial t-designs has attracted a lot of attention. It is well known that the supports of all codewords of a fixed Hamming weight in a linear code may hold a t-design. In this paper, we first settle the weight distributions of two classes of linear codes, and then determine the parameters of infinite families of 2-designs held in these codes.

Appendix

Table 5 The value distribution of S(a, b, c) when m is odd

Table 6 The value distribution of S(a, b, c) when m is even

Proof

(Proof of Theorem 3) For each nonzero codeword ${\mathbf {c}}(a,b,c,h)=(c_0, c_1,\ldots , c_n)$ in ${\overline{{{\mathcal {C}}_1}^{\bot }}}^{\bot },$ the Hamming weight of ${\mathbf {c}}(a,b,c,h)$ is

$$\begin{aligned} w_H({\mathbf {c}}(a,b,c,h))=n+1-T(a,b,c,h)=p^m-T(a,b,c,h), \end{aligned}$$

(6)

where

$$\begin{aligned} T(a,b,c,h)=|\{x: Tr(ax^{p^l+1}+bx^2+cx)+h=0,\, x,a,b,c \in {\mathbb {F}}_q, h\in {\mathbb {F}}_p\}|. \end{aligned}$$

Then

$$\begin{aligned} T(a,b,c,h)= & {} \frac{1}{p}\sum \limits _{y \in {\mathbb {F}}_p}\sum \limits _{x \in {\mathbb {F}}_q}\zeta _p^{y(Tr(ax^{p^l+1}+bx^2+cx)+h)}\\= & {} \frac{1}{p}\sum \limits _{y \in {\mathbb {F}}_p}\zeta _p^{yh}\sum \limits _{x \in {\mathbb {F}}_q}\zeta _p^{yTr(ax^{p^l+1}+bx^2+cx)}\\= & {} p^{m-1}+\frac{1}{p}\sum \limits _{y \in {\mathbb {F}}_p^*}\zeta _p^{yh}\sigma _y(S(a,b,c)). \end{aligned}$$

By Lemma 7, for $\varepsilon =\pm 1, j \in {\mathbb {F}}_p^*$ and $0\le i\le 2,$ we have $\ell =\frac{m+i}{2}$ if $m-i$ is even, and $\ell =\frac{m+i-1}{2}$ if $m-i$ is odd, and then

$$\begin{aligned} S(a,b,c)=\{\varepsilon p^\ell , \varepsilon \sqrt{p^*}p^\ell , 0 , \varepsilon p^\ell \zeta _p^j, \varepsilon \sqrt{p^*}\zeta _p^jp^\ell , p^m\}. \end{aligned}$$

Hence from Lemma 8, it follows that

$$\begin{aligned} \sigma _y(S(a,b,c))=\left\{ \begin{array}{ll} 0 &{} \mathrm {if}\,\ S(a,b,c)=0 ,\\ \varepsilon p^\ell &{} \mathrm {if}\,\ S(a,b,c)=\varepsilon p^\ell ,\\ \varepsilon p^\ell \sqrt{p^*}\eta '(y) &{} \mathrm {if}\,\ S(a,b,c)=\varepsilon p^\ell \sqrt{p^*} ,\\ \varepsilon p^\ell \zeta _p^{yj} &{} \mathrm {if}\,\ S(a,b,c)=\varepsilon p^\ell \zeta _p^j ,\\ \varepsilon p^\ell \sqrt{p^*}\eta '(y) \zeta _p^{yj}&{} \mathrm {if}\,\ S(a,b,c)=\varepsilon \sqrt{p^*}p^\ell \zeta _p^j ,\\ p^m &{} \mathrm {if}\,\ S(a,b,c)=p^m. \end{array} \right. \end{aligned}$$

That is,

$$\begin{aligned} T(a,b,c,h)=\left\{ \begin{array}{ll} p^{m-1} &{} \mathrm {if}\,\ S(a,b,c)=0 ,\\ p^{m-1}+\varepsilon p^{\ell -1}(p-1) &{} \mathrm {if}\,\ S(a,b,c)=\varepsilon p^\ell \, \mathrm {and}\,\ h=0,\\ p^{m-1}+\varepsilon p^{\ell -1}(-1) &{} \mathrm {if}\,\ S(a,b,c) =\varepsilon p^\ell \, \mathrm {and}\,\ h\ne 0 ,\\ p^{m-1}+\varepsilon p^{\ell -1}\sqrt{p^*}\eta '(y)G(\eta ', \chi '_1) &{} \mathrm {if}\,\ S(a,b,c)=\varepsilon \sqrt{p^*}p^\ell ,\\ p^{m-1}+\varepsilon p^{\ell -1}(p-1)&{} \mathrm {if}\,\ S(a,b,c)=\varepsilon p^\ell \zeta _p^j\, \mathrm {and}\,\ h+j=0, \\ p^{m-1}+\varepsilon p^{\ell -1}(-1)&{} \mathrm {if}\,\ S(a,b,c)=\varepsilon p^\ell \zeta _p^j \, \mathrm {and}\,\ h+j\ne 0, \\ p^{m-1}+\varepsilon p^{\ell -1}\sqrt{p^*}\eta (h+j)G(\eta ',\chi '_1) &{} \mathrm {if}\,\ S(a,b,c)=\varepsilon \sqrt{p^*}p^\ell \zeta _p^j ,\\ p^m&{} \mathrm {if}\,\ S(a,b,c)=p^m \, \mathrm {and}\,\ h=0 ,\\ 0&{} \mathrm {if}\,\ S(a,b,c)=p^m\, \mathrm {and}\,\ h \ne 0 . \end{array} \right. \end{aligned}$$

Obviously, when m is odd, by Lemmas 3, 4, 7 and Eq. (6) we have

$w_1=p^m-p^{m-1},$

$A_{w_1}=pw+2n_{1,0,0}+2n_{1,2,0}+(p-1)[(n_{1,0,1}+n_{-1,0,1})+(n_{1,2,1}+n_{-1,2,1})],$

$w_2=p^m-[p^{m-1}+p^{\frac{m-1}{2}}(p-1)],$

$A_{w_2}=n_{1,1,0}+(p-1)n_{1,1,1},$

$w_3=p^m-[p^{m-1}-p^{\frac{m-1}{2}}(p-1)],$

$A_{w_3}=n_{-1,1,0}+(p-1)n_{-1,1,1},$

$w_4=p^m-(p^{m-1}-p^{\frac{m-1}{2}}),$

$A_{w_4}=(p-1)n_{1,1,0}+(p-1)^2n_{1,1,1}+{\frac{p-1}{2}}(n_{1,0,0}+n_{-1,0,0})+{\frac{p-1}{2}}{(p-1)}(n_{1,0,1}+n_{-1,0,1}),$

$w_5=p^m-(p^{m-1}+p^{\frac{m-1}{2}}),$

$A_{w_5}=(p-1)n_{-1,1,0}+(p-1)^2n_{-1,1,1}+2\cdot {\frac{p-1}{2}}n_{\pm 1,0,0}+2\cdot {\frac{(p-1)^2}{2}}n_{\pm 1,0,1},$

$w_6=p^m-(p^{m-1}-p^{\frac{m+1}{2}}),$

$A_{w_6}=2\cdot {\frac{p-1}{2}}n_{\pm 1,2,0}+2\cdot {\frac{(p-1)^2}{2}}n_{\pm 1,2,1},$

$w_7=p^m-(p^{m-1}+p^{\frac{m+1}{2}}),$

$A_{w_7}=A_{w_6},$

$A_{p^m}=p-1.$ When m is even, by Lemmas 3, 4, 7 and Eq. (6) we get

$w_1=p^m-p^{m-1},$

$A_{w_1}=pw+2n_{\pm 1,1,0}+2(p-1)n_{\pm 1,1,1},$

$w_2=p^m-[p^{m-1}+p^{\frac{m}{2}-1}(p-1)],$

$A_{w_2}=n_{1,0,0}+(p-1)n_{1,0,1},$

$w_3=p^m-[p^{m-1}+p^{\frac{m}{2}}(p-1)],$

$A_{w_3}=n_{1,2,0}+(p-1)n_{1,2,1},$

$w_4=p^m-[p^{m-1}-p^{\frac{m}{2}-1}(p-1)],$

$A_{w_4}=n_{-1,0,0}+(p-1)n_{-1,0,1},$

$w_5=p^m-[p^{m-1}-p^{\frac{m}{2}}(p-1)],$

$A_{w_5}=n_{-1,2,0}+(p-1)n_{-1,2,1},$

$w_6=p^m-[p^{m-1}-p^{\frac{m}{2}}(p-1)],$

$A_{w_6}=(p-1)n_{1,0,0}+(p-1)^2n_{1,0,1},$

$w_7=p^m-(p^{m-1}-p^{\frac{m}{2}}),$

$A_{w_7}=(p-1)n_{1,2,0}+(p-1)^2n_{1,2,1}+2\cdot {\frac{p-1}{2}}n_{\pm 1,1,0}+2\cdot {\frac{(p-1)^2}{2}}n_{\pm 1,1,1},$

$w_8=p^m-(p^{m-1}+p^{\frac{m}{2}-1}),$

$A_{w_8}=(p-1)n_{-1,0,0}+(p-1)^2n_{-,0,1},$

$w_9=p^m-(p^{m-1}+p^{\frac{m}{2}}),$

$A_{w_9}=(p-1)n_{-1,2,0}+(p-1)^2n_{-1,2,1}+2\cdot {\frac{p-1}{2}}n_{\pm 1,1,0}+2\cdot {\frac{(p-1)^2}{2}}n_{\pm 1,1,1},$

$A_{p^m}=p-1.$

Thus we complete the proof of Theorem 3. $\square$

Proof

(Proof of Theorem 4) For each nonzero codeword ${\mathbf {c}}(a,b,h)=(c_0,\ldots ,c_n)$ in ${\overline{{{\mathcal {C}}_2}^{\bot }}}^{\bot },$ the Hamming weight of ${\mathbf {c}}(a,b,h)$ is

$$\begin{aligned} w_H({\mathbf {c}}(a,b,h))=p^m-T(a,b,h), \end{aligned}$$

(7)

where

$$\begin{aligned} T(a,b,h)=|\{x:Tr(ax^{p^l+1}+bx)+h=0,\, x,a,b \in {\mathbb {F}}_q, h \in {\mathbb {F}}_p\}|. \end{aligned}$$

Then

$$\begin{aligned} T(a,b,h)= & {} {\frac{1}{p}}\sum \limits _{y \in {\mathbb {F}}_p}\sum \limits _{x \in {\mathbb {F}}_q}\zeta _p^{yTr\left(ax^{p^l+1}+bx\right)+hy}\\= & {} p^{m-1}+{\frac{1}{p}}\sum \limits _{y\in {\mathbb {F}}_p^*}\zeta _p^{hy}\sum \limits _{x \in {\mathbb {F}}_q}\zeta _p^{yTr\left(ax^{p^l+1}+bx\right)}. \end{aligned}$$

If $a=b=h=0$, then ${\mathbf {c}}(a,b,h)$ is the zero codeword.

If $a=b=0, h\ne 0$, then $T(a,b,h)=p^{m-1}+p^{m-1}\sum \limits _{y \in {\mathbb {F}}_p^*}\zeta _p^{hy}=0.$

If $a=0, b\ne 0$, then $T(a,b,h)=p^{m-1}+{\frac{1}{p}}\sum \limits _{y \in {\mathbb {F}}_p^*}\zeta _p^{hy}\sum \limits _{x\in {\mathbb {F}}_q}\zeta _p^{yTr(bx)}=p^{m-1}.$

If $a\ne 0$, then

$$\begin{aligned} T(a,b,h)= & {} p^{m-1}+{\frac{1}{p}}\sum \limits _{y \in {\mathbb {F}}_p^*}\zeta _p^{hy}\sigma _y(\sum \limits _{x\in {\mathbb {F}}_q}\zeta _p^{Tr(ax^{p^l+1}+bx)})\\= & {} p^{m-1}+{\frac{1}{p}}\sum \limits _{y \in {\mathbb {F}}_p^*}\zeta _p^{hy}\sigma _y(S(a,b))\\= & {} p^{m-1}+{\frac{1}{p}}\sum \limits _{y \in {\mathbb {F}}_p^*}\zeta _p^{hy}S(ay,by). \end{aligned}$$

When m is odd, by Lemmas 3–5, we have

$$\begin{aligned} T(a,b,h)=\left\{ \begin{array}{ll} p^{m-1} &{} \mathrm {if}\,\ h=Tr(ax_{a,b}^{p^l+1}),\\ p^{m-1}+p^{\frac{m-1}{2}}(-1)^\frac{(p-1)(m+1)}{4}\eta (a)\eta '(h-Tr(ax_{a,b}^{p^l+1})) &{} \mathrm {if}\,\ h\ne Tr(ax_{a,b}^{p^l+1}).\\ \end{array} \right. \end{aligned}$$

Obviously, for $a\in {\mathbb {F}}_q^*,$ we get that $T(a,b,h)=p^{m-1}$ appears $p^m(p^m-1)$ times, both $T(a,b,h)=p^{m-1}+p^{\frac{m-1}{2}}$ and $T(a,b,h)=p^{m-1}-p^{\frac{m-1}{2}}$ appear $\frac{p^m(p-1)(p^m-1)}{2}$ times, respectively.

When m is even and $a^{\frac{q-1}{p+1}}\ne (-1)^{\frac{m}{2}},$ by Lemma 5, we have

$$\begin{aligned} T(a,b,h)=\left\{ \begin{array}{ll} p^{m-1}+(-1)^{\frac{m}{2}}p^{\frac{m}{2}-1}(p-1) &{} \mathrm {if}\,\ h=Tr\left(ax_{a,b}^{p^l+1}\right),\\ p^{m-1}-(-1)^{\frac{m}{2}}p^{\frac{m}{2}-1} &{} \mathrm {if}\,\ h\ne Tr\left(ax_{a,b}^{p^l+1}\right)\,. \end{array} \right. \end{aligned}$$

Clearly, there exist $p^m-1-{\frac{p^m-1}{p+1}}={\frac{p(p^m-1)}{p+1}}$ elements $a\in {\mathbb {F}}_q^*$ such that $a^{\frac{q-1}{p+1}}\ne (-1)^{\frac{m}{2}}.$ Then

$$\begin{aligned} T(a,b,h)=p^{m-1}+(-1)^{\frac{m}{2}}p^{\frac{m}{2}-1}(p-1) \end{aligned}$$

appears ${\frac{p^{m+1}(p^m-1)}{p+1}}$ times, and

$$\begin{aligned} T(a,b,h)=p^{m-1}-(-1)^{\frac{m}{2}}p^{\frac{m}{2}-1} \end{aligned}$$

appears ${\frac{p^{m+1}(p-1)(p^m-1)}{p+1}}$ times.

When m is even and $a^{\frac{q-1}{p+1}}=(-1)^{\frac{m}{2}},$ from Lemma 5, we get

$$\begin{aligned} T(a,b,h)=\left\{ \begin{array}{ll} p^{m-1}-(-1)^{\frac{m}{2}}p^{\frac{m}{2}}(p-1) &{} \mathrm {if}\,\ f(x)=-b^{p^l}\, \ \mathrm {is\, solvable}\, \mathrm {and}\,\ h=Tr(ax_{a,b}^{p^l+1}),\, \\ p^{m-1}+(-1)^{\frac{m}{2}}p^{\frac{m}{2}} &{} \mathrm {if}\,\ f(x)=-b^{p^l}\, \ \mathrm {is \,solvable},\,\ b\ne 0\,\mathrm {and}\\ {} &{}\ h\ne Tr\left(ax_{a,b}^{p^l+1}\right)\ ,\ \mathrm {or}\,\ b=0\,\mathrm {and}\,\ h\ne 0,\\ p^{m-1} &{} \mathrm {if}\,\ f(x)=-b^{p^l}\, \ \mathrm {is \, no\, solvable}. \end{array} \right. \end{aligned}$$

By Lemma 6, there are $\frac{q-1}{p+1}$ elements $a\in {\mathbb {F}}_q^*$ such that $a^{\frac{q-1}{p+1}}=(-1)^{\frac{m}{2}},$ and $p^{m-2}$ elements $b\in {\mathbb {F}}_q$ such that $f(x)=-b^{p^l}$ is solvable. Therefore,

$$\begin{aligned} T(a,b,h)=p^{m-1}+(-1)^{\frac{m}{2}+1}p^{\frac{m}{2}}(p-1) \end{aligned}$$

appears ${\frac{p^{m-2}(p^m-1)}{p+1}}$ times,

$$\begin{aligned} T(a,b,h)=p^{m-1}+(-1)^{\frac{m}{2}}p^{\frac{m}{2}} \end{aligned}$$

appears ${\frac{p^{m-2}(p^m-1)(p-1)}{p+1}}$ times, and $T(a,b,h)=p^{m-1}$ appears $(p^m-1)p^{m-1}(p-1)$ times.

By all the discussions above, the proof is completed. $\square$