Mutually disjoint Steiner systems from BCH codes

Abstract

Liu et al. (IEEE Trans Inf Theory 68:3096–3107, 2022) investigated a class of BCH codes $\mathcal {C}_{(q,q+1,\delta ,1)}$ with $q=\delta ^m$ a prime power and proved that the set $\mathcal {B}_{\delta +1}$ of supports of the minimum weight codewords supports a Steiner system ${{\text {S}}}(3,\delta +1,q+1)$. In this paper, we give an equivalent formulation of $\mathcal {B}_{\delta +1}$ in terms of elementary symmetric polynomials and then construct a number of mutually disjoint Steiner systems S$(3,\delta +1,\delta ^m+1)$ when m is even and a number of mutually disjoint G-designs G$\big ({\frac{\delta ^m+1}{\delta +1}},\delta +1,\delta +1,3\big )$ when m is odd. In particular, the existence of three mutually disjoint Steiner systems ${{\text {S}}}(3,5,4^m+1)$ or three mutually disjoint G-designs G$\big ({\frac{4^m+1}{5}},5,5,3\big )$ is established.

A Appendix

Proof of Lemma 2.2

We use induction on the order n of the determinant $|\textbf{A}|$, which is also the length of the subsequence $\{a_{t_{i}}\}_{i=1}^{n}$. The base case $n=1$ is trivial. Suppose that the conclusion holds for any subsequence of length $n-1$. Then we prove that it also holds for a subsequence $\{a_{t_{1}},a_{t_{2}},\ldots ,a_{t_{n}}\}$. Consider the subsequence $\{a_{t_{1}},\ldots ,a_{t_{j-1}},a_{t_{j+1}}\ldots ,a_{t_{n}}\}$ and the submatrix $\textbf{D}$ of $\textbf{A}$ by deleting the n-th row and j-th column. By inductive hypothesis, one obtains $|\textbf{D}|=\sum \limits _{i=1}^{n}E_{i}\cdot x^{i-1},$ where $E_{i}$ is the determinant of the submatrix $\textbf{E}_{i}$ of $\textbf{E}$ by deleting the i-th row and

$$\begin{aligned} \textbf{E}=\left[ \begin{array}{cccccc} a_{t_1} &{}\cdots &{}a_{t_{j-1}} &{}a_{t_{j+1}} &{}\cdots &{}a_{t_{n}} \\ a_{t_{1}+1} &{}\cdots &{}a_{t_{j-1}+1} &{}a_{t_{j+1}+1} &{}\cdots &{}a_{t_{n}+1} \\ \vdots &{} \cdots &{}\vdots &{} \vdots &{} \cdots &{} \vdots \\ a_{t_{1}+n-1} &{} \cdots &{} a_{t_{j-1}+n-1} &{}a_{t_{j+1}+n-1} &{}\cdots &{} a_{t_{n}+n-1} \\ \end{array} \right] _{n\times (n-1)}. \end{aligned}$$

It is not difficult to see that $E_i=(B_{i})_{n,j}$, namely the complement minor of the (n, j)-entry of the matrix $\textbf{B}_{i}$. Let $A_{n,j}$ $(1\le j\le n)$ denote the complement minor of the (n, j)-entry of $\textbf{A}$. Then

$$\begin{aligned} A_{n,j}=|\textbf{D}|=\sum \limits _{i=1}^{n}(B_{i})_{n,j}\cdot x^{i-1}. \end{aligned}$$

Thus expanding $|\textbf{A}|$ along the last row yields

$$\begin{aligned} \begin{array}{l} |\textbf{A}|=\sum \limits _{j=1}^{n}(-1)^{n+j}(a_{t_{j}+n}+xa_{t_{j}+n-1})A_{n,j}\\ \quad =\sum \limits _{j=1}^{n}(-1)^{n+j}(a_{t_{j}+n}+xa_{t_{j}+n-1})\sum \limits _{i=1}^{n}(B_{i})_{n,j}\cdot x^{i-1} \\ \quad =\sum \limits _{i=1}^{n}\sum \limits _{j=1}^{n}(-1)^{n+j}\Big (a_{t_{j}+n}(B_{i})_{n,j}\cdot x^{i-1}+a_{t_{j}+n-1}(B_{i})_{n,j}\cdot x^{i}\Big )\\ \quad =\sum \limits _{j=1}^{n}(-1)^{n+j}a_{t_{j}+n}(B_{1})_{n,j}\cdot x^{0}+\sum \limits _{j=1}^{n}(-1)^{n+j}a_{t_{j}+n-1} (B_{n+1})_{n,j}\cdot x^{n} \\ \quad \quad +\sum \limits _{i=1}^{n-1}\Big (\sum \limits _{j=1}^{n}(-1)^{n+j}a_{t_{j}+n}(B_{i+1})_{n,j}+ \sum \limits _{j=1}^{n}(-1)^{n+j}a_{t_j+n-1}(B_{i})_{n,j}\Big )\cdot x^{i}\\ \quad =B_1+B_{n+1}\cdot x^{n}+\sum \limits _{i=1}^{n-1}\big (B_{i+1}+0\big )\cdot x^{i}\\ \quad =\sum \limits _{i=1}^{n+1}B_{i} \cdot x^{i-1},\\ \end{array} \end{aligned}$$

where the penultimate equation holds because $\sum \limits _{j=1}^{n}(-1)^{n+j}a_{t_j+n-1}(B_{i})_{n,j}$ is the determinant by replacing the n-th row of $B_{i}$ by its $(n-1)$-th row, which equals zero. This completes the proof. $\square$

Proof of Lemma 4.4

We apply induction on the order n of the determinant $|\textbf{A}_{n}|$, which is also the length of the sequence $\{a_{1},a_{2},\ldots ,a_{n}\}$. The base case $n = 2$ holds clearly. Thus we suppose that the claim has already been proven for all smaller n. In particular from this inductive hypothesis we already have

$$\begin{aligned} |\textbf{A}_{n-1}|=\sum \limits _{i=0}^{\lceil \frac{n-1}{2}\rceil -1}(-1)^{n-i}(i+1)a_{1}^{i}a_{n-1-i}+a_{1}^{n-1}, \ n \ge 3. \end{aligned}$$

(34)

Then we prove that it also holds for a sequence $\{a_{1},a_{2},\ldots ,a_{n}\}$. If $a_{2}=a_{3}=\cdots =a_{\lfloor \frac{n}{2}\rfloor }=0$, then expanding $|\textbf{A}_{n}|$ along the first column yields

$$\begin{aligned} \begin{array}{rl} |\textbf{A}_{n}|= &{} a_{1}|\textbf{A}_{n-1}|+\sum \limits _{i=0}^{n-2}(-1)^{n-i+1}a_{n-i}A_{n-i,1}\\ =&{}a_{1}|\textbf{A}_{n-1}|+\sum \limits _{i=0}^{\lceil \frac{n}{2}\rceil -1}(-1)^{n-i+1}a_{n-i}A_{n-i,1}.\\ \end{array} \end{aligned}$$

(35)

where $A_{i,1}$ denotes the complement minor of the (i, 1)-entry of $\textbf{A}_{n}$. For $0\le i\le \lceil \frac{n}{2}\rceil -1$, we have

(36)

Note that $0\le i\le \lceil \frac{n}{2}\rceil -1\le \lfloor \frac{n}{2}\rfloor$. Then $a_{2}=a_{3}=\cdots =a_{i}=0$. Thus (36) is the same as

$$\begin{aligned} A_{n-i,1}=\left| \begin{array}{ccccc} a_{1} &{} 1&{} \cdots &{} 0 &{} 0 \\ 0 &{} a_{1}&{} \cdots &{} 0 &{} 0 \\ \vdots &{} \vdots &{} \cdots &{} \vdots &{}\vdots \\ 0 &{} 0&{} \cdots &{} a_{1} &{} 1 \\ 0 &{} 0&{} \cdots &{} 0 &{} a_{1} \\ \end{array} \right| _{i\times i}=a_{1}^{i}. \end{aligned}$$

(37)

Plugging (34) and (37) into (35) yields

$$\begin{aligned} |\textbf{A}_{n}|= & {} a_{1}\cdot \left[ \sum \limits _{i=0}^{\lceil \frac{n-1}{2}\rceil -1}(-1)^{n-i}(i+1)a_{1}^{i}a_{n-1-i}+a_{1}^{n-1}\right] +\sum \limits _{i=0}^{\lceil \frac{n}{2}\rceil -1}(-1)^{n-i+1}a_{n-i}\cdot a_{1}^{i}\\= & {} \sum \limits _{i=0}^{\lceil \frac{n-1}{2}\rceil -1}(-1)^{n-i}(i+1)a_{1}^{i+1}a_{n-1-i}+ \sum \limits _{i=0}^{\lceil \frac{n}{2}\rceil -1}(-1)^{n-i+1}a_{1}^{i}a_{n-i}+a_{1}^{n}\\= & {} \sum \limits _{i=0}^{\lceil \frac{n-1}{2}\rceil }(-1)^{n+1-i}ia_{1}^{i}a_{n-i}+ \sum \limits _{i=0}^{\lceil \frac{n}{2}\rceil -1}(-1)^{n-i+1}a_{1}^{i}a_{n-i}+a_{1}^{n}\\= & {} \left\{ \begin{array}{ll} \sum \limits _{i=0}^{\lceil \frac{n}{2}\rceil -1}(-1)^{n-i+1}(i+1)a_{1}^{i}a_{n-i}+a_{1}^{n}, &{}\text {if}\ n\ \text {is odd}; \\ \sum \limits _{i=0}^{\lceil \frac{n}{2}\rceil -1}(-1)^{n-i+1}(i+1)a_{1}^{i}a_{n-i}+a_{1}^{n}+ (-1)^{n-\frac{n}{2}+1}\frac{n}{2} a_{1}^{\frac{n}{2}}a_{n-\frac{n}{2}}, &{}\text {if}\ n\ \text {is even}. \\ \end{array} \right. \end{aligned}$$

If n is even, then $a_{\frac{n}{2}}=0$ by assumption. Therefore, the conclusion is valid for any n. This completes the proof. $\square$