To determine the stable homotopy groups of spheres is one of the most important problems in algebraic topology. So far, several methods were found to determine the stable homotopy groups of spheres. For example, we have the classical Adams spectral sequence (ASS) (see [1]) based on the Eilenberg-MacLane spectrum $K\mathbb{Z}_p$, whose $E_2$-term is ${\rm Ext}_A^{s, t}(\mathbb{Z}_p, \mathbb{Z}_p)$ and the Adams differential is given by $\tilde{d}_r: E_r^{s, t}\rightarrow E_r^{s+r, t+r-1}$, where $A$ denotes the mod $p$ Steenrod algebra. There are three problems in using the ASS: calculation of $E_2$-term ${\rm Ext}_A^{\ast, \ast}(\mathbb{Z}_p, \mathbb{Z}_p)$, computation of the differentials and determination of the nontrivial extensions from $E_{\infty}$ to the stable homotopy groups of spheres. So, for computing the stable homotopy groups of spheres with the classical ASS, we must compute the $E_2$-term of the ASS, ${\rm Ext}_A^{\ast, \ast}(\mathbb{Z}_p, \mathbb{Z}_p)$.

Throughout this paper, $p$ denotes an odd prime and $q=2(p-1)$. The known results on ${\rm Ext}_A^{\ast, \ast}(\mathbb{Z}_p, \mathbb{Z}_p)$ are as follows. ${\rm Ext}_A^{0, \ast}(\mathbb{Z}_p, \mathbb{Z}_p)$ is trivial by its definition. From [2], ${\rm Ext}_A^{1, \ast}(\mathbb{Z}_p, \mathbb{Z}_p)$ has $\mathbb{Z}_p$-basis consisting of $a_0\in {\rm Ext}_A^{1, 1}(\mathbb{Z}_p , \mathbb{Z}_p)$, $h_i\in {\rm Ext}_A^{1, p^iq}(\mathbb{Z}_p, \mathbb{Z}_p)$ for all $i\geqslant 0$ and ${\rm Ext}_A^{2, \ast}(\mathbb{Z}_p, \mathbb{Z}_p)$ has $\mathbb{Z}_p$-basis consisting of $\alpha_2$, $a_0^2$, $a_0h_i(i>0)$, $g_i(i\geqslant 0)$, $k_i(i\geqslant 0)$, $b_i(i\geqslant 0)$, and $h_ih_j(j\geqslant i+2, i\geqslant 0)$ whose internal degrees are $2q+1$, $2$, $p^iq+1, p^{i+1}q+2p^iq$, $2p^{i+1}q+p^iq$, $p^{i+1}q$ and $p^iq+p^jq$, respectively. In 1980, Aikawa [3] determined ${\rm Ext}_A^{3, \ast}(\mathbb{Z}_p, \mathbb{Z}_p)$ by $\lambda$-algebra.

Studying higher-dimensional cohomology of the mod $p$ Steenrod algebra $A$ was an interesting subject and studied by several authors. For example, Liu and Zhao [4] proved the following theorems, respectively.

Theorem 1.1 For $p\geqslant 11$ and $4\leqslant s <p$, the product $h_0b_0\tilde{\delta}_s\not=0$ in the classical Adams spectral sequence, where $\tilde{\delta}_{s}$ was given in [5].

In this paper, our main result can be stated as follows.

Theorem 1.2 Let $p\geqslant 7$, and $0\leqslant s <p-5$. Then in the cohomology of the mod $p$ Steenrod algebra $A$, the product $ b_0^3 \tilde \delta _{s + 4}\in{\rm Ext}_A^{s+10, t(s)}(\mathbb{Z}_p, \mathbb{Z}_p)$ is nontrivial, where

$ t(s)=q[(s+1) +(s + 5)p + (s + 3)p^2 + (s + 4)p^3 ]+s. $

The main method of proof is the (modified) May spectral sequence, so we will recall some knowledge on the May spectral sequence in Section 2. After detecting the generators of some May $E_1$-terms in Section 3, we will prove Theorem 1.2.

As we know, the most successful method to compute ${\rm Ext}_A^{\ast, \ast}(\mathbb{Z}_p, \mathbb{Z}_p)$ is the MSS. From [6], there is a May spectral sequence(MSS) $\{E_r^{s, t, \ast}, d_r\}$ which converges to ${\rm Ext}_A^{s, t}(\mathbb{Z}_p, \mathbb{Z}_p)$ with $E_1$-term

$ {E}_1^{\ast, \ast, \ast}= E(h_{m, i}|m>0, i\geqslant 0)\otimes P(b_{m, i}|m>0, i\geqslant 0)\otimes P(a_n|n\geqslant 0), $

(2.1)

where $E(~)$ is the exterior algebra, $P(~)$ is the polynomial algebra, and

$ h_{m, i}\in E_1^{1, 2(p^m-1)p^i, 2m-1}, b_{m, i}\in E_1^{2, 2(p^m-1)p^{i+1}, p(2m-1)}, a_n\in E_1^{1, 2p^n-1, 2n+1}. $

One has

$ d_r:E_r^{s, t, u}\rightarrow E_r^{s+1, t, u-r} $

(2.2)

and if $x\in E_r^{s, t, \ast}$ and $y\in E_r^{s^{\prime}, t^{\prime}, \ast}$, then

$ d_r(x\cdot y)=d_r(x)\cdot y+(-1)^sx\cdot d_r(y). $

(2.3)

In particular, the first May differential $d_1$ is given by

$ d_1(h_{i, j})=\sum\limits_{0 <k <i}^{}h_{i-k, k+j}h_{k, j}, ~ d_1(a_i)=\sum\limits_{0\leqslant k <i}h_{i-k, k}a_k, ~ d_1(b_{i, j})=0. $

(2.4)

There also exists a graded commutativity in the MSS:

$ x\cdot y=(-1)^{ss^{\prime}+tt^{\prime}}y\cdot x~~ \mbox{for}~~ x, y=h_{m, i}, b_{m, i}~~ \mbox{or}~~ a_n. $

For each element $x\in E_{1}^{s, t, u}$, we define $\hbox{dim}~x=s$, $\hbox{deg}~x=t$, $M(x)=u$. Then we have that

$ \left\{\begin{array}{l} \hbox{dim}~h_{i, j}=\hbox{dim}~a_{i}=1, \\\hbox{dim}~b_{i, j}=2, \hbox{deg}~a_0=1, \\ \hbox{deg}~h_{i, j}=q(p^{i+j-1}+\cdots+p^j), \\ \hbox{deg}~b_{i, j}=q(p^{i+j}+\cdots+p^{j+1}), \\ \hbox{deg}~a_i=q(p^{i-1}+\cdots+1)+1, \\ M(h_{i, j})=M(a_{i-1})=2i-1, \\M(b_{i, j})=(2i-1)p, \end{array}\right. $

(2.5)

where $i\geqslant 1$, $j\geqslant 0$.

Note that by the knowledge on the $p$-adic expression in number theory, for each integer $t\geqslant 0$, it can be expressed uniquely as $t=q(c_np^n+c_{n-1}p^{n-1}+\cdots+c_1p+c_0)+e, $ where $0\leqslant c_i <p$ ($0\leqslant i <n$), $p>c_n>0$, $0\leqslant e <q$.

Before showing Theorem 1.2, we first give some important lemmas which will be used in the proof of it. The first one is a lemma on the representative of $\tilde\delta _{s + 4}$ in the May spectral sequence.

Lemma 3.1 For $p\geqslant 7$ and $0 \leqslant s <p-4$. Then the fourth Greek letter element $\tilde \delta _{s + 4} \in {\rm Ext}_A^{s+4, t_1(s)}(\mathbb{Z}_p, \mathbb{Z}_p)$ is represented by

$ a_4^sh_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}\in E_1^{s+4, t_1(s), \ast} $

in the $E_1$-term of the May spectral sequence, where $\tilde \delta _{s + 4}$ is actually $\tilde\alpha_{s+4}^{(4)}$ described in [6] and $t_1(s)=q[(s+1)+(s+2)p+(s+3)p^2+(s+4)p^3]+s.$

By (2.2), we know that to prove the non-triviality of $b_0^3\tilde{\delta}_{s+4}\in {\rm Ext}_A^{s+10, t(s)}(\mathbb{Z}_p, \mathbb{Z}_p)$, we have to show that the representative of the product cannot be hit by any May differential. For doing it, we give the following two lemmas.

Lemma 3.2 Let $p\geqslant 7$, $0\leqslant s <p-5$. Then we have the May $E_1$-term

$ E_1^{s+9, t(s), \ast}=\mathbb{Z}_p\left\{G_1, G_2, \cdots, G_{11}\right\}, $

where $t(s, n)=q[(s+1) +(s+ 5)p + (s + 3)p^2 + (s + 4)p^3 ]+s$, and

$ \begin{eqnarray*}&G_1=a_4^{s-1}a_2b_{3, 0}^2b_{1, 0}h_{4, 0}h_{3, 1}h_{1, 3}, ~~ G_2=a_4^{s-1}a_{2}b_{3, 0}^3h_{4, 0}h_{1, 3}h_{1, 1}, \\ &G_3=a_4^{s-1}a_2b_{3, 0}^2b_{1, 2}h_{4, 0}h_{3, 1}h_{1, 1}, ~~ G_4=a_4^sb_{3, 0}b_{1, 0}^2h_{4, 0}h_{3, 1}h_{1, 3}, \\ &G_5=a_4^sb_{3, 0}^2b_{1, 0}h_{4, 0}h_{1, 3}h_{1, 1}, ~~ G_6=a_4^sb_{3, 0}b_{1, 2}b_{1, 0}h_{4, 0}h_{3, 1}h_{1, 1}, \\ &G_7=a_4^{s-2}a_2^2b_{3, 0}^2h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}h_{1, 1}, ~~ G_8=a_4^{s-1}a_2b_{3, 0}b_{1, 0}h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}h_{1, 1}, \\ &G_9=a_4^{s-1}a_2b_{3, 0}b_{1, 2}h_{4, 0}h_{3, 1}h_{2, 1}h_{1, 3}h_{1, 1}, ~~ G_{10}=a_4^sb_{1, 0}^2h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}h_{1, 1}, \\ &G_{11}=a_4^sb_{1, 2}b_{1, 0}h_{4, 0}h_{3, 1}h_{2, 1}h_{1, 3}h_{1, 1}.\end{eqnarray*} $

For the convenience of writing, we make the following rules:

(ⅰ) ${\rm if}~i>j, ~{\rm we~ put}~a_i~{\rm on~ the~ left~ side~ of}~a_j;$

(ⅱ) ${\rm if}~j <k, ~{\rm we~ put}~h_{i, j}~{\rm on~ the~ left~ side~ of}~ h_{w, k};$

(ⅲ) ${\rm if}~i>w, ~ {\rm we~ put}~h_{i, j}~{\rm on~the~left~side~of}~h_{w, j};$

(ⅳ) $ {\rm apply~ the~ rules~ (ii) ~and~ (iii)~ to} ~b_{i, j}.$

Now we give the proof of the above lemma.

Proof The proof of this lemma is divided into the following six cases. Consider

$ h=x_1x_2\cdots x_m\in E_1^{s+9, t(s), \ast} $

in the MSS, where $x_i$ is one of $a_{k}$, $h_{r, j}$ or $b_{u, z}$, $0\leqslant k\leqslant 4$, $0\leqslant r+j\leqslant 4$, $0\leqslant u+z\leqslant 3$, $r>0$, $j\geqslant 0$, $u>0$, $z\geqslant 0$. By (2.5), we can assume that $\hbox{deg}~x_i=q(c_{i, 3}p^3+c_{i, 2}p^2+c_{i, 1}p+c_{i, 0})+e_i$, where $c_{i, j}=0$ or $1$, $e_i=1$ if $x_i=a_{k_i}$ or $e_i=0$. It follows that $\hbox{dim}~h=\sum\limits_{i=1}^{m}\hbox{dim}~x_i=s+9$ and

$ \begin{array}{ccl} \hbox{deg}~h&=&\sum\limits_{i=1}^{m}\hbox{deg}~x_i =q[(\sum\limits_{i=1}^{m}c_{i, 3})p^3+(\sum\limits_{i=1}^{m}c_{i, 2})p^2+ (\sum\limits_{i=1}^{m}c_{i, 1})p+(\sum\limits_{i=1}^{m}c_{i, 0})]+(\sum\limits_{i=1}^{m}e_i)\\ &=&q[(s+4)p^3+(s+3)p^2+(s+5)p+(s+1)]+s. \end{array} $

Note that

$ \hbox{dim}~h_{i, j}=\hbox{dim}~a_{i}=1, ~~ \hbox{dim}~b_{i, j}=2~~ \mbox{and}~~ 0\leqslant s <p-5. $

From $\hbox{dim}~h=\sum\limits_{i=1}^{m}\hbox{dim}~x_i=s+9$, we can have $m\leqslant s+9\leqslant p+3.$

Using $0\leqslant s+5, s+4, s+3, s+1, s <p$ and the knowledge on the $p$-adic expression in number theory, we have that

$ {\left\{ \begin{array}{l} \sum\limits_{i=1}^{m}e_i=s;\\ \sum\limits_{i=1}^{m}c_{i, 0}=s+1;\\ \sum\limits_{i=1}^{m}c_{i, 1}=s+5;\\ \sum\limits_{i=1}^{m}c_{i, 2}=s+3;\\ \sum\limits_{i=1}^{m}c_{i, 3}=s+4. \end{array} \right.} $

(3.1)

By $c_{i, 2}=0$ or 1, one has $m\geqslant s+4$ from $\sum\limits_{i=1}^{m}c_{i, 3}=s+4$. Note that $m\leqslant s+9$. Thus $m$ may equal $s+4$, $s+5$, $s+6$, $s+7$, $s+8$ or $s+9$. Since $\sum\limits_{i=1}^{m}e_i=s$, ${\rm deg}~h_{i, j}\equiv 0(\hbox{mod}~q)$ ($i>0, j\geqslant 0$), $\hbox{deg}~a_i\equiv 1(\hbox{mod}~q)$ ($i\geqslant 0$) and $\hbox{deg}~b_{i, j}\equiv 0(\hbox{mod}~q)$ ($i>0, j\geqslant 0$), then by the graded commutativity of $E_1^{\ast, \ast, \ast}$ and degree reasons, we can assume that $h= a_0^{x} a_1^{y} a_2^{z} a_3^k a_4^lh^{\prime}$ with $h^{\prime}=x_{s+1}x_{s+2}\cdots x_{m}$, where $0\leqslant x, y, z, k, l\leqslant s$, $x+y+z+k+l=s$. Consequently, we have

$ h^{\prime}=x_{s+1}x_{s+2}\cdots x_{m}\in E_1^{9, t_2(s), \ast}, $

where $t_2(s)=q[(s+4-l)p^3+(s+3-l-k)p^2+(s+5-l-k-z)p+(s+1-l-k-z-y)]$. From (3.1), we have

$ {\left\{ \begin{array}{l} \sum\limits_{i=s+1}^{m}e_i=0;\\ \sum\limits_{i=s+1}^{m}c_{i, 0}=s+1-l-k-z-y;\\ \sum\limits_{i=s+1}^{m}c_{i, 1}=s+5-l-k-z;\\ \sum\limits_{i=s+1}^{m}c_{i, 2}=s+3-l-k;\\ \sum\limits_{i=s+1}^{m}c_{i, 3}=s+4-l. \end{array} \right.} $

(3.2)

By the reason of dimension, all the possibilities of $h^{\prime}$ can be listed as

$ {y_1z_1\cdots z_4}, ~~ {y_1y_2y_3z_1z_2z_3}, ~~ {y_1\cdots y_5z_1z_2}, ~~ {y_1\cdots y_7z_1}, ~~ {y_1\cdots y_9}, $

where $y_i$ is in the form of $h_{r, j}$with $0\leqslant r+j\leqslant 4$, $r>0$, $j\geqslant 0$ and $z_i$is in the form of $b_{u, z}$ with $0\leqslant u+z\leqslant 3$, $u>0$, $z\geqslant 0$.

Case 1 $m=s+4$. So $h^{\prime}=x_{s+1}x_{s+2}x_{s+3}x_{s+4}\in E_1^{9, q(4p^3+3p^2+5p+1), \ast}$ and it is impossible to exist. Then $h$ doesn't exist either.

Case 2 $m=s+5$. From $\sum\limits_{i=s+1}^{s+5}c_{i, 3}=s+4-l$ in (3.2), we have that $l=s+4-\sum\limits_{i=s+1}^{s+5}c_{i, 3}\geqslant s-1$. Thus $l=s-1$ or $s$ and $h^{\prime}=y_1z_1\cdots z_4\in E_1^{9, t_2(s), \ast}$. We list all the possibilities in Table 1.

Case 3 $m=s+6$. From $\sum\limits_{i=s+1}^{s+6}c_{i, 3}=s+4-l$ in (3.2), we have that $l=s+4-\sum\limits_{i=s+1}^{s+6}c_{i, 3}\geqslant s-2$. Thus $l=s-2$, $s-1$ or $s$ and $h^{\prime}=y_1y_2y_3z_1z_2z_3\in E_1^{9, t_2(s), \ast}$. We list all the possibilities in Table 2.

In the table, $b_{3, 0}^2b_{1, 0}h_{4, 0}h_{3, 1}h_{1, 3}$, $b_{3, 0}^3h_{4, 0}h_{1, 3}h_{1, 1}$, $b_{3, 0}^2b_{1, 2}h_{4, 0}h_{3, 1}h_{1, 1}$, $b_{3, 0}b_{1, 0}^2h_{4, 0}h_{3, 1}h_{1, 3}$, $b_{3, 0}^2b_{1, 0}h_{4, 0}h_{1, 3}h_{1, 1}$, $b_{3, 0}b_{1, 2}b_{1, 0}h_{4, 0}h_{3, 1}h_{1, 1}$, denoted by $\mathbf{g}_1$, $\mathbf{g}_2$, $\mathbf{g}_3$, $\mathbf{g}_4$, $\mathbf{g}_5$, $\mathbf{g}_6$, respectively. Consequently, in this case up to sign $h=a_4^{s-1}a_{2}\mathbf{g}_1$, $a_4^{s-1}a_{2}\mathbf{g}_2$, $a_4^{s-1}a_{2}\mathbf{g}_3$, $a_4^{s}\mathbf{g}_4$, $a_4^{s}\mathbf{g}_5$, $a_4^{s}\mathbf{g}_6$denoted by $\mathbf{G}_1$, $\mathbf{G}_2$, $\mathbf{G}_3$, $\mathbf{G}_4$, $\mathbf{G}_5$, $\mathbf{G}_6$, respectively.

Case 4 $m=s+7$. From $\sum\limits_{i=s+1}^{s+7}c_{i, 3}=s+4-l$ in (3.2), we have that $l=s+4-\sum\limits_{i=s+1}^{s+7}c_{i, 3}\geqslant s-3$. Thus $l=s-3$, $s-2$, $s-1$ or $s$, and $h^{\prime}=y_1\cdots y_5z_1z_2\in E_1^{9, t_2(s), \ast}$. When $l=s-3$, we have that $t_2(s)=q[7p^3+\cdots]$. In this case, $h^{\prime}$ is impossible to exist. Then $h$ doesn't exist either. Next we list all the rest of possibilities in Table 3.

In the table, $b_{3, 0}^2h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}h_{1, 1}$, $b_{3, 0}b_{1, 0}h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}h_{1, 1}$, $b_{3, 0}b_{1, 2}h_{4, 0}h_{3, 1}h_{2, 1}h_{1, 3}h_{1, 1}$, $b_{1, 0}^2h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}h_{1, 1}$, $b_{1, 2}b_{1, 0}h_{4, 0}h_{3, 1}h_{2, 1}h_{1, 3}h_{1, 1}$, denoted by $\mathbf{g}_7$, $\mathbf{g}_8$, $\mathbf{g}_9$, $\mathbf{g}_{10}$, $\mathbf{g}_{11}$, respectively. Consequently, in this case up to sign $h=a_4^{s-2}a_{2}^2\mathbf{g}_7$, $a_4^{s-1}a_{2}\mathbf{g}_8$, $a_4^{s-1}a_{2}\mathbf{g}_9$, $a_4^{s}\mathbf{g}_{10}$, $a_4^{s}\mathbf{g}_{11}$ denoted by $\mathbf{G}_7$, $\mathbf{G}_8$, $\mathbf{G}_9$, $\mathbf{G}_{10}$, $\mathbf{G}_{11}$, respectively.

Case 5 $m=s+8$. From $\sum\limits_{i=s+1}^{s+8}c_{i, 3}=s+4-l$ in (3.2), we have that $l=s+4-\sum\limits_{i=s+1}^{s+8}c_{i, 3}\geqslant s-4$. Thus $l=s-4$, $s-3$, $s-2$, $s-1$ or $s$, and $h^{\prime}=y_1\cdots y_7z_1\in E_1^{9, t_2(s), \ast}$. When $l\leqslant s-2$, the coefficient of $P^3\in t_2(s)$is $>5$. In these cases, $h^{\prime}$ is impossible to exist. Then $h$ doesn't exist either. Next we list all the other possibilities in Table 4.

Case 6 $m=s+9$. From $\sum\limits_{i=s+1}^{s+9}c_{i, 3}=s+4-l$ in (3.2), we have that $l=s+4-\sum\limits_{i=s+1}^{s+9}c_{i, 3}\geqslant s-5$. Thus $l=s-5$, $s-4$, $s-3$, $s-2$, $s-1$ or $s$, and $h^{\prime}=y_1\cdots y_9\in E_1^{9, t_2(s), \ast}$. When $l\leqslant s-1$, the coefficient of $P^3\in t_2(s)$ is $\geqslant5$. In these cases, $h^{\prime}$ is impossible to exist. Then $h$ doesn't exist either. In the last possibility, $t_2(s)=4p^3+3p^2+5p+1$, so $h_{4, 0}, h_{3, 1}, h_{2, 2}, h_{1, 3}\in h^{\prime} $, $h^{\prime}$is impossible to exist in this case by the reason of dimension. Then $h$ doesn't exist either.

Combining Cases 1-6 above, we obtain that $E_1^{s+9, t(s), \ast}= \mathbb{Z}_p\{\mathbf{G}_1, \mathbf{G}_2, \cdots, \mathbf{G}_{11}\}$. This completes the proof of Lemma 3.2.

Lemma 3.3(1) $b_0^3\tilde{\delta}_{s+4}\in {\rm Ext}_A^{s+10, t(s)}(\mathbb{Z}_p, \mathbb{Z}_p)$ is represented by $b_{1, 0}^3a_4^{s}h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}\in E_1^{s+10, t(s), \ast}$ in the MSS, where $t(s)=q[(s+4)p^3+(s+3)p^2+(s+5)p+(s+1)]+s$.

(2) For the eleven generators of $E_1^{s+9, t(s), \ast}$, we have that

$ \begin{eqnarray*}&M(\mathbf{G}_1)=M(\mathbf{G}_3)=M(\mathbf{G}_5)=11p+9s+9, \\ &M(\mathbf{G}_2)=15p+9s+5, ~~ M(\mathbf{G}_4)=M(\mathbf{G}_6)=7p+9s+13, \\ &M(\mathbf{G}_7)=10p+9s++9, ~~ M(\mathbf{G}_8)=M(\mathbf{G}_9)=6p+9s+13, \\ &M(\mathbf{G}_{10})=M(\mathbf{G}_{11})=2p+9s+17.\end{eqnarray*} $

Moreover, we have that $M(b_{1, 0}^3a_4^{s}h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3})=3p+9s+16$.

Proof (1) Since it is known that $b_{1, i}$ and $a_4^{s}h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}\in E_1^{\ast, \ast, \ast}$ are all permanent cycles in the MSS as [7] and converge nontrivially to $b_i$, ${\tilde{\delta}}_{s+4}\in {\rm Ext}_A^{\ast, \ast} (\mathbb{Z}_p, \mathbb{Z}_p)$ for $0\leqslant s <p-5$ and $i\geqslant 0$, respectively (cf. Lemma 3.1), then $b_{1, 0}^3a_4^{s}h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}\in E_1^{s+10, t(s), 3p+9s+16}$ is a permanent cycle in the MSS and converges to $b_0^3\tilde{\delta}_{s+4}\in {\rm Ext}_A^{s+10, t(s)}(\mathbb{Z}_p, \mathbb{Z}_p)$.

(2) From (2.5), the result follows by direct calculation.

Now we give the proof of Theorem 1.2.

Proof of Theorem 1.2 From Lemma 3.3 (1), $b_0^3\tilde{\delta}_{s+4}\in {\rm Ext}_A^{s+10, t(s)}(\mathbb{Z}_p, \mathbb{Z}_p)$ is represented by $b_{1, 0}^3a_4^{s}h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}\in E_1^{s+10, t(s), 3p+9s+16}$ in the MSS. Now we will show that nothing hits the permanent cycle $b_{1, 0}^3a_4^{s}h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}$ under the May differential $d_r$ for $r\geqslant 1$. From Lemma 3.2, we have $E_1^{s+9, t(s), \ast}= \mathbb{Z}_p\{\mathbf{G}_1, \mathbf{G}_2, \cdots, \mathbf{G}_11\}$.

For the generators $\mathbf{G}_1$, $\mathbf{G}_3$ and $\mathbf{G}_5$ whose May filtration are

$ M(\mathbf{G}_1)=M(\mathbf{G}_3)=M(\mathbf{G}_5)=11p+9s+9 $

(see Lemma 3.3), by the reason of May filtration, from (2.2) we see that

$ b_{1, 0}^3a_4^{s}h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}\in E_1^{s+10, t(s), 3p+9s+16}, $

which represents $b_0^3\tilde{\delta}_{s+4}\in {\rm Ext}_A^{s+10, t(s)}(\mathbb{Z}_p, \mathbb{Z}_p)$ in the MSS is not in $d_1(E_1^{s+9, t(s), 11p+9s+9})$. Now we will show $E_r^{s+9, t(s), 11p+9s+9}=0$ for $r\geqslant 2$. By an easy calculation, from (2.3) and (2.4), one can have the first May differentials of $\mathbf{G}_1$, $\mathbf{G}_3$ and $\mathbf{G}_5$ as follows

$ \begin{array}{l} d_1(\mathbf{G}_1)=(-1)^{s+8}a_4^{s-1}a_2b_{3, 0}^2b_{1, 0}h_{3, 1}h_{2, 2}h_{2, 0}h_{1, 3}+\cdots\not=0, \\ d_1(\mathbf{G}_3)=(-1)^{s+8}a_4^{s-1}a_2b_{3, 0}^2b_{1, 2}h_{3, 1}h_{2, 2}h_{2, 0}h_{1, 1}+\cdots\not=0, \\ d_1(\mathbf{G}_5)=(-1)^{s+8}a_4^{s}b_{3, 0}^2b_{1, 0}h_{2, 2}h_{2, 0}h_{1, 3}h_{1, 1}+\cdots\not=0. \end{array} $

It is easy to see that the first May differentials of $\mathbf{G}_1$, $\mathbf{G}_3$and $\mathbf{G}_5$ are linearly independent.Consequently, the cocycle of $E_1^{s+9, t(s), 11p+9s+9}$ must be zero. This means that $E_r^{s+9, t(s), 11p+9s+9}=0$ for $r\geqslant 2$, from which we have that

$ b_{1, 0}^3a_4^{s}h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}\notin d_r(E_r^{s+9, t(s), 11p+9s+9}) $

for $r\geqslant 2$. In all, $b_{1, 0}^3a_4^{s}h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}\notin d_r(E_r^{s+9, t(s), 11p+9s+9})$ for $r\geqslant 1$.

For the generator $\mathbf{G}_2$ with May filtration $M(\mathbf{G}_2)=15p+9s+5$ (see Lemma 3.3), by an easy calculation, from (2.3) and (2.4), we have the first May differentials of $\mathbf{G}_2$ as follows

$ d_1(\mathbf{G}_2)=(-1)^{s+8}a_4^{s-1}a_2b_{3, 0}^2h_{3, 1}h_{1, 3}h_{1, 1}h_{1, 0}+\cdots\not=0. $

Thus $E_r^{s+9, t(s), 15p+9s+5}=0$ for $r\geqslant 2$. At the same time, we also have that up to nonzero scalar $d_1(\mathbf{G}_2)\not=b_{1, 0}^3a_4^{s}h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}.$

In summary, $b_{1, 0}^3a_4^{s}h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}\notin d_r(E_r^{s+9, t(s), 15p+9s+5})$ for $r\geqslant 1$.

For the generators $\mathbf{G}_4$ and $\mathbf{G}_6$ whose May filtration are $M(\mathbf{G}_4)=M(\mathbf{G}_6)=7p+9s+13$ (see Lemma 3.3), by the reason of May filtration, from (2.2) we see that

$ b_{1, 0}^3a_4^{s}h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}\in E_1^{s+10, t(s), 3p+9s+16}, $

which represents $b_0^3\tilde{\delta}_{s+4}\in {\rm Ext}_A^{s+10, t(s)}(\mathbb{Z}_p, \mathbb{Z}_p)\notin d_1(E_1^{s+9, t(s), 7p+9s+13})$. Now we will show $E_r^{s+9, t(s), 7p+9s+13}=0$ for $r\geqslant 2$. By an easy calculation, from (2.3) and (2.4) one can have the first May differentials of $\mathbf{G}_4$ and $\mathbf{G}_6$ as follows

$ \begin{array}{l} d_1(\mathbf{G}_4)=(-1)^{s+8}a_4^{s}b_{3, 0}b_{1, 0}^2h_{3, 1}h_{2, 2}h_{2, 0}h_{1, 3}+\cdots\not=0, \\ d_1(\mathbf{G}_6)=(-1)^{s+8}a_4^{s}b_{3, 0}b_{1, 2}b_{1, 0}h_{3, 1}h_{2, 2}h_{2, 0}h_{1, 1}+\cdots\not=0. \end{array} $

It is easy to see that the first May differentials of $\mathbf{G}_4$ and $\mathbf{G}_6$ are linearly independent. Consequently, the cocycle of $E_1^{s+9, t(s), 7p+9s+13}$ must be zero. This means that

$ E_r^{s+9, t(s), 7p+9s+13}=0 $

for $r\geqslant 2$, from which we have that $b_{1, 0}^3a_4^{s}h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}\notin d_r(E_r^{s+9, t(s), 7p+9s+13})$ for $r\geqslant 2$. In all, $b_{1, 0}^3a_4^{s}h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}\notin d_r(E_r^{s+9, t(s), 7p+9s+13})$ for $r\geqslant 1$.

For the generator $\mathbf{G}_7$ with May filtration $M(\mathbf{G}_7)=10p+9s+9$ (see Lemma 3.3), by an easy calculation, from (2.3) and (2.4) we have the first May differentials of $\mathbf{G}_7$ as follows

$ d_1(\mathbf{G}_7)=(-1)^{s+8}a_4^{s-2}a_2a_0b_{3, 0}^2h_{4, 0}h_{3, 1}h_{2, 2}h_{2, 0}h_{1, 3}h_{1, 1}+\cdots\not=0. $

Thus $E_r^{s+9, t(s), 10p+9s+9}=0$ for $r\geqslant 2$. At the same time, we also have that up to nonzero scalar $d_1(\mathbf{G}_7)\not=b_{1, 0}^3a_4^{s}h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}.$

In summary, $b_{1, 0}^3a_4^{s}h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}\notin d_r(E_r^{s+9, t(s), 10p+9s+9})$ for $r\geqslant 1$.

Finally, for the generators $\mathbf{G}_8$ and $\mathbf{G}_9$ whose May filtration are $M(\mathbf{G}_8)=M(\mathbf{G}_9)=6p+9s+13$ (see Lemma 3.3), by the reason of May filtration, from (2.2) we see that $b_{1, 0}^3a_4^{s}h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}\in E_1^{s+10, t(s), 3p+9s+16}, $ which represents $b_0^3\tilde{\delta}_{s+4}\in {\rm Ext}_A^{s+10, t(s)}(\mathbb{Z}_p, \mathbb{Z}_p)$ in the MSS is not in $d_r(E_1^{s+9, t(s), 6p+9s+13})$ for $r\geqslant 1$.

The discussion of $\mathbf{G}_{10}$ and $\mathbf{G}_{11}$ whose May filtration are $M(\mathbf{G}_{10})=M(\mathbf{G}_{11})=2p+9s+17$ is just like the analysis about $\mathbf{G}_{10}$ and $\mathbf{G}_{11}$.

From the above discussion, we see the permanent cycle $b_{1, 0}^3a_4^{s}h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}$ cannot be hit by any May differential in the MSS. Thus, $b_{1, 0}^3a_4^{s}h_{4, 0}h_{3, 1}h_{2, 2}h_{1, 3}\in E_1^{s+10, t(s), 3p+9s+16}$ converges nontrivially to $b_0^3\tilde{\delta}_{s+4}\in {\rm Ext}_A^{s+10, t(s)}(\mathbb{Z}_p, \mathbb{Z}_p)$ in the MSS. Consequently, $b_0^3\tilde{\delta}_{s+4}\not=0$. This finishes the proof of Theorem 1.2.

Remark For further study on the typesetting based on English-Chinese LATEX and some special techniques, we may refer to [1-7].