1_{K})+(\bar{c}_{2}\wedge 1_{X})\tilde{g}_{0}+(\bar{a}_{3}\wedge
1_{X})f^{\prime\prime}_{1}$ for some
$f^{\prime\prime}_{1}\in[\Sigma^{p^{n}q+q+1}K,E_{4}\wedge X]$ and so we have
$(\bar{a}_{2}\wedge
1_{X})f^{\prime\prime}_{0}=\lambda^{\prime}(\bar{a}_{2}\wedge
1_{X})(1_{E_{3}}\wedge w)(\zeta_{n-1,3}\wedge
1_{K})+(\bar{a}_{2}\bar{a}_{3}\wedge 1_{X})f^{\prime\prime}_{1}$ and (9.7.12)
becomes
(9.7.13) $(1_{E_{2}}\wedge w)(\eta^{\prime\prime}_{n,2})_{Y}\cdot
r=(\bar{c}_{1}\wedge 1_{X})g^{\prime\prime}$

$\qquad\qquad+\lambda^{\prime}(\bar{a}_{2}\wedge 1_{X})(1_{E_{3}}\wedge
w)(\zeta_{n-1,3}\wedge 1_{K})+(\bar{a}_{2}\bar{a}_{3}\wedge
1_{X})f^{\prime\prime}_{1}$
with $g^{\prime\prime}\in[\Sigma^{p^{n}q+q-1}K,KG_{1}\wedge X]$,
$f^{\prime\prime}_{1}\in[\Sigma^{p^{n}q+q+1}K,E_{4}\wedge X]$ and
$\lambda^{\prime}\in Z_{p}$.

To prove the Lemma, it suffices to prove the scalar $\lambda^{\prime}$ in
(9.7.13) is nonzero. Suppose in contrast that $\lambda^{\prime}$ = 0, then by
(9.7.13)(9.1.4) we have
(9.7.14) $(\bar{a}_{2}\bar{a}_{3}\wedge
1_{X})f^{\prime\prime}_{1}i^{\prime}i=-(\bar{c}_{1}\wedge
1_{X})g^{\prime\prime}i^{\prime}i$
This will yield a contradiction as shown below.

Note that the $d_{1}$-cycle
$g^{\prime\prime}i^{\prime}i\in\pi_{p^{n}q+q-1}KG_{1}\wedge X$ represents an
element in $Ext_{A}^{1.p^{n}q+q-1}(H^{*}X,Z_{p})\cong
Z_{p}\\{\tau_{*}(h_{n})\\}$ ( cf. Prop. 9.7.7(3)). Then
$g^{\prime\prime}i^{\prime}i=\lambda_{0}(1_{KG_{1}}\wedge\tau)(h_{n})$ , where
$h_{n}\in\pi_{p^{n}q}KG_{1}\cong Ext_{A}^{1,p^{n}q}(Z_{p},Z_{p})$ and
$\lambda_{0}\in Z_{p}$. Consequently, (9.7.14) becomes
(9.7.15) $(\bar{a}_{2}\bar{a}_{3}\wedge
1_{X})f^{\prime\prime}_{1}i^{\prime}i=-\lambda_{0}(\bar{c}_{1}\wedge
1_{X})(1_{KG_{1}}\wedge\tau)(h_{n})$

The equation (9.7.15) means the secondary differential
$-\lambda_{0}d_{2}(\tau_{*}(h_{n}))$ = 0. However, by [12] p.11 Theorem
1.2.14, $d_{2}(h_{n})=a_{0}b_{n-1}\neq 0\in
Ext_{A}^{3,p^{n}q+1}(Z_{p},Z_{p})$, where
$d_{2}:Ext_{A}^{1,p^{n}q}(Z_{p},Z_{p})\rightarrow
Ext_{A}^{3,p^{n}q+1}(Z_{p},Z_{p})$ is the secondary differential in the ASS.
This implies that
$d_{2}(\tau_{*}(h_{n}))=\tau_{*}d_{2}(h_{n})=\tau_{*}(a_{0}b_{n-1})=w_{*}(i^{\prime}i)_{*}(h_{0}b_{n-1})\neq
0\in Ext_{A}^{3,p^{n}q+q}(H^{*}X,Z_{p})$ ( cf. (9.7.8)). This shows that
$\lambda_{0}$ = 0 and so by (9.7.15) we have $(\bar{a}_{2}\bar{a}_{3}\wedge
1_{X})f^{\prime\prime}_{1}i^{\prime}i$ = 0.

It follows that $(\bar{a}_{3}\wedge
1_{X})f^{\prime\prime}_{1}i^{\prime}i=(\bar{c}_{2}\wedge
1_{X})g^{\prime\prime}_{2}$ = 0 , where the $d_{1}$-cycle
$g^{\prime\prime}_{2}\in\pi_{p^{n}q+q}KG_{2}\wedge X$ represents an element in
$Ext_{A}^{2,p^{n}q+q}(H^{*}X,Z_{p})\cong Z_{p}\\{w_{*}(i^{\prime}i)_{*}\\\
(h_{0}h_{n})\\}$ (cf. Prop. 9.7.9(3)) and the generator of this group is a
permanent cycle in the ASS (cf. Theorem 9.5.1) so that we have
$(\bar{c}_{2}\wedge 1_{X})g^{\prime\prime}_{2}=0$. Then
$f^{\prime\prime}_{1}i^{\prime}i=(\bar{c}_{3}\wedge
1_{X})g^{\prime\prime}_{3}=(\bar{c}_{3}\wedge
1_{X})g^{\prime\prime}_{4}i^{\prime}i$ for some
$g^{\prime\prime}_{3}\in\pi_{p^{n}q+q+1}KG_{3}\wedge X$ and
$g^{\prime\prime}_{4}\in[\Sigma^{p^{n}q+q+1}K,KG_{3}\wedge X]$ , this is
because $g^{\prime\prime}_{3}\cdot\epsilon$ = 0 which is obtained by the fact
that $\epsilon:Y\rightarrow\Sigma S$ induces zero homomorphism in
$Z_{p}$-cohomology. Consequently we have
$f^{\prime\prime}_{1}=(\bar{c}_{3}\wedge
1_{X})g^{\prime\prime}_{4}+f^{\prime\prime}_{2}r$ with
$f^{\prime\prime}_{2}\in[\Sigma^{p^{n}q+q+1}Y,E_{4}\wedge X]$ and
$(\bar{a}_{2}\bar{a}_{3}\wedge
1_{X})f^{\prime\prime}_{1}=(\bar{a}_{2}\bar{a}_{3}\wedge
1_{X})f^{\prime\prime}_{2}r$. Hence, if $\lambda^{\prime}$ = 0, (9.7.13)
becomes

$(1_{E_{2}}\wedge w)(\eta^{\prime\prime}_{n,2})_{Y}\cdot
r=(\bar{a}_{2}\bar{a}_{3}\wedge 1_{X})f^{\prime\prime}_{2}r+(\bar{c}_{1}\wedge
1_{X})g^{\prime\prime}_{5}r$

where $g^{\prime\prime}_{5}\in[\Sigma^{p^{n}q+q-1}Y,KG_{2}\wedge X]$ such that
$g^{\prime\prime}_{5}r=g^{\prime\prime}$. Moreover, by the above equation we
have
(9.7.16) $(1_{E_{2}}\wedge
w)(\eta^{\prime\prime}_{n,2})_{Y}=(\bar{a}_{2}\bar{a}_{3}\wedge
1_{X})f^{\prime\prime}_{2}+(\bar{c}_{1}\wedge
1_{X})g^{\prime\prime}_{5}+f^{\prime\prime}_{3}\epsilon$
with $f^{\prime\prime}_{3}\in\pi_{p^{n}q+q}E_{2}\wedge X$. Since
$\epsilon:Y\rightarrow\Sigma S$ induces zero homomorphism in
$Z_{p}$-cohomology , then the right hand side of (9.7.16) has filtration $\geq
3$. However, $(\eta^{\prime\prime}_{n})_{Y}=(\bar{a}_{0}\bar{a}_{1}\wedge
1_{K})(\eta^{\prime\prime}_{n,2})_{Y}$ has filtration 2, this is because it is
represented by $(h_{0}h_{n})^{\prime\prime}_{Y}\in
Ext_{A}^{2,p^{n}q+q-1}(H^{*}K,H^{*}Y)$ in the ASS. Moreover, by the following
exact sequence

$0=Ext_{A}^{1,p^{n}q}(H^{*}K,H^{*}Y)\stackrel{{\scriptstyle\alpha^{\prime\prime}_{*}}}{{\longrightarrow}}Ext_{A}^{2,p^{n}q+q-1}(H^{*}K,H^{*}Y)$

$\qquad\qquad\quad\stackrel{{\scriptstyle
w_{*}}}{{\longrightarrow}}Ext_{A}^{2,p^{n}q+q-1}(H^{*}X,H^{*}Y)\stackrel{{\scriptstyle(\alpha^{\prime\prime})_{*}}}{{\longrightarrow}}$
induced by (9.3.7) we know that $w_{*}(h_{0}h_{n})^{\prime\prime}_{Y}\neq 0$,
where the left group is zero by Prop. 9.7.5(3). That is to say,
$(1_{E_{2}}\wedge w)(\eta^{\prime\prime}_{n})_{Y}$ has filtration 2 which is
represented by $w_{*}(h_{0}h_{n})^{\prime\prime}_{Y}$ in the ASS. This shows
that the equation (9.7.16) is a contradiction so that the scalar
$\lambda^{\prime}$ must be nonzero. Q.E.D.

Lemma 9.7.17 Let $w:K\rightarrow X$ be the map in the cofibration (9.3.7) and
$W$ is the cofibre of $wi^{\prime}i:S\rightarrow X$ given by the cofibration
$S\stackrel{{\scriptstyle
wi^{\prime}i}}{{\longrightarrow}}X\stackrel{{\scriptstyle
w_{1}}}{{\longrightarrow}}W\stackrel{{\scriptstyle
u_{1}}}{{\longrightarrow}}\Sigma S$ , then

(1) $Ext_{A}^{s-1,p^{n}q+q+s-3}(H^{*}W,Z_{p})$ = 0 for $s=1,3$ and has unique
generator $(w_{1})_{*}\tau_{*}(h_{n})=(\tau)^{*}(w_{1})_{*}[h_{n}\wedge
1_{X}]$ for $s=2$.

(2) $Ext_{A}^{1,p^{n}q}(H^{*}W,H^{*}X)\cong
Z_{p}\\{(w_{1})_{*}(h_{n})^{\prime}_{X}=(w_{1})_{*}[h_{n}\wedge 1_{X}]\\}$.

(3) $(w_{1})_{*}[a_{0}b_{n-1}\wedge 1_{X}]\neq 0\in
Ext_{A}^{3,p^{n}q+1}(H^{*}W,H^{*}X)$.

Proof : (1) Note that $W$ also is the cofibre of
$r\alpha^{\prime\prime}:\Sigma^{q-2}K\rightarrow Y$, this can be seen by the
following homotopy commutative diagram of $3\times 3$-Lemma.

$\qquad\qquad\quad S\qquad\stackrel{{\scriptstyle
wi^{\prime}i}}{{\longrightarrow}}\qquad X\qquad\stackrel{{\scriptstyle
u}}{{\longrightarrow}}\quad\Sigma^{q-1}K$

$\qquad\qquad\qquad\searrow i^{\prime}i\qquad\nearrow w\quad\searrow
w_{1}\qquad\nearrow u_{2}$

$\qquad\qquad\qquad\qquad K\quad\qquad\qquad\qquad W$

$\qquad\qquad\qquad\nearrow\alpha^{\prime\prime}\qquad\searrow r\quad\nearrow
w_{2}\qquad\searrow u_{1}$

$\qquad\qquad\Sigma^{q-2}K\quad\stackrel{{\scriptstyle
r\alpha^{\prime\prime}}}{{\longrightarrow}}\qquad
Y\qquad\stackrel{{\scriptstyle\epsilon}}{{\longrightarrow}}\qquad\Sigma S$
That is , we have a cofibration
$\qquad\Sigma^{q-2}K\stackrel{{\scriptstyle
r\alpha^{\prime\prime}}}{{\longrightarrow}}Y\stackrel{{\scriptstyle
w_{2}}}{{\longrightarrow}}W\stackrel{{\scriptstyle
u_{2}}}{{\longrightarrow}}\Sigma^{q-1}K$ and it induces the following exact
sequence

$Ext_{A}^{s-1,p^{n}q+q+s-3}(H^{*}Y,Z_{p})\stackrel{{\scriptstyle(w_{2})_{*}}}{{\longrightarrow}}Ext_{A}^{s-1,p^{n}q+q+s-3}(H^{*}W,Z_{p})$

$\qquad\qquad\quad\stackrel{{\scriptstyle(u_{2})_{*}}}{{\longrightarrow}}Ext_{A}^{s-1,p^{n}q+s-2}(H^{*}K,Z_{p})\stackrel{{\scriptstyle(r\alpha^{\prime\prime})_{*}}}{{\longrightarrow}}$
The left group is zero for $s=1,2,3$ ( cf. Prop. 9.7.5(2)). The right group
also is zero for $s=1,3$ (cf. Prop. 9.7.5(1)) and has unique generator
$(i^{\prime}i)_{*}(h_{n})=u_{*}\tau_{*}(h_{n})=(u_{2})_{*}(w_{1})_{*}(\tau)_{*}(h_{n})$
for $s=2$. Then the result follows.

(2) Consider the following exact sequence

$Ext_{A}^{1,p^{n}q}(H^{*}X,H^{*}X)\stackrel{{\scriptstyle(w_{1})_{*}}}{{\longrightarrow}}Ext_{A}^{1,p^{n}q}(H^{*}W,H^{*}X)$

$\qquad\qquad\quad\stackrel{{\scriptstyle(u_{1})_{*}}}{{\longrightarrow}}Ext_{A}^{2,p^{n}q}(Z_{p},H^{*}X)=0$
The right group is zero by Prop. 9.7.7(1) and by Prop. 9.7.9(2)(1) we know
that the left group has unique generator $(h_{n})^{\prime}_{X}=[h_{n}\wedge
1_{X}]$ which satisfies $u_{*}(h_{n})^{\prime}_{X}=u^{*}(h_{n})^{\prime}\in
Ext_{A}^{1,p^{n}q-q+1}(H^{*}K,H^{*}X)$ . Then the middle group has unique
generator $(w_{1})_{*}(h_{n})^{\prime}_{X}$.

(3) Since $\alpha^{\prime\prime}:\Sigma^{q-2}K\to K$ is not an $M$-module map,
then , as the cofibre of $\alpha^{\prime\prime}$, the spectrum $X$ is not an
$M$-moduld spectrum, that is, the map $p\wedge 1_{X}\neq 0\in[X,X]$. So
$[a_{0}\wedge 1_{X}]=(p\wedge 1_{X})_{*}[\tau\wedge 1_{X}]\neq 0\in
Ext_{A}^{1,1}(H^{*}X,X^{*}X)$ ( where $\tau$ is the unit in $\pi_{0}KG_{0}$),
and so $[a_{0}b_{n-1}\wedge 1_{X}]=(p\wedge 1_{X})_{*}[b_{n-1}\wedge
1_{X}]=[a_{0}\wedge 1_{X}][b_{n-1}\wedge 1_{X}]\neq 0\in
Ext_{A}^{3,p^{n}q+1}(H^{*}X,X^{*}X)$ which can be obtained by by knowledge of
Yoneda products and $a_{0}b_{n-1}\neq 0\in Ext_{A}^{3,p^{n}q+1}(Z_{p},Z_{p})$.
Note to the following exact sequence

$0=Ext_{A}^{3,p^{n}q+1}(Z_{p},H^{*}X)\stackrel{{\scriptstyle(wi^{\prime}i)_{*}}}{{\longrightarrow}}Ext_{A}^{3,p^{n}q+1}(H^{*}X,H^{*}X)$

$\qquad\qquad\quad\stackrel{{\scriptstyle(w_{1})_{*}}}{{\longrightarrow}}Ext_{A}^{3,p^{n}q+1}(H^{*}W,H^{*}X)$
where the left group is zero by Prop. 9.7.7(1), then $(w_{1})_{*}$ is monic
and so the reslt follows. Q.E.D.

Remark The result on $[a_{0}b_{n-1}\wedge 1_{X}]\neq 0$ in Lemma 9.7.17(3)
also can be proved by some computation in Ext groups as follows. Suppose in
contrast that $(p\wedge 1_{X})_{*}[b_{n-1}\wedge 1_{X}]=[a_{0}b_{n-1}\wedge
1_{X}]=0$, then by (9.1.1), $[b_{n-1}\wedge 1_{X}]=(j\wedge 1_{X})_{*}(x_{1})$
with $x_{1}\in Ext_{A}^{2,p^{n}q+1}(H^{*}M\wedge X,H^{*}X)$. Recall that $X$
is the spectrum in (9.3.7), then we have $w^{*}(1_{M}\wedge u)_{*}(x_{1})\in
Ext_{A}^{2,p^{n}q-q+1}(H^{*}M\wedge K,H^{*}K)$ = 0 which can be obtained by
$Ext_{A}^{2,p^{n}q-q+1}\\\ (H^{*}M\wedge K,H^{*}M)$ = 0 and
$Ext_{A}^{2,p^{n}q+2}(H^{*}M\wedge K,H^{*}M)$ = 0. By the Ext exact sequence
induced by (9.3.7) we have $(1_{M}\wedge u)_{*}(x_{1})\in
u^{*}Ext_{A}^{2,p^{n}q+1}(H^{*}M\wedge K,H^{*}K)$. However, this group has
unique generator $(\overline{m}_{K})_{*}(b_{n-1})^{\prime}$, then
$(1_{M}\wedge u)_{*}(x_{1})=\lambda
u^{*}(\overline{m}_{K})_{*}(b_{n-1})^{\prime}$ for some $\lambda\in Z_{p}$. By
applying $(1_{M}\wedge\alpha^{\prime\prime})_{*}$ we have $\lambda
u^{*}(1_{M}\wedge\alpha^{\prime\prime})_{*}(\overline{m}_{K})_{*}(b_{n-1})^{\prime}$
= 0 and so
$\lambda(1_{M}\wedge\alpha^{\prime\prime})_{*}(\overline{m}_{K})_{*}(b_{n-1})^{\prime}\in(\alpha^{\prime\prime})^{*}Ext_{A}^{2,p^{n}q+1}(H^{*}M\wedge
K,H^{*}K)$. Then $\lambda(\alpha_{1}\wedge
1_{K})_{*}(b_{n-1})^{\prime}=\lambda(m_{K})_{*}(1_{M}\wedge\alpha^{\prime\prime})_{*}(\overline{m}_{K})_{*}(b_{n-1})^{\prime}$
= 0 which shows that $\lambda=0$. So $x_{1}=(1_{M}\wedge w)_{*}(x_{2})$ for
some $x_{2}\in Ext_{A}^{2,p^{n}q+1}(H^{*}M\wedge K,H^{*}X)$. Similarly we can
prove that $w^{*}(x_{2})$ = 0. Then $x_{1}\in u^{*}(1_{M}\wedge
w)_{*}Ext_{A}^{2,p^{n}q+q}(H^{*}M\wedge K,H^{*}K)$. However,
$Ext_{A}^{2,p^{n}q+q}(H^{*}M\wedge K,H^{*}K)$ has two generators $(i\wedge
1_{K})_{*}(h_{0}h_{n})^{\prime}$ ,
$(\overline{m}_{K})_{*}(h_{0}h_{n})^{\prime\prime}$ and
$u^{*}(h_{0}h_{n})^{\prime\prime}=u^{*}(\alpha^{\prime\prime})^{*}(h_{n})^{\prime}$
= 0, then we have $x_{1}=u^{*}(1_{M}\wedge w)_{*}(i\wedge
1_{K})_{*}(h_{0}h_{n})^{\prime}$( up to scalar) and so $[b_{n-1}\wedge
1_{X}]=(j\wedge 1_{X})_{*}(x_{1})$ = 0. This is a contradiction and shows that
$[a_{0}b_{n-1}\wedge 1_{X}]\neq 0\in Ext_{A}^{3,p^{n}q+1}(H^{*}X,X^{*}X)$.

Proof of Theorem 9.7.4 : The result for $n=0,1$ is wellknown, then we assume
that $n\geq 2$. By Lemma 9.7.11 and (9.1.4) we have
(9.7.18)
$\lambda^{\prime}wi^{\prime}i\zeta_{n-1}=\lambda^{\prime}w(\zeta_{n-1}\wedge
1_{K})i^{\prime}i=-(\bar{a}_{0}\bar{a}_{1}\bar{a}_{2}\bar{a}_{3}\wedge
1_{X})f^{\prime\prime}_{1}i^{\prime}i.$

Moreover, by the cofibration in Lemma 9.7.17 we have
$(\bar{a}_{0}\bar{a}_{1}\bar{a}_{2}\bar{a}_{3}\wedge 1_{W})(1_{E_{4}}\wedge
w_{1})f^{\prime\prime}_{1}i^{\prime}i$ = 0 and so
$(\bar{a}_{0}\bar{a}_{1}\bar{a}_{2}\bar{a}_{3}\wedge 1_{W})(1_{E_{4}}\wedge
w_{1})f^{\prime\prime}_{1}u=k\tau^{\prime}$ for some
$k\in[\Sigma^{p^{n}q-2}X^{\prime},W]$ , where
$i^{\prime}i=u\tau:\Sigma^{q-1}S\stackrel{{\scriptstyle\tau}}{{\rightarrow}}X\stackrel{{\scriptstyle
u}}{{\rightarrow}}\Sigma^{q-1}K$ which is obtained by
$\alpha^{\prime\prime}i^{\prime}i$ = 0 and (9.3.7) and
$\tau^{\prime}:X\rightarrow X^{\prime}$ is the map in the following
cofibration
(9.7.19)
$\Sigma^{q-1}S\stackrel{{\scriptstyle\tau}}{{\rightarrow}}X\stackrel{{\scriptstyle\tau^{\prime}}}{{\rightarrow}}X^{\prime}\stackrel{{\scriptstyle\tau^{\prime\prime}}}{{\rightarrow}}\Sigma^{q}S.$

We claim that $k\in[\Sigma^{p^{n}q-2}X^{\prime},W]$ has filtration $\geq 4$,
this can be proved as follows.

By Lemma 9.7.17(1) and (9.7.18) we have
$(\tau^{\prime\prime})^{*}Ext_{A}^{s-1,p^{n}q+q+s-3}(H^{*}W,\\\
Z_{p})=0\subset Ext_{A}^{s,p^{n}q+s-2}(H^{*}W,H^{*}X^{\prime})$ and so
$(\tau^{\prime})^{*}:Ext_{A}^{s,p^{n}q+s-2}(H^{*}W,\\\
H^{*}X^{\prime})\rightarrow Ext_{A}^{s,p^{n}q+s-2}$ $(H^{*}W,H^{*}X)$ for
$s=1,2,3$ is monic. Then, the fact that $k\tau^{\prime}$ has filtration$\geq
4$ implies that $k\in[\Sigma^{p^{n}q-2}X^{\prime},W]$ also has filtration
$\geq 4$. This shows the above claim and so
$k=(\bar{a}_{0}\bar{a}_{1}\bar{a}_{2}\bar{a}_{3}\wedge 1_{W})k_{3}$ for some
$k_{3}\in[\Sigma^{p^{n}q+2}X^{\prime},E_{4}\wedge W]$ and
$(\bar{a}_{0}\bar{a}_{1}\bar{a}_{2}\bar{a}_{3}\wedge 1_{W})(1_{E_{4}}\wedge
w_{1})f^{\prime\prime}_{1}u=(\bar{a}_{0}\bar{a}_{1}\bar{a}_{2}\bar{a}_{3}\wedge
1_{W})k_{3}\tau^{\prime}$. It follows that
(9.7.20) $(\bar{a}_{2}\bar{a}_{3}\wedge 1_{W})(1_{E_{4}}\wedge
w_{1})f^{\prime\prime}_{1}u=(\bar{a}_{2}\bar{a}_{3}\wedge
1_{W})k_{3}\tau^{\prime}+(\bar{c}_{1}\wedge 1_{W})\bar{g}$

$\quad=(\bar{a}_{2}\bar{a}_{3}\wedge
1_{W})k_{3}\tau^{\prime}+\lambda_{1}(\bar{c}_{1}\wedge 1_{W})(1_{KG_{1}}\wedge
w_{1})(h_{n}\wedge 1_{X})$
where the $d_{1}$-cycle $\bar{g}=\lambda_{1}(1_{KG_{1}}\wedge
w_{1})(h_{n}\wedge 1_{X})\in[\Sigma^{p^{n}q}X,KG_{1}\wedge W]$ with
$\lambda_{1}\in Z_{p}$ which is obtained by Lemma 9.7.17(2).

The equation (9.7.20) means that the differential
$d_{2}(\lambda_{1}(w_{1})_{*}[h_{n}\wedge 1_{X}])$ = 0. However,
$d_{2}((w_{1})_{*}[h_{n}\wedge 1_{X}])=(w_{1})_{*}[a_{0}b_{n-1}\wedge
1_{X}]\neq 0$ ( cf. Lemma 9.7.17(3)). Then the scalar $\lambda_{1}$ = 0 and we
have $\bar{g}$ = 0, $(\bar{a}_{2}\bar{a}_{3}\wedge 1_{W})(1_{E_{4}}\wedge
w_{1})f^{\prime\prime}_{1}u=(\bar{a}_{2}\bar{a}_{3}\wedge
1_{W})k_{3}\tau^{\prime}$ and $(\bar{a}_{2}\bar{a}_{3}\wedge
1_{K})(1_{E_{4}}\wedge
u)f^{\prime\prime}_{1}i^{\prime}i=(\bar{a}_{2}\bar{a}_{3}\wedge
1_{K})(1_{E_{4}}\wedge u_{2}w_{1})f^{\prime\prime}_{1}u\tau$ = 0. Consequently
we have $(\bar{a}_{3}\wedge 1_{K})(1_{E_{4}}\wedge
u)f^{\prime\prime}_{1}i^{\prime}i=(\bar{c}_{2}\wedge 1_{K})\bar{g}_{2}=0$ ,
this is because the $d_{1}$-cycle $\bar{g}_{2}\in\pi_{p^{n}q+1}KG_{2}\wedge K$
represents an element in $Ext_{A}^{2,p^{n}q+1}(H^{*}K,Z_{p})$ = 0 (cf.
9.7.5(1)) . Then, $(1_{E_{4}}\wedge
u)f^{\prime\prime}_{1}i^{\prime}i=(\bar{c}_{3}\wedge 1_{K})\bar{g}_{3}$ for
some $\bar{g}_{3}\in[\Sigma^{p^{n}q+2}S,KG_{3}\wedge K]$. Since
$(1_{KG_{3}}\wedge\alpha^{\prime\prime})\bar{g}_{3}$ = 0, then
$\bar{g}_{3}=(1_{KG_{3}}\wedge u)\bar{g}_{4}$ for some
$\bar{g}_{4}\in[\Sigma^{p^{n}q+q+1}S,KG_{3}\wedge X]$ and so we have
$(1_{E_{4}}\wedge u)f^{\prime\prime}_{1}i^{\prime}i=(\bar{c}_{3}\wedge
1_{K})(1_{KG_{3}}\wedge u)\bar{g}_{4}$ ,
$f^{\prime\prime}_{1}i^{\prime}i=(\bar{c}_{3}\wedge
1_{X})\bar{g}_{4}+(1_{E_{4}}\wedge w)\bar{f}_{2}$ with
$\bar{f}_{2}\in[\Sigma^{p^{n}q+q+1}S,E_{4}\wedge K]$.

Hence, by (9.7.18) we have
$\lambda^{\prime}wi^{\prime}i\zeta_{n-1}=-(\bar{a}_{0}\bar{a}_{1}\bar{a}_{2}\bar{a}_{3}\wedge
1_{X})f^{\prime\prime}_{1}i^{\prime}i=-(\bar{a}_{0}\bar{a}_{1}\bar{a}_{2}\\\
\bar{a}_{3}\wedge 1_{X})(1_{E_{4}}\wedge w)\bar{f}_{2}$ and by (9.3.7),
$\lambda^{\prime}i^{\prime}i\zeta_{n-1}=-(\bar{a}_{0}\bar{a}_{1}\bar{a}_{2}\bar{a}_{3}\wedge
1_{K})\bar{f}_{2}+\alpha^{\prime\prime}\omega_{n}$ with
$\omega_{n}\in\pi_{p^{n}q-1}K$. Since $\lambda^{\prime}i^{\prime}i\zeta_{n-1}$
is a map of filtration 3 which is represented by
$\lambda^{\prime}(i^{\prime}i)_{*}(h_{0}b_{n-1})\in
Ext_{A}^{3,p^{n}q+q}(H^{*}K,Z_{p})$ in the ASS, then
$\alpha^{\prime\prime}\omega_{n}$ has filtration 3 and so
$\omega_{n}\in\pi_{p^{n}q-1}K$ has filtration $\leq 2$. However, by Prop.
9.7.5(1) we have $Ext_{A}^{2,p^{n}q+1}(H^{*}K,Z_{p})$ = 0 , then
$\omega_{n}\in\pi_{p^{n}q-1}K$ must be represented by the unique generator
$(i^{\prime}i)_{*}(h_{n})\in Ext_{A}^{1,p^{n}q}(H^{*}K,Z_{p})$ (up to nonzero
scalar). This shows the Theorem. Q.E.D.

Remark The element $\omega_{n}\in\pi_{p^{n}q-1}K$ obtained in Theorem 9.7.4
can be extended to $(\omega_{n})^{\prime}\in[\Sigma^{p^{n}q-1}K,K]$ such that
$(\omega_{n})^{\prime}i^{\prime}i=\omega_{n}$. Then, $(\omega_{n})^{\prime}$
is represented by $(h_{n})^{\prime}\in Ext_{A}^{1,p^{n}q}(H^{*}K,H^{*}K)$ in
the ASS and $\alpha^{\prime\prime}(\omega_{n})^{\prime}$,
$(\omega_{n})^{\prime}\alpha^{\prime\prime}\in[\Sigma^{p^{n}q+q-3}K,K]$ is
represented by
$\alpha^{\prime\prime}_{*}(h_{n})^{\prime}=(\alpha^{\prime\prime})^{*}(h_{n})^{\prime}=(h_{0}h_{n})^{\prime\prime}\\\
\in Ext_{A}^{2,p^{n}q+q-1}(H^{*}K,H^{*}K)$. By Theorem 9.7.4 and Lemma 9.7.11
we have
$\alpha^{\prime\prime}(\omega_{n})^{\prime}=(\omega_{n})^{\prime}\alpha^{\prime\prime}+\lambda^{\prime}\zeta_{n-1}\wedge
1_{K}$ (modulo higher filtration).

By [10] p.511, there is a map $\phi_{*}\phi:BP_{*}BP\rightarrow A_{*}$ such
that $t_{n}\mapsto$ the conjugate of $\xi_{n}$, where
$A_{*}=E[\tau_{0},\tau_{1},\tau_{2},...]\otimes P[\xi_{1},\xi_{2},...]$ is the
dual of the Steenrod algebra $A$. Then $\phi_{*}\phi$ induces the Thom map
$\Phi:Ext_{BP_{*}BP}^{1,p^{n}q}(BP_{*},\\\ BP_{*}K)\rightarrow
Ext_{A}^{1,p^{n}q}(H^{*}K,Z_{p})$ such that the image of $h_{n}\in
Ext_{BP_{*}BP}^{1,p^{n}q}(BP_{*},\\\ BP_{*}K)$ is
$\Phi(h_{n})=(i^{\prime}i)_{*}(h_{n})\in Ext_{A}^{1,p^{n}q}(H^{*}K,Z_{p})$.
Then, the element $\omega_{n}\in\pi_{p^{n}q-1}K$ obtained in Theorem 9.7.4 is
represented by $h_{n}$ \+ other terms $\in
Ext_{BP_{*}BP}^{1,p^{n}q}(BP_{*},\\\ BP_{*}K)$ in the Adams-Novikov spectral
sequence. To know what the elements in the other terms, we first prove the
following Lemma.

Lemma 9.7.21 By degree reason, $Ext_{BP_{*}BP}^{1,p^{n}q}(BP_{*},BP_{*}K)$ is
generated (additively) by the following $v_{2}$-torsion elements
$c_{2}(p^{n-2})$ and $v_{2}$-torsion free elements
$h_{n},v_{2}^{p^{n-2}(p-1)}h_{n-2},v_{2}^{a_{i}p^{i}}h_{i}$ , where $i\geq
0,a_{i}=(p^{2k}-1)/(p+1)$ , $n-i=2k\geq 4$. In addition, there is a relation
$h_{n}=c_{2}(p^{n-2})+v_{2}^{p^{n-2}(p-1)}h_{n-2}\in
Ext_{BP_{*}BP}^{1,p^{n}q}(BP_{*},BP_{*}K)$.

Proof : By [19] Theorem 1.1 and 1.5, $Ext^{1,*}(BP_{*},BP_{*}K)$ is a
$Z_{p}[v_{2}]$-module which is generated by $v_{2}$-torsion elements
$c_{2}(ap^{s})$ and $v_{2}$-torsion free elements $w_{2},h_{i}$, where $a\neq
0$ (mod $p$), $s\geq 0$ and $i\geq 0$. Moreover, the internal degree $\mid
h_{i}\mid=p^{i}q$, $\mid c_{2}(ap^{s})\mid=ap^{s}(p^{2}+p+1)q-q(ap^{s})(p+1)q$
and $\mid w_{2}\mid=(p+1)^{2}q$.

Since $\mid v_{2}^{b}w_{2}\mid\equiv 0$ (mod $(p+1)q)$, then $\mid
v_{2}^{b}w_{2}\mid\neq p^{n}q$. If $\mid v_{2}^{b}h_{i}\mid=p^{n}q$, then
$b(p+1)q+p^{i}q=p^{n}q,b(p+1)=p^{i}(p^{n-i}-1)$ and so $(p^{n-i}-1)$ must be
divisible by $p+1$. Hence $b=0$ and $i=n$ or $b=a_{i}p^{i}$ with
$a_{i}=(p^{2k}-1)/(p+1)$ and $n-i=2k\geq 2$. Then
$h_{n},v_{2}^{p^{n-2}(p-1)}h_{n-2}$ and $v_{2}^{a_{i}p^{i}}h_{i}(0\leq i<n-2$)
are the only torsion free elements of $Ext^{1,p^{n}q}(BP_{*},BP_{*}K)$.

If $\mid v_{2}^{b}c_{2}(ap^{s})\mid=p^{n}q$, then
$p^{n}q=ap^{s}(p^{2}+p+1)q-(q(ap^{s})-b)(p+1)q$,
$ap^{s}(p^{2}+p+1)=p^{n}+(q(ap^{s})-b)(p+1)$ and so the right hand side must
be divisible by $p^{2}+p+1$. So we have
(9.7.22) $ap^{s}=p^{n-2}+\frac{(q(ap^{s})-b-p^{n-2})(p+1)}{p^{2}+p+1}.$
We claim that $s\leq n-2$ which will be proved below , then $q(ap^{s})-b$ must
be divisible by $p^{s}$. However, by [19] p.132, $q(ap^{s})=p^{s}$ for $a=1$
and $q(ap^{s})=p^{s}$ \+ other terms for $a\geq 2$. Then, the only possibility
is $q(ap^{s})=p^{s},a=1,b=0$ and $s=n-2$. That is to say, the only
$v_{2}$-torsion elements in $Ext^{1,p^{n}q}(BP_{*},BP_{*}K)$ is
$c_{2}(p^{n-2})$.

Now we prove the above claim. Suppose in contrast that $s\geq n-1$ , then, by
(9.7.22) we have
$\frac{(q(ap^{s})-b-p^{n-2})(p+1)}{p^{2}+p+1}=ap^{s}-p^{n-2}\geq
p^{s}-p^{n-2},$

$2p^{s}>q(ap^{s})-b-p^{n-2}\geq\frac{(p^{s}-p^{n-2})(p^{2}+p+1)}{p+1}>p^{s+1}-p^{n-1}$

and this is a contradiction which shows the above claim. Q.E.D.

Proof of Theorem 9.7.1 For the Thom
$\Phi:Ext_{BP_{*}BP}^{1,p^{n}q}(BP_{*},BP_{*}K)\\\ \rightarrow
Ext_{A}^{1,p^{n}q}(H^{*}K,Z_{p})$ we have
$\Phi(h_{n})=(i^{\prime}i)_{*}(h_{n})$. By this we know that the element
$\omega_{n}\in\pi_{p^{n}q-1}K$ obtained in Theorem 9.7.4 is represented by
$h_{n}$ \+ ( other terms) $\in Ext_{BP_{*}BP}^{1,p^{n}q}(BP_{*},BP_{*}K)$ in
the Adams-Novikov spectral sequence. By Lemma 9.7.21, the other terms are the
linear combination of $v_{2}^{p^{n-2}(p-1)}h_{n-2}$ and
$v_{2}^{a_{i}p^{i}}h_{i}$, where $i\geq 0,n-i=2k\geq 4$ and
$a_{i}=(p^{2k}-1)/(p+1)$. Let $\beta\in[\Sigma^{(p+1)q}K,K]$ be the known
$v_{2}$-map, then $i^{\prime}i\alpha_{1}\in\pi_{q-1}K$ and
$i^{\prime}j^{\prime}\beta i^{\prime}i\in\pi_{pq-1}K$ is represented by
$h_{0},h_{1}\in Ext_{BP_{*}BP}^{1,*}(BP_{*},BP_{*}K)$ respectively. That is,
$h_{0},h_{1}\in Ext_{BP_{*}BP}^{1,*}(BP_{*},BP_{*}K)$ are permanent cycles in
the Adams-Novikov spectral sequence. Suppose inductively that $h_{i}\in
Ext_{BP_{*}BP}^{1,p^{i}q}(BP_{*},BP_{*}K)$ for $i\leq n-1(n\geq 2)$ are
permanent cycles in the Adams-Novikov spectral sequence. Since
$\omega_{n}\in\pi_{p^{n}q-1}K,\omega_{n+1}\in\pi_{p^{n+1}q-1}K$ are
represented by the linear combination of

$h_{n}+v_{2}^{p^{n-2}(p-1)}h_{n-2}$ and $v_{2}^{a_{i}p^{i}}h_{i}\in
Ext_{BP_{*}BP}^{1,p^{n}q}(BP_{*},BP_{*}K),$

$h_{n+1}+v_{2}^{p^{n-1}(p-1)}h_{n-1}$ and $v_{2}^{a_{i}p^{i}}h_{i}\in
Ext_{BP_{*}BP}^{1,p^{n+1}q}(BP_{*},BP_{*}K)$

then $h_{n},h_{n+1}\in Ext_{BP_{*}BP}^{1,*}(BP_{*},BP_{*}K)$ also are
permanent cycles. This completes the induction and the result of the Theorem
follows. Q.E.D.

Conjecture 9.7.22 Theorem 9.7.4 can be generalzed to be the following general
result. Let $p\geq 5,s\leq 4$ , $Ext_{A}^{s,tq}(Z_{p},Z_{p})\cong
Z_{p}\\{x\\},Ext_{A}^{s+1,tq+q}\\\ (Z_{p},Z_{p})\cong
Z_{p}\\{h_{0}x\\},Ext_{A}^{s+2,tq+2q+1}(Z_{p},Z_{p})\cong
Z_{p}\\{\widetilde{\alpha}_{2}x\\}$ and some supposition on vanishes of some
Ext groups. If the secondary differential $d_{2}(x)=a_{0}x^{\prime}\in\\\
Ext_{A}^{s+2,tq+1}(Z_{p},Z_{p})$ with $x^{\prime}\in
Ext_{A}^{s+1,tq}(Z_{p},Z_{p})$, that is, $x$ and $x^{\prime}$ is a pair of
$a_{0}$-related elements, then there exists $\omega\in\pi_{tq-s}K$ such that
$i^{\prime}i\xi=\alpha^{\prime\prime}\cdot\omega$ (mod $F^{s+2}\pi_{*}K)$ and
$\omega\in\pi_{tq-s}K$ is represented by $(i^{\prime}i)_{*}(x)\in
Ext_{A}^{s,tq}(H^{*}K,Z_{p})$ in the ASS, where $\xi\in\pi_{tq+q-s-2}S$ is the
homotopy element which is represented by $h_{0}x^{\prime}\in
Ext_{A}^{s+2,tq+q}(Z_{p},Z_{p})$ in the ASS and $F^{s+2}\pi_{*}K$ denotes the
group consisting of all elements in $\pi_{*}K$ filtration $\geq s+2$.

§8. Second periodicity families in the stable homotopy groups of spheres

By Theorem 8.1.2 in chapter 8, $Ext_{BP_{*}BP}^{1,*}(BP_{*},BP_{*})$ is
generated by $\alpha_{tp^{n}/n+1}(n\geq 0,p$ not divisible by $t\geq 1$) and
It was proved by Novikov that all these first periodicity families converge to
the im $J\subset\pi_{*}S$. In this section, using the $h_{0}h_{n+1}$-element
obtained in Theorem 9.5.1 and the elements $\beta_{p/r},1\leq r\leq p-1$ and
$\beta_{tp/r},t\geq 2,1\leq r\leq p$ as our geometric input, we prove the
following Theorem on the convergence of second periodicity families
$\beta_{tp^{n}/r}$ in the Adams-Novikov spectral sequence.

Theorem 9.8.1 Let $p\geq 5,n\geq 1,1\leq s\leq p^{n}-1$ if $t\geq 1$ is not
divisible by $p$ or $1\leq s\leq p^{n}$ if $t\geq 2$ is not divisible by $p$ ,
then The elements

$\beta_{tp^{n}/s}\in Ext_{BP_{*}BP}^{2,tp^{n}(p+1)q-sq}(BP_{*},BP_{*})$

in Theorem 8.1.3 are permanent cycles in the Adams-Novikov spectral sequence
and they converge to the corresponding homotopy elements of order $p$ in
$\pi_{tp^{n}(p+1)q-sq-2}S$.

We will prove Theorem 9.8.1 in case $t\geq 1$ or $t\geq 2$ separately. The
proof will be done by some arguments processing in the cannical Adams-Novikov
resolution. We first do some preminilaries as follows.

Let $M$ be the Moore spectrum whose $BP_{*}$-homology are
$BP_{*}(M)=BP_{*}/(p)$. Let $\alpha:\Sigma^{q}M\rightarrow M$ be the Adams map
which induces $BP_{*}$-homomorphisms are $v_{1}:BP_{*}/(p)\rightarrow
BP_{*}/(p)$. Let $K_{r}$ be the cofibre of $\alpha^{r}:\Sigma^{rq}M\rightarrow
M$ given by the cofibration
(9.8.2)$\qquad\qquad\Sigma^{rq}M\stackrel{{\scriptstyle\alpha^{r}}}{{\longrightarrow}}M\stackrel{{\scriptstyle
i^{\prime}_{r}}}{{\longrightarrow}}K_{r}\stackrel{{\scriptstyle
j^{\prime}_{r}}}{{\longrightarrow}}\Sigma^{rq+1}M$
The cofibration (9.8.2) induces a short exact sequence of $BP_{*}$-homology

$0\rightarrow BP_{*}/(p)\stackrel{{\scriptstyle
v_{1}^{r}}}{{\longrightarrow}}BP_{*}/(p)\longrightarrow
BP_{*}/(p,v_{1}^{r})\rightarrow 0$

Recall from §5 in chapter 6, $K_{r}$ is a $M$-module spectrum and we have the
following derivations
(9.8.3) $d(i^{\prime}_{r})$ = 0, $d(j^{\prime}_{r})$ = 0, $d(\alpha)=0,\quad
d(ij)=-1_{M}$.

Moreover, the cofibre of
$i^{\prime}_{s}j^{\prime}_{r}:K_{r}\rightarrow\Sigma^{rq+1}K_{s}$ is $\Sigma
K_{r+s}$ given by the cofibration
(9.8.4)
$\qquad\Sigma^{rq}K_{s}\stackrel{{\scriptstyle\psi_{s,s+r}}}{{\longrightarrow}}K_{s+r}\stackrel{{\scriptstyle\rho_{s+r,r}}}{{\longrightarrow}}K_{r}\stackrel{{\scriptstyle
i^{\prime}_{s}j^{\prime}_{r}}}{{\longrightarrow}}\Sigma^{rq+1}K_{s}$
This can be seen by the following homotopy commutative diagram of $3\times
3$-Lemma

$\qquad\qquad K_{r}\quad\stackrel{{\scriptstyle
i^{\prime}_{s}j^{\prime}_{r}}}{{\longrightarrow}}\quad\Sigma^{rq+1}K_{s}\quad\stackrel{{\scriptstyle
j^{\prime}_{s}}}{{\longrightarrow}}\quad\Sigma^{(r+s)q+2}M$

$\qquad\qquad\quad\searrow j^{\prime}_{r}\quad\nearrow
i^{\prime}_{s}\quad\searrow\psi_{s,r+s}\quad\nearrow
j^{\prime}_{r+s}\quad\searrow\alpha^{s}$
(9.8.5) $\Sigma^{rq+1}M\qquad\qquad\Sigma
K_{r+s}\qquad\qquad\qquad\Sigma^{rq+2}M$

$\qquad\qquad\quad\nearrow\alpha^{s}\quad\searrow\alpha^{r}\quad\nearrow
i^{\prime}_{r+s}\quad\searrow\rho_{r+s,r}\quad\nearrow j^{\prime}_{r}$

$\qquad\quad\Sigma^{(r+s)q+1}M\stackrel{{\scriptstyle\alpha^{r+s}}}{{\longrightarrow}}\Sigma
M\quad\quad\stackrel{{\scriptstyle
i^{\prime}_{r}}}{{\longrightarrow}}\qquad\Sigma K_{r}$
Moreover, the cofibration (9.8.4) induces a short exact sequence of
$BP_{*}$-homology

$0\rightarrow BP_{*}/(p,v_{1}^{s})\stackrel{{\scriptstyle
v_{1}^{r}}}{{\longrightarrow}}BP_{*}/(p,v_{1}^{s+r})\longrightarrow
BP_{*}/(p,v_{1}^{r})\rightarrow 0$

and by the homotopy commutative diagram (9.8.5) , we have the following
relations
(9.8.6) $\qquad\psi_{s,s+r}i^{\prime}_{s}=i^{\prime}_{s+r}\alpha^{r},\qquad
j^{\prime}_{r}\rho_{s+r,r}=\alpha^{s}j^{\prime}_{s+r}$

$\qquad\quad
j^{\prime}_{s+r}\psi_{s,s+r}=j^{\prime}_{s},\qquad\rho_{s+r,r}i^{\prime}_{s+r}=i^{\prime}_{r}$.

Proposition 9.8.7 Let $p\geq 5$ and $f\in[\Sigma^{t}K_{r},S]$ be any map, then
$f=jj^{\prime}_{r}\overline{f}$ for some
$\overline{f}\in[\Sigma^{t+rq+2}K_{r},K_{r}]$.

Proof : By Theorem 6.5.16(A) in chapter 6, there is
$\nu_{r}:\Sigma^{rq+2}K_{r}\rightarrow K_{r}\wedge K_{r}$ such that
$(jj^{\prime}_{r}\wedge 1_{K_{r}})\nu_{r}=1_{K_{r}}$. Let $K^{\prime}_{r}$ be
the cofibre of $jj^{\prime}_{r}:\Sigma^{-1}K_{r}\rightarrow\Sigma^{rq+1}S$
given by the cofibration $\Sigma^{-1}K_{r}\stackrel{{\scriptstyle
jj^{\prime}_{r}}}{{\longrightarrow}}\Sigma^{rq+1}S\stackrel{{\scriptstyle
z_{r}}}{{\longrightarrow}}K^{\prime}_{r}\rightarrow K_{r}$, then $z_{r}\wedge
1_{K_{r}}=(z_{r}jj^{\prime}_{r}\wedge 1_{K_{r}})\nu_{r}$ = 0
$\in[\Sigma^{rq+1}K_{r},K^{\prime}_{r}\wedge K_{r}]$. Consequently,
$z_{r}f=(1_{K^{\prime}_{r}}\wedge f)(z_{r}\wedge 1_{K_{r}})$ = 0 and so
$f=jj^{\prime}_{r}\overline{f}$ for some
$\overline{f}\in[\Sigma^{t+rq+1}K_{r},K_{r}]$. Q.E.D.

Let

(9.8.8)$\quad\qquad\cdots\stackrel{{\scriptstyle\tilde{a}_{2}}}{{\longrightarrow}}\quad\Sigma^{-2}\widetilde{E}_{2}\stackrel{{\scriptstyle\tilde{a}_{1}}}{{\longrightarrow}}\quad\Sigma^{-1}\widetilde{E}_{1}\quad\stackrel{{\scriptstyle\tilde{a}_{0}}}{{\longrightarrow}}\qquad\widetilde{E}_{0}=S$

$\quad\qquad\qquad\qquad\qquad\qquad\big{\downarrow}\tilde{b}_{2}\qquad\qquad\quad\big{\downarrow}\tilde{b}_{1}\qquad\qquad\qquad\big{\downarrow}\tilde{b}_{0}$

$\qquad\qquad\qquad\qquad\Sigma^{-2}BP\wedge\widetilde{E}_{2}\qquad\Sigma^{-1}BP\wedge\widetilde{E}_{1}\qquad
BP\wedge\widetilde{E}_{0}=BP$
be the canonnical Adams-Novikov resolution of the sphere spectrum $S$, where
$\widetilde{E}_{s}\stackrel{{\scriptstyle\tilde{b}_{s}}}{{\longrightarrow}}BP\wedge\widetilde{E}_{s}\stackrel{{\scriptstyle\tilde{c}_{s}}}{{\longrightarrow}}\widetilde{E}_{s+1}\stackrel{{\scriptstyle\tilde{a}_{s}}}{{\longrightarrow}}\Sigma\widetilde{E}_{s}$
are cofibrations for all $s\geq 0$ such that
$\widetilde{E}_{0}=S,\tilde{b}_{s}=\tau\wedge 1_{\widetilde{E}_{s}}$( $s>$0)
and $\tilde{b}_{0}=\tau:S\rightarrow BP$ is the injection of the bottom cell.
Then $\pi_{t}BP\wedge\widetilde{E}_{s}$ is the $E_{1}^{s,t}$-term of the
Adams-Novikov spectral sequence,
$(\tilde{b}_{s+1}\tilde{c}_{s})_{*}:\pi_{t}BP\wedge\widetilde{E}_{s}\rightarrow\pi_{t}BP\wedge\widetilde{E}_{s+1}$
are the $d_{1}^{s,t}$-differential and

$E_{2}^{s,t}=Ext_{BP_{*}BP}^{s,t}(BP_{*},BP_{*})\Longrightarrow(\pi_{t-s}S)_{p}$

Proposition 9.8.9 Let $p\geq 3,r\geq 1,s\geq 0$ and $\widetilde{E}_{s}$ be the
spectrum in the Adams-Novikov resolution (9.8.8),
$\wedge^{s}BP=BP\wedge\cdots\wedge BP$ be the smash products of $s$ copies of
$BP$, then
$(BP\wedge\widetilde{E}_{s})^{*},(BP\wedge\widetilde{E}_{s})^{*}(M),(BP\wedge\widetilde{E}_{s})^{*}(K_{r})$
are the direct summand of
$(\wedge^{s+1}BP)^{*},(\wedge^{s+1}BP)^{*}(M),(\wedge^{s+1}BP)^{*}(K_{r})$ and
we have
$[\Sigma^{t}M,BP\wedge\widetilde{E}_{s}]=(BP\wedge\widetilde{E}_{s})^{-t}(M)$
= 0 for $t\neq-1$ (mod q),
$[\Sigma^{t}K_{r},BP\wedge\widetilde{E}_{s}]=(BP\wedge\widetilde{E}_{s})^{-t}(K_{r})$
= 0 for $t\neq-2$ (mod q).

Proof : We first consider the $BP^{*}$-cohomology. It is known that
$\pi_{t}BP=BP_{t}=BP^{-t}$, then $BP^{*}=Z_{(p)}[v_{1},v_{2},\cdots]$ , where
$\mid v_{i}\mid=-2(p^{i}-1)$ and
$I_{n}=(p,v_{1},\cdots,v_{n-1}),(p,v_{1}^{r})$ is the invariant ideal of
$BP^{*}$. Clearly, there are two exact sequences on $BP^{*}$-cohomology as
follows

$\qquad 0\rightarrow BP^{*}\stackrel{{\scriptstyle
p}}{{\longrightarrow}}BP^{*}\stackrel{{\scriptstyle\rho_{0}}}{{\longrightarrow}}BP^{*}/(p)\rightarrow
0$ $\qquad 0\rightarrow\Sigma^{-rq}BP^{*}/(p)\stackrel{{\scriptstyle
v_{1}^{r}}}{{\longrightarrow}}BP^{*}/(p)\stackrel{{\scriptstyle\rho_{1}}}{{\longrightarrow}}BP^{*}/(p,v_{1}^{r})\rightarrow
0$

where $\rho_{0},\rho_{1}$ are the projections.

Note that
$(\wedge^{s}BP)^{*}=\pi_{*}(\wedge^{s}BP)=BP_{*}(\wedge^{s-1}BP)=BP_{*}BP\otimes\cdots\otimes
BP_{*}BP$ with $s-1$ copies of $BP_{*}BP$ and $s\geq 2$. Then we have the
follwing short exact sequences ($s\geq 1$)
(9.8.10) $\qquad 0\rightarrow(\wedge^{s}BP)^{*}\stackrel{{\scriptstyle
p}}{{\longrightarrow}}(\wedge^{s}BP)^{*}\longrightarrow(\wedge^{s}BP)^{*}/(p)\rightarrow
0$

$\quad 0\rightarrow\Sigma^{-rq}(\wedge^{s}BP)^{*}/(p)\stackrel{{\scriptstyle
v_{1}^{r}}}{{\longrightarrow}}(\wedge^{s}BP)^{*}/(p)\longrightarrow(\wedge^{s}BP)^{*}/(p,v_{1}^{r})\rightarrow
0$

For any $f\in[\Sigma^{t}M,\wedge^{s}BP]=(\wedge^{s}BP)^{-t}(M)$, if $t\neq
0$(mod q) , then by the sparseness of
$(\wedge^{s}BP)^{*}=BP_{*}BP\otimes\cdots\otimes BP_{*}BP$ ( that is,
$BP_{r}BP$ = 0 for $r\neq 0$ (mod $q$)) we have $fi\in(\wedge^{s}BP)^{-t}$ = 0
; if $t=0$ (mod q), then $fi$ is an element of order $p$ in $Z_{(p)}$-module
$(\wedge^{s}BP)^{-t}$ so that we have $fi$ = 0. This shows that
$i^{*}=0:(\wedge^{s}BP)^{*}(M)\rightarrow(\wedge^{s}BP)^{*}$. Similarly we
have
$(i^{\prime}_{r})^{*}=0:(\wedge^{s}BP)^{*}(K_{r})\rightarrow(\wedge^{s}BP)^{*}(M)$
( $r\geq 1$). Then, the cofibration (9.1.1) (9.8.2) induces respectively the
following short exact sequences for all $(r\geq 1)$

$\qquad 0\rightarrow(\wedge^{s}BP)^{*}\stackrel{{\scriptstyle
p}}{{\longrightarrow}}(\wedge^{s}BP)^{*}\stackrel{{\scriptstyle
j^{*}}}{{\longrightarrow}}(\wedge^{s}BP)^{*}(M)\rightarrow 0$ $\qquad
0\rightarrow(\wedge^{s}BP)^{*}(M)\stackrel{{\scriptstyle(\alpha^{r})^{*}}}{{\longrightarrow}}(\wedge^{s}BP)^{*}(M)\stackrel{{\scriptstyle(j^{\prime}_{r})^{*}}}{{\longrightarrow}}(\wedge^{s}BP)^{*}(K_{r})\rightarrow
0$

where the degrees $\mid j^{*}\mid=-1$, $\mid(j^{\prime}_{r})^{*}\mid=-(rq+1)$.
By comparison to the above two short exact sequences with (9.8.10) we have
(9.8.11) $\qquad(\wedge^{s}BP)^{*}(M)\cong\Sigma(\wedge^{s}BP)^{*}/(p),$

$\qquad\quad(\wedge^{s}BP)^{*}(K_{r})\cong\Sigma^{rq+2}(\wedge^{s}BP)^{*}/(p,v_{1}^{r})$.

Let $\mu:BP\wedge BP\rightarrow BP$ be the multiplication of the ring spectrum
$BP$ and $\tau:S\rightarrow BP$ be the injection of the bottom cell, , then we
have $\mu(1_{BP}\wedge\tau)=1_{BP}=\mu(\tau\wedge 1_{BP})$ so that the
cofibration
$\widetilde{E}_{s-1}\stackrel{{\scriptstyle\tilde{b}_{s-1}}}{{\longrightarrow}}BP\wedge\widetilde{E}_{s-1}\stackrel{{\scriptstyle\tilde{c}_{s-1}}}{{\longrightarrow}}\widetilde{E}_{s}\stackrel{{\scriptstyle\tilde{a}_{s-1}}}{{\longrightarrow}}\Sigma\widetilde{E}_{s-1}$
induces a split short exact sequence

$BP\wedge\widetilde{E}_{s-1}\stackrel{{\scriptstyle
1_{BP}\wedge\tilde{b}_{s-1}}}{{\longrightarrow}}BP\wedge
BP\wedge\widetilde{E}_{s-1}\stackrel{{\scriptstyle
1_{BP}\wedge\tilde{c}_{s-1}}}{{\longrightarrow}}BP\wedge\widetilde{E}_{s}$

this is because $(\mu\wedge
1_{\widetilde{E}_{s-1}})(1_{BP}\wedge\tilde{b}_{s-1})=(\mu\wedge
1_{\widetilde{E}_{s-1}})(1_{BP}\wedge\tau\wedge
1_{\widetilde{E}_{s-1}})=1_{BP\wedge\widetilde{E}_{s-1}}$. That is to say,
$BP\wedge\widetilde{E}_{s}$ is the direct summand of $BP\wedge
BP\wedge\widetilde{E}_{s-1}$ and by induction we have
$BP\wedge\widetilde{E}_{s}$ is the direct summand of $\wedge^{s+1}BP$. Hence,
$(BP\wedge\widetilde{E}_{s})^{*},(BP\wedge\widetilde{E}_{s})^{*}(M),(BP\wedge\widetilde{E}_{s})^{*}(K_{r})$
are the direct summand of
$(\wedge^{s+1}BP)^{*},(\wedge^{s+1}BP)^{*}(M),(\wedge^{s+1}BP)^{*}(K_{r})$
respectively and the last result can be obtained by (9.8.11). Q.E.D.

Proposition 9.8.12 Let $p\geq 3,n\geq 1$, then

$\qquad\qquad
Ext_{BP_{*}BP}^{0,p^{n}(p+1)q}(BP_{*},BP_{*}/(p,v_{1}^{p^{n}-1}))$
is generated additively by the generators
$v_{2}^{p^{n}},v_{1}^{p^{n}-p^{n-2r}}\tilde{c}_{1}(t_{r}p^{n-2r})$ ( $r\geq
1$), where $t_{r}=(p^{2r+1}+1)/(p+1)$ and $\tilde{c}_{1}(ap^{s})$ is the
generator in Theorem 8.1.7 in chapter 8 which has degree $sp^{s}(p+1)q$.

Proof : By Theorem 8.1.7, the desired generators are of the form
$v_{1}^{b}\tilde{c}_{1}(ap^{s})$ with degrees $bq+ap^{s}(p+1)q=p^{n}(p+1)q$,
$a\geq 1$ is not divisible by $p$, $0\leq b<p^{n}-1$ and $b\geq$ max
$\\{0,p^{n}-1-q_{1}(ap^{s})\\}$, where $q_{1}(ap^{s})=p^{s}$ if $a=1$,
$q_{1}(ap^{s})=p^{s}+p^{s-1}-1$ if $a\geq 2$ is not divisible by $p$.

If $b=0$, then the generator is $v_{2}^{p^{n}}$. Since $b<p^{n}-1$, then $s<n$
and $b\equiv 0$ ( mod $p^{s}$) and so $b\geq p^{n}-1-q_{1}(ap^{s})\geq
p^{n}-p^{n-1}$ if $b\geq 1$. Let
$b=(p-1)p^{n-1}+c_{n-2}p^{n-2}+\cdots+c_{s}p^{s}$ be the p-adic expasion of
$b$ such that $0\leq c_{i}\leq p-1$. By $b\geq(p^{n}-1)-(p^{s}+p^{s-1}-1)$ or
$b\geq(p^{n}-1)-p^{s}$ we have $c_{n-2}p^{n-2}+\cdots+c_{s}p^{s}\geq
p^{n-1}-p^{s}-p^{s-1}$ or $p^{n-1}-p^{s}-1$. Consequently we have
$c_{n-2}=\cdots=c_{s}=p-1$. On the other hand, $b$ is divisible by $p+1$, then
$(p-1)-c_{n-2}+\cdots+(-1)^{n-1-s}c_{s}$ = 0 so that $n-1-s$ must be odd. Let
$n-1-s=2r-1$, then we have $s=n-2r,b=p^{n}-p^{n-2r}$ as desired and
$a=(p^{2r+1}+1)/(p+1)$. Q.E.D.

Proposition 9.8.13 Let $p\geq 3,n\geq 1$, then

(1) $Ext_{BP_{*}BP}^{2,p^{n+1}q+q}(BP_{*},BP_{*})$ is generated additively by
the generators
$\beta_{p^{n}/p^{n}-1},\beta_{t_{r}p^{n-2r}/p^{n-2r}-1}$ for all $r\geq 1$),
where $t_{r}=(p^{2r+1}+1)/(p+1)$.

(2) $Ext_{BP_{*}BP}^{1,p^{n+1}q+q}(BP_{*},BP_{*}M)$ is generated additively by
the generators
$\beta^{\prime}_{p^{n}/p^{n}-1},\beta^{\prime}_{t_{r}p^{n-2r}/p^{n-2r}-1}$ for
all $r\geq 1$), where $t_{r}=(p^{2r+1}+1)/(p+1)$ and
$\beta^{\prime}_{tp^{n}/s}$ is the generator in
$Ext_{BP_{*}BP}^{1,*}(BP_{*},BP_{*}M)$ such that
$j_{*}(\beta^{\prime}_{tp^{n}/s})=\beta_{tp^{n}/s}\in
Ext_{BP_{*}BP}^{2,*}(BP_{*},BP_{*})$.

Proof : By Theorem 8.1.3 in chapter 8, $Ext_{BP_{*}BP}^{2,*}(BP_{*},BP_{*})$
is generated additively by the generators $\beta_{ap^{s}/b,c+1}\in
Ext_{BP_{*}BP}^{2,ap^{s}(p+1)q-bq}(BP_{*},BP_{*})$ , where $s\geq 0$, $a\geq
1$ is not divisible by $p$, $b\geq 1,c\geq 0$ and subject to

(i) $b\leq s$ if $a$ = 1.

(ii) $p^{c}\mid b\leq p^{s-c}+p^{s-c-1}-1$

(iii) $p^{s-c-1}+p^{s-c-2}-1<b$ if $p^{c+1}\mid b$,
and $\beta_{ap^{s}/b,1}=\beta_{ap^{s}/b}$. Then, for $\beta_{ap^{s}/b,c+1}\in
Ext_{BP_{*}BP}^{2,p^{n+1}q+q}(BP_{*},BP_{*})$ we have
$ap^{s}(p+1)q-bq=p^{n+1}q+q=p^{n}(p+1)q-(p^{n}-1)q$ so that
$ap^{s}(p+1)q+(p^{n}-1-b)q=p^{n}(p+1)q$. Similar to that in the proof of Prop.
9.8.12 we have $a=1,s=n,b=p^{n}-1$ or
$a=(p^{2r+1}+1)/(p+1),b=p^{n-2r}-1,s=p^{n-2r}$( $r\geq 1$) and consequently
$c=0$. This shows (1) and the proof of (2) is similar. Q.E.D.

After finishing the proof of the above Proposition, we proceed to prove
Theorem 9.8.1 in case $t\geq 1$. The proof will be done by some argument
processing in the Adams-Novikov resolution of some spectra and using the
$h_{0}h_{n+1}$-map in Theorem 9.5.1 as our geometric input. We first prove the
following Lemma.

Lemma 9.8.14 If $g^{\prime\prime}$ is the element in $\pi_{p^{n}(p+1)q}BP$
such that $\tilde{b}_{1}\tilde{c}_{0}g^{\prime\prime}j\\\
j^{\prime}_{p^{n}-1}=0\in[\Sigma^{p^{n+1}q+q-2}K_{p^{n}-1},BP\wedge E_{1}]$,
then there exists $\bar{g}=px_{1}+v_{1}^{p^{n}-1}x_{2}\in\pi_{p^{n}(p+1)q}BP$
such that
$\tilde{b}_{1}\tilde{c}_{0}(g^{\prime\prime}-\bar{g})=0\in\pi_{p^{n}(p+1)q}BP\wedge\widetilde{E}_{1}$,
where $x_{1},x_{2}$ is some elements in $\pi_{*}BP$.

Proof : Let $\mu:BP\wedge BP\rightarrow BP$ be the multiplication of the ring
spectrum $BP$, then $\mu(\tilde{b}_{0}\wedge
1_{BP})=1_{BP}=\mu(1_{BP}\wedge\tilde{b}_{0})$, where
$\tilde{b}_{0}=\tau:S\rightarrow BP$ is the injection of the bottom cell as
stated above. Then we have the following split cofibration
$\quad BP\stackrel{{\scriptstyle
1_{BP}\wedge\tilde{b}_{0}}}{{\longrightarrow}}BP\wedge
BP\stackrel{{\scriptstyle
1_{BP}\wedge\tilde{c}_{0}}}{{\longrightarrow}}BP\wedge\widetilde{E}_{1}\stackrel{{\scriptstyle
1_{BP}\wedge\tilde{a}_{0}=0}}{{\longrightarrow}}\Sigma BP$

$BP\wedge\widetilde{E}_{1}\stackrel{{\scriptstyle
1_{BP}\wedge\tilde{b}_{0}\wedge
1_{\widetilde{E}_{1}}}}{{\longrightarrow}}BP\wedge
BP\wedge\widetilde{E}_{1}\stackrel{{\scriptstyle
1_{BP}\wedge\tilde{c}_{1}}}{{\longrightarrow}}BP\wedge\widetilde{E}_{2}\stackrel{{\scriptstyle
1_{BP}\wedge\tilde{a}_{0}=0}}{{\longrightarrow}}\Sigma
BP\wedge\widetilde{E}_{1}$
and there is $\mu^{\prime}:BP\wedge\widetilde{E}_{1}\rightarrow BP\wedge BP$
such that
$(1_{BP}\wedge\tilde{b}_{0})\mu+\mu^{\prime}(1_{BP}\wedge\tilde{c}_{0})=1_{BP\wedge
BP}$.

By $\tilde{b}_{1}\tilde{c}_{0}g^{\prime\prime}jj^{\prime}_{p^{n}-1}$ = 0 we
have
$\tilde{c}_{0}g^{\prime\prime}jj^{\prime}_{p^{n}-1}=\tilde{a}_{1}g^{\prime}$
with $g^{\prime}\in$
$[\Sigma^{p^{n+1}q+q-1}K_{p^{n}-1},\widetilde{E}_{2}]$. Then
$(1_{BP}\wedge\tilde{c}_{0}g^{\prime\prime}jj^{\prime}_{p^{n}-1})=(1_{BP}\wedge\tilde{a}_{1})(1_{BP}\wedge
g^{\prime})$ = 0 so that$1_{BP}\wedge
g^{\prime\prime}jj^{\prime}_{p^{n}-1}=[(1_{BP}\wedge\tilde{b}_{0})\mu+\mu^{\prime}(1_{BP}\wedge\tilde{c}_{0})](1_{BP}\wedge
g^{\prime\prime}jj^{\prime}_{p^{n}-1})$ =
$(1_{BP}\wedge\tilde{b}_{0})\mu(1_{BP}\wedge
g^{\prime\prime}jj^{\prime}_{p^{n}-1})$ and we have
(9.8.15) $\qquad(\tilde{b}_{0}\wedge
1_{BP})g^{\prime\prime}jj^{\prime}_{p^{n}-1}=(1_{BP}\wedge
g^{\prime\prime}jj^{\prime}_{p^{n}-1})(\tilde{b}_{0}\wedge 1_{K_{p^{n}-1}})$

$\qquad\qquad\quad=(1_{BP}\wedge\tilde{b}_{0})\mu(1_{BP}\wedge
g^{\prime\prime}jj^{\prime}_{p^{n}-1})(\tilde{b}_{0}\wedge 1_{K_{p^{n}-1}})$

$\qquad\qquad\quad=(1_{BP}\wedge\tilde{b}_{0})g^{\prime\prime}jj^{\prime}_{p^{n}-1}$.

Note that $(\tilde{b}_{0}\wedge
1_{BP})_{*},(1_{BP}\wedge\tilde{b}_{0})_{*}:BP_{*}\rightarrow BP_{*}BP$ are
the right and left unit $\eta_{R},\eta_{L}:BP_{*}\rightarrow BP_{*}BP$
respectively, then by (9.8.15) we have
$\eta_{R}(g^{\prime\prime})=\eta_{L}(g^{\prime\prime})$ mod
$(p,v_{1}^{p^{n}-1})$. This means that $(p,v_{1}^{p^{n}-1},g^{\prime\prime})$
is a $BP_{*}$ invariant ideal , or equivalently, $g^{\prime\prime}\in
Ext_{BP_{*}BP}^{0.p^{n}(p+1)q}(BP_{*},BP_{*}/(p,v_{1}^{p^{n}-1}))$. Then by
Prop. 9.8.12 we have
(9.8.16)$\qquad g^{\prime\prime}=\lambda
v_{2}^{p^{n}}+\Sigma\lambda_{r}v_{1}^{p^{n}-p^{n-2r}}\tilde{c}_{1}(t_{r}p^{n-2r})+px_{1}+v_{1}^{p^{n}-1}x_{2}\in
BP_{*}$
where $1\leq\lambda,\lambda_{r}\leq p-1,t_{r}=(p^{2r+1}+1)/(p+1)$ and
$x_{1},x_{2}$ are some elements in $BP_{*}$.

Let $\bar{g}=px_{1}+v_{1}^{p^{n}-1}x_{2}$, then $(\tilde{b}_{0}\wedge
1_{BP})(g^{\prime\prime}-\bar{g})=\eta_{R}(g^{\prime\prime}-\bar{g})=\eta_{L}(g^{\prime\prime}-\bar{g})=(1_{BP}\wedge\tilde{b}_{0})(g^{\prime\prime}-\bar{g})$
so that
$\tilde{b}_{1}\tilde{c}_{0}(g^{\prime\prime}-\bar{g})=(1_{BP}\wedge\tilde{c}_{0})(\tilde{b}_{0}\wedge
1_{BP})(g^{\prime\prime}-\bar{g})$ = 0. Q.E.D.

Proof of Theorem 9.8.1 in case $t\geq 1$ By Theorem 9.5.1, there is
$\tilde{\eta}_{n+1}\in\pi_{p^{n+1}q+q-1}M$ such that
$\eta_{n+1}=j\tilde{\eta}_{n+1}\in\pi_{p^{n+1}q+q-2}S$ is represented in the
ASS by $h_{0}h_{n+1}\in Ext_{A}^{2,p^{n+1}q+q}(Z_{p},Z_{p})$. By Theorem 8.1.5
in chapter 8, $\Phi(\beta_{p^{n}/p^{n}-1})=h_{0}h_{n+1}$, where
$\Phi:Ext_{BP_{*}BP}^{2,*}(BP_{*},BP_{*})\rightarrow
Ext_{A}^{2,*}(Z_{p},Z_{p})$ is the Thom map. Then
$\eta_{n+1}=j\tilde{\eta}_{n+1}$ is represented by $\beta_{p^{n}/p^{n}-1}+x\in
Ext_{BP_{*}BP}^{2,p^{n+1}q+q}(BP_{*},BP_{*})$ in the Adams-Novikov spectral
sequence, where $x=\Sigma_{r\geq
1}\lambda_{r}\beta_{t_{r}p^{n-2r}/p^{n-2r}-1}$ with $\lambda_{r}\in Z_{(p)}$ (
cf. Prop. 9.8.13). Moreover, $\tilde{\eta}_{n+1}\in\pi_{p^{n+1}q+q-1}M$ is
represented by $\beta^{\prime}_{p^{n}/p^{n}-1}+x^{\prime}+i_{*}(y)$ in the
Adams spectral sequence, where $y\in Ext_{BP_{*}BP}^{1,*}(BP_{*},BP_{*})$, and
$x^{\prime}=\Sigma_{r\geq
1}\lambda_{r}\beta^{\prime}_{t_{r}p^{n-2r}/p^{n-2r}-1}$ ,
$\beta^{\prime}_{tp^{n}/s}$ are the elements in
$Ext_{BP_{*}BP}^{1,*}(BP_{*},BP_{*}(M))$ such that
$j_{*}\beta^{\prime}_{tp^{n}/s}=\beta_{tp^{n}/s}\in
Ext_{BP_{*}BP}^{2,*}(BP_{*},BP_{*})$ . It is known that all the generators in
$Ext_{BP_{*}BP}^{1,*}(BP_{*},BP_{*})$ are permanent cycles in the Adams-
Novikov spectral sequence , then there exists
$\tilde{f}\in\pi_{p^{n+1}q+q-1}M$ such that it is represented by
$\beta^{\prime}_{p^{n}/P^{n}-1}+x^{\prime}$. In addition, $\tilde{f}$ can be
extended by $f\in[\Sigma^{p^{n+1}q+q-1}M,M]\cap kerd$ such that
$\tilde{f}=fi$. Recall from (9.8.8)

$\qquad\cdots\stackrel{{\scriptstyle\tilde{a}_{2}\wedge
1_{M}}}{{\longrightarrow}}\Sigma^{-2}\widetilde{E}_{2}\wedge
M\stackrel{{\scriptstyle\tilde{a}_{1}\wedge
1_{M}}}{{\longrightarrow}}\Sigma^{-1}\widetilde{E}_{1}\wedge
M\stackrel{{\scriptstyle\tilde{a}_{0}\wedge
1_{M}}}{{\longrightarrow}}\widetilde{E}_{0}\wedge M=M$

$\qquad\quad\qquad\qquad\qquad\Big{\downarrow}\tilde{b}_{2}\wedge
1_{M}\qquad\qquad\Big{\downarrow}\tilde{b}_{1}\wedge
1_{M}\quad\qquad\Big{\downarrow}\tilde{b}_{0}\wedge 1_{M}$

$\qquad\qquad\Sigma^{-2}BP\wedge\widetilde{E}_{2}\wedge
M\qquad\Sigma^{-1}BP\wedge\widetilde{E}_{1}\wedge M\qquad BP\wedge M$
is the Adams-Novikov resolution of the Moore spectrum $M$. Then $fi$ can be
lifted to $f_{1}i\in\pi_{p^{n+1}q+q}(\widetilde{E}_{1}\wedge M)$ with
$f_{1}\in[\Sigma^{p^{n+1}q+q}M,E_{1}\wedge M]\cap kerd$ such that
$\tilde{a}_{0}\wedge 1_{M})f_{1}i=fi$ and the $d_{1}$-cycle
$(\tilde{b}_{1}\wedge
1_{M})f_{1}i\in\pi_{p^{n+1}q+q}BP\wedge\widetilde{E}_{1}\wedge M$ represents
$\beta^{\prime}_{p^{n}/p^{n}-1}+x^{\prime}\in
Ext_{BP_{*}BP}^{1,p^{n+1}q+q}(BP_{*},BP_{*}(M))$. By applying $d$ to the
equation $(\tilde{a}_{0}\wedge 1_{M})f_{1}ij=fij$ we have
$(\tilde{a}_{0}\wedge 1_{M})f_{1}=f$.

Since $(\tilde{a}_{0}\wedge 1_{M})f_{1}i=fi\in\pi_{p^{n+1}q+q-1}M$ is
represented by $\beta^{\prime}_{p^{n}/p^{n}-1}+x^{\prime}\in
Ext_{BP_{*}BP}^{1,p^{n+1}q+q}(BP_{*},BP_{*}(M))$ in the Adams-Novikov spectral
sequence and $v_{1}^{p^{n}-1}(\beta^{\prime}_{p^{n}/p^{n}-1}+x^{\prime})$ = 0,
then $(\tilde{a}_{0}\wedge
1_{M})f_{1}\alpha^{p^{n}-1}i=f\alpha^{p^{n}-1}i=\alpha^{p^{n}-1}fi$ has $BP$\-
filtration $>1$ so that $(\tilde{b}_{1}\wedge 1_{M})f_{1}\alpha^{p^{n}-1}i$ is
a $d_{1}$-boundary and it equals to $(\tilde{b}_{1}\tilde{c}_{0}\wedge
1_{M})gi$ for some $g\in[\Sigma^{p^{n}(p+1)q}M,BP\wedge M]$. Hence,
$(\tilde{b}_{1}\wedge
1_{M})f_{1}\alpha^{p^{n}-1}=(\tilde{b}_{1}\tilde{c}_{0}\wedge 1_{M})g$ , this
is because $\pi_{p^{n}(p+1)q+1}BP\wedge\widetilde{E}_{1}\wedge M$ = 0 which is
obtained by the sparseness fo $BP_{*}(\widetilde{E}_{1}\wedge M)$.
Consequently we have

$f_{1}\alpha^{p^{n}-1}=(\tilde{c}_{0}\wedge 1_{M})g+(\tilde{a}_{1}\wedge
1_{M})f_{2}$ with $f_{2}\in[\Sigma^{p^{n}(p+1)q+1}M,\widetilde{E}_{2}\wedge
M]$
and

$(1_{\widetilde{E}_{1}}\wedge
j)f_{1}\alpha^{p^{n}-1}=\tilde{c}_{0}g^{\prime\prime}j+\tilde{a}_{1}(1_{\widetilde{E}_{2}}\wedge
j)f_{2}$ with $g^{\prime\prime}\in\pi_{p^{n}(p+1)q}BP$,
where $g^{\prime\prime}j=(1_{BP}\wedge j)g$, this is because $(1_{BP}\wedge
j)gi\in\pi_{p^{n}(p+1)q-1}BP$ = 0. In addition, by
$\tilde{b}_{2}(1_{\widetilde{E}_{2}}\wedge
j)f_{2}\in[\Sigma^{p^{n}(p+1)q}M,BP\wedge\widetilde{E}_{2}]$= 0 and
$[\Sigma^{p^{n}(p+1)q+1}M,BP\\\ \wedge\widetilde{E}_{3}]$ = 0 ( cf. Prop.
9.8.9), then $(1_{\widetilde{E}_{2}}\wedge
j)f_{2}=\tilde{a}_{2}\tilde{a}_{3}f_{3}$ for some
$f_{3}\in[\Sigma^{p^{n}(p+1)q+2}M,E_{4}]$ and we have
(9.8.17) $\qquad(1_{E_{1}}\wedge
j)f_{1}\alpha^{p^{n}-1}=\bar{c}_{0}g^{\prime\prime}j+\tilde{a}_{1}\tilde{a}_{2}\tilde{a}_{3}f_{3}$.

By (9.8.17) we have
$\tilde{b}_{1}\tilde{c}_{0}g^{\prime\prime}jj^{\prime}_{p^{n}-1}$ = 0, then by
Lemma 9.8.14, there is
$\bar{g}=px_{1}+v_{1}^{p^{n}-1}x_{2}\in\pi_{p^{n}(p+1)q}BP$ with
$x_{1},x_{2}\in\pi_{*}BP$ such that
$\tilde{b}_{1}\tilde{c}_{0}(g^{\prime\prime}-\bar{g})$ = 0. Consequently,
$\tilde{c}_{0}(g^{\prime\prime}-\bar{g})=\tilde{a}_{1}f_{4}$ for some
$f_{4}\in\pi_{p^{n}(p+1)q+1}\widetilde{E}_{2}$ and
$f_{4}=\tilde{a}_{2}\tilde{a}_{3}f_{5}$ with
$f_{5}\in\pi_{p^{n}(p+1)q+3}\widetilde{E}_{4}$ which is obtained by the
sparseness of $\pi_{*}BP\wedge\widetilde{E}_{s}$. So, (9.8.17) becomes
(9.8.18) $\qquad(1_{\widetilde{E}_{1}}\wedge
j)f_{1}\alpha^{p^{n}-1}=\tilde{c}_{0}\bar{g}j+\tilde{a}_{1}\tilde{a}_{2}\tilde{a}_{3}f_{5}j+\tilde{a}_{1}\tilde{a}_{2}\tilde{a}_{3}f_{3}$.
Note that $\bar{g}=px_{1}+v_{1}^{p^{n}-1}x_{2}$, then ,
$\bar{g}jj^{\prime}_{p^{n}-1}$ = 0 $\in
BP^{*}(K_{p^{n}-1})\cong\Sigma^{-(p^{n}-1)q-2}\\\ BP^{*}/(p,v_{1}^{p^{n}-1})$
so that $\bar{g}j=\tilde{g}\alpha^{p^{n}-1}$ for some
$\tilde{g}\in[\Sigma^{p^{n+1}q+q-1}M,BP]$. Consequently, by (9.8.4), the
equation (9.8.18) becomes
(9.8.19) $\qquad(1_{\widetilde{E}_{1}}\wedge
j)f_{1}j^{\prime}_{1}\rho_{p^{n},1}=(1_{\widetilde{E}_{1}}\wedge
j)f_{1}\alpha^{p^{n}-1}j^{\prime}_{p^{n}}$

$\qquad\qquad=\tilde{c}_{0}\tilde{g}\alpha^{p^{n}-1}j^{\prime}_{p^{n}}+\tilde{a}_{1}\tilde{a}_{2}\tilde{a}_{3}(f_{5}j+f_{3})j^{\prime}_{p^{n}}$

$\qquad\qquad=\tilde{c}_{0}\tilde{g}j^{\prime}_{1}\rho_{p^{n},1}+\tilde{a}_{1}\tilde{a}_{2}\tilde{a}_{3}(f_{5}j+f_{3})j^{\prime}_{p^{n}}$
Moreover, by (9.8.4)(9.8.6) we have
$\tilde{a}_{1}\tilde{a}_{2}\tilde{a}_{3}(f_{5}j+f_{3})j^{\prime}_{p^{n}-1}=\tilde{a}_{1}\tilde{a}_{2}\tilde{a}_{3}(f_{5}j+f_{3})j^{\prime}_{p^{n}}\psi_{p^{n}-1,p^{n}}$
= 0 and so $(f_{5}j+f_{3})j^{\prime}_{p^{n}-1}$ = 0 , this is because
$[\Sigma^{p^{n+1}q+q+r}\\\ K_{p^{n}-1},BP\wedge\widetilde{E}_{2+r}]$ = 0 for
$r=-1,0,1$( cf. Prop. 9.8.9). This shows that
$(f_{5}j+f_{3})=f_{6}\alpha^{p^{n}-1}$ with
$f_{6}\in[\Sigma^{p^{n+1}q+q+2}M,\widetilde{E}_{4}]$. Hence, the equation
(9.8.19) becomes
(9.8.20) $\qquad(1_{\widetilde{E}_{1}}\wedge
j)f_{1}j^{\prime}_{1}\rho_{p^{n},1}=\tilde{c}_{0}\tilde{g}j^{\prime}_{1}\rho_{p^{n},1}+\tilde{a}_{1}\tilde{a}_{2}\tilde{a}_{2}f_{6}\alpha^{p^{n}-1}j^{\prime}_{p^{n}}$

$\qquad\qquad\qquad=\tilde{c}_{0}\tilde{g}j^{\prime}_{1}\rho_{p^{n},1}+\tilde{a}_{1}\tilde{a}_{2}\tilde{a}_{3}f_{6}j^{\prime}_{1}\rho_{p^{n},1}$
and by (9.8.6) we have
(9.8.21) $\qquad(1_{\widetilde{E}_{1}}\wedge
j)f_{1}j^{\prime}_{1}=\tilde{c}_{0}\tilde{g}j^{\prime}_{1}+\tilde{a}_{1}\tilde{a}_{2}\tilde{a}_{3}f_{6}j^{\prime}_{1}+\epsilon
i^{\prime}_{p^{n}-1}j^{\prime}_{1}$
for some $\epsilon\in[\Sigma^{p^{n+1}q+q-1}K_{p^{n}-1},\widetilde{E}_{1}]$. By
composing $\tilde{a}_{0}$ to (9.8.21) we have
$jfj^{\prime}_{1}=\tilde{a}_{0}\tilde{a}_{1}\tilde{a}_{2}\tilde{a}_{3}f_{6}j^{\prime}_{1}+\tilde{a}_{0}\epsilon
i^{\prime}_{p^{n}-1}j^{\prime}_{1}=\tilde{a}_{0}\tilde{a}_{1}\tilde{a}_{2}\tilde{a}_{3}f_{6}j^{\prime}_{1}+jj^{\prime}_{p^{n}-1}\overline{\epsilon}i^{\prime}_{p^{n}-1}j^{\prime}_{1}$
, this is because
$\tilde{a}_{0}\epsilon=jj^{\prime}_{p^{n}-1}\overline{\epsilon}$ for some
$\overline{\epsilon}\in[\Sigma^{p^{n}(p+1)q}K_{p^{n}-1},K_{p^{n}-1}]$ (cf.
Prop. 9.8.2). Consequently we have
(9.8.22) $\qquad
jfi=\tilde{a}_{0}\tilde{a}_{1}\tilde{a}_{2}\tilde{a}_{3}f_{6}i+jj^{\prime}_{p^{n}-1}\overline{\epsilon}i^{\prime}_{p^{n}-1}i+f_{7}\alpha
i$
with $f_{7}\in[\Sigma^{p^{n+1}q-2}M,S]$.

We claim that the map $f_{7}\alpha i$ in (9.8.22) has filtration $\geq 3$ so
that by (9.8.22) we obtain that
$jj^{\prime}_{p^{n}-1}\tilde{\epsilon}i^{\prime}_{p^{n}-1}i\in\pi_{p^{n+1}q+q-2}S$
is represented by $h_{0}h_{n+1}\in$ $Ext_{A}^{2,p^{n+1}q+q}(Z_{p},Z_{p})$ in
the Adams spectral sequence. This claim will be proved in the last.

Then, $jj^{\prime}_{p^{n}-1}\tilde{\epsilon}i^{\prime}_{p^{n}-1}i$ is
represented by $\lambda_{0}\beta_{p^{n}/p^{n}-1}+\Sigma_{r\geq
1}\lambda_{r}\beta_{t_{r}p^{n-2r}/p^{n-2r}-1}\\\ \in
Ext_{BP_{*}BP}^{2,*}(BP_{*},BP_{*})$ in the Adams-Novikov spectral sequence so
that by Prop. 9.8.12 we know that
$\tilde{\epsilon}i^{\prime}_{p^{n}-1}i\in\pi_{p^{n}(p+1)q}K_{p^{n}-1}$ is
represented by $v_{2}^{p^{n}}+\Sigma_{r\geq
1}\lambda_{r}v_{1}^{p^{n}-p^{n-2r}}\tilde{c}_{1}(t_{r}p^{n-2r})\in
Ext_{BP_{*}BP}^{0,p^{n}(p+1)q}(BP_{*},BP_{*}K_{p^{n}-1})$, where
$\lambda_{r}\in Z_{p}$ and $t_{r}=(p^{2r+2}+1)/(p+1)$.

By [22] Theorem C,D, it is known that $v_{2}^{tp}\in
Ext_{BP_{*}BP}^{0,*}(BP_{*},BP_{*}K_{r})$ , for $t\geq 1,1\leq r\leq p-1$, is
a permanent cyce in the Adams-Novikov spectral sequence. Suppose inductively
that $v_{2}^{tp^{s}}\in Ext_{BP_{*}BP}^{0,*}(BP_{*},BP_{*}K_{r})$ (for $t\geq
1,1\leq r\leq p^{s}-1$ and $s\leq n-1$) is a permanent cycle in the Adams-
Novikov spectral sequence, then we know that
$v_{1}^{p^{n}-p^{n-2r}}\tilde{c}_{1}(t_{r}p^{n-2r})\in
Ext_{BP_{*}BP}^{0,p^{n}(p+1)q}\\\ (BP_{*},BP_{*}K_{p^{n}-1})$ also is a
permanent cycle for all $r\geq 1$. Moreover, by the representation of the
above $\tilde{\epsilon}i^{\prime}_{p^{n}-1}i$ we obtain that $v_{2}^{p^{n}}\in
Ext_{BP_{*}BP}^{0,p^{n}(p+1)q}(BP_{*},\\\ BP_{*}K_{p^{n}-1})$ is a permanent
cycle. Hence, by (9.8.6), there exists
$k\in[\Sigma^{p^{n}(p+1)q}K_{p^{n}-1},K_{p^{n}-1}]$ such that the induced
$BP_{*}$-homomorphism $k_{*}=v_{2}^{p^{n}}$. In addition, the map
$\rho_{p^{n}-1,r}:K_{p^{n}-1}\rightarrow K_{r}$ in (9.8.4) for all $r\leq
p^{n}-1$ is a projection, then
$\rho_{p^{n}-1,r}k^{t}i^{\prime}_{p^{n}-1}i\in\pi_{tp^{n}(p+1)q}K_{r}$ is
represented by $v_{2}^{tp^{n}}\in Ext_{BP_{*}BP}^{0,*}(BP_{*},BP_{*}K_{r})$ in
the Adams-Novikov spectral sequence. This completes the induction and
$jj^{\prime}_{r}\rho_{p^{n}-1,r}k^{t}i^{\prime}_{p^{n}-1}i\in\pi_{*}S$ is just
the $\beta_{tp^{n}/r}$-element of the Theorem.

Now our remaining work is to prove the above claim. We turn to an argument in
the ASS and let $A$ be the mod $p$ Steenrod algebra. By
$Ext_{A}^{1,p^{n+1}q-1}(Z_{p},H^{*}M)\cong Z_{p}\\{j^{*}(h_{n+1})\\}$ and the
result on $\beta_{p^{n}/p^{n}}\in Ext_{BP_{*}BP}^{2,p^{n+1}q}\\\
(BP_{*},BP_{*})$ support a nontrivial differential in the Adams-Novikov
spectral sequence in [12] p.106 Thoerem 5.4.8(i), we know that
(9.8.23) $\qquad j^{*}(h_{n+1})\in Ext_{A}^{1,p^{n+1}q-1}(Z_{p},H^{*}M)$ dies
in the ASS
Then, the map $f_{7}\in[\Sigma^{p^{n+1}q-2}M,S]$ in (9.8.22) has filtration
$\geq 2$ in the ASS and so $f_{7}\alpha i$ has filtration $\geq 3$. Moreover,
by (9.8.21) we know that
$jj^{\prime}_{p^{n}-1}\overline{\epsilon}i^{\prime}_{p^{n}-1}i$ and
$jfi\in\pi_{p^{n+1}q+q-2}S$ must have the same filtration so that it is
repesented by $h_{0}h_{n+1}\in Ext_{A}^{2,p^{n+1}q+q}(Z_{p},Z_{p})$ in the
ASS. This shows the above claim and the Theorem is proved. Q.E.D.

Remark 9.8.24 We give a detail proof of the result in (9.8.23) as follows. It
will be done by some argument processing in the Adams resolution (9.2.9).
Suppose in contrast that the map $j^{*}h_{n+1}\in
Ext_{A}^{1,p^{n+1}q-1}(Z_{p},H^{*}M)$ is a permanent cycle in the ASS, then we
have $\bar{c}_{1}h_{n+1}\cdot j$ = 0 , where
$h_{n+1}\in\pi_{p^{n+1}q}KG_{1}\cong Ext_{A}^{1,p^{n+1}q}(Z_{p},Z_{p})$.
Consequently $\bar{c}_{1}h_{n+1}=\bar{f}\cdot p$ for some
$\bar{f}\in\pi_{p^{n+1}q}E_{2}$. On the other hand,
$\bar{b}_{2}\bar{f}\in\pi_{p^{n+1}q}KG_{2}\cong
Ext_{A}^{2,p^{n+1}q}(Z_{p},Z_{p})\cong Z_{p}\\{b_{n}\\}$ so that we have
$\bar{b}_{2}\bar{f}=\lambda\cdot b_{n}$ with $\lambda\in Z_{p}$. However, the
scalar $\lambda$ must be zero, this is because $b_{n}$ support a nontrivial
differential $d_{2p-1}(b_{n})=d_{2p-1}\Phi(\beta_{p^{n}/p^{n}})=\Phi
d_{2p-1}(\beta_{p^{n}/p^{n}})=\Phi(\alpha_{1}\beta_{p^{n-1}/p^{n-1}}^{p})=h_{0}b_{n-1}^{p}\neq
0$ (cf. [12] p.206 Theorem 5.4.8(i) ). Hence $\bar{f}=\bar{a}_{2}\bar{f}_{1}$
for some $\bar{f}_{1}\in\pi_{p^{n+1}q+1}E_{3}$ and we have
$\bar{c}_{1}h_{n+1}=\bar{a}_{2}\bar{f}_{1}\cdot
p=\bar{a}_{2}\bar{a}_{3}\bar{f}_{2}$ with
$\bar{f}_{2}\in\pi_{p^{n+1}q+2}E_{4}$. This means that the secondary
differential $d_{2}(h_{n+1})$ = 0 which contradicts with the following known
nontrivial differential $d_{2}(h_{n+1})=a_{0}b_{n}\neq 0\in
Ext_{A}^{3,p^{n+1}q+1}(Z_{p},Z_{p})$ ( cf. [12] p.11 Theorem 1.2.14). So we
have $\tilde{c}_{1}h_{n+1}\cdot j\neq 0$ and so (9.8.23) holds.

Now we proceed to prove Theorem 9.8.1 in case $t\geq 2$. We first prove the
following Lemmas and Propositions.

Lemma 9.8.25 Let $p\geq 3$ and $v_{1}x\in
Ext_{BP_{*}BP}^{0,tp^{n}(p+1)q}(BP_{*},BP_{*}K_{p^{n}})$, then
$v_{1}x=\Sigma_{r=1}^{[n/2]}\lambda_{r}v_{1}^{p^{n}-p^{n-2r}}\tilde{c}_{1}(a_{r}p^{n-2r})$,
where $\lambda_{r}\in Z_{p},a_{r}=(tp^{2r+1}+tp^{2r}-p^{2r}+1)/(p+1)$ and
$\tilde{c}_{1}(ap^{s})$ is the generator in Theorem 8.1.7 in chapter 6 which
has degree $ap^{s}(p+1)q$.

Proof : By Theorem 8.1.7 in chapter 8, $v_{1}x$ is a linear combination of the
following generators $v_{1}^{b}\tilde{c}_{1}(ap^{s})$, where $a\geq 1$ is not
divisible by $p$, $1\leq b<p^{n},b\geq max\\{0,p^{n}-q_{1}(ap^{s}\\}$ and
$q_{1}(ap^{s})=p^{s}$ if $a=1,q_{1}(ap^{s})=p^{s}+p^{s-1}-1$ if $a\geq 2$.

By degree reasons we have $bq+ap^{s}(p+1)q=tp^{n}(p+1)q$, then $s<n,b\geq
p^{n}-(p^{s}+p^{s-1}-1)>0$ and so $b\geq p^{n}-p^{n-1}$ if $s\leq n-2$. If
$s=n-1$, then $b$ is divisible by $p^{n-1}(p+1)$ so that $b\geq
p^{n}+p^{n-1}$. So, in any case we have $b\geq p^{n-1}(p-1)$ and the remaining
steps is similar to that given in the proof of Prop. 9.8.12. Q.E.D.

Proposition 9.8.26 Let $r>s$ and $\rho_{r,s}:K_{r}\rightarrow K_{s}$ be the
map in (9.8.4), then $d(\rho_{r,s})=i^{\prime}_{s}\xi j^{\prime}_{r}$ with
$\xi\in[\Sigma^{rq+1}M,M]\cap kerd$.

Proof : By (9.8.6)(9.8.3) we have
$j^{\prime}_{s}d(\rho_{r,s})=d(j^{\prime}_{s}\rho_{r,s})=d(\alpha^{r-s}j^{\prime}_{r})$
= 0 , then $d(\rho_{r,s})=i^{\prime}_{s}\overline{\xi}$ for some
$\overline{\xi}\in[\Sigma K_{r},M]$ and $\overline{\xi}=\xi j^{\prime}_{r}$
with $\xi\in[\Sigma^{rq+1}M,M]$ , this is because
$\overline{\xi}i^{\prime}_{r}\in[\Sigma M,M]$ = 0. By Theorem 6.4.14 in
chapter 6, we may assume $\xi=\xi_{1}+\xi_{2}ij$ with $\xi_{1},\xi_{2}\in
kerd\cap[\Sigma^{*}M,M]$. Then
$d(\rho_{r,s})=i^{\prime}_{s}\xi_{1}j^{\prime}_{r}+i^{\prime}_{s}\xi_{2}ijj^{\prime}_{r}$
and by applying the derivation $d$ using (9.8.3) we have
$i^{\prime}_{s}\xi_{2}j^{\prime}_{r}$ = 0. Consequently we have
$i^{\prime}_{s}\xi_{2}=\xi_{3}\alpha^{r}=0$, this is because
$\xi_{3}\in[\Sigma^{2}M,K_{s}]$ = 0. Then
$d(\rho_{r,s})=i^{\prime}_{s}\xi_{1}j^{\prime}_{r}$ with $\xi_{1}\in
kerd\cap[\Sigma^{rq+1}M,M]$. Q.E.D.

Proof of Thoerem 9.8.1 in case $t\geq 2$: By Theorem 9.8.1 in case $t\geq 1$,
there exists $f^{\prime}\in[\Sigma^{p^{n}(p+1)q}K_{a},K_{a}]$ such that the
induced $BP_{*}$-homomorphism $f^{\prime}_{*}=v_{2}^{p^{n}}$, where we briefly
write $p^{n}-1$ as $a$. By Theorem 6.5.22 in chapter 6, we may assume $f\in
Mod_{*}$ , this is because the components of $f^{\prime}$ in $Der_{*}$ and
$Mod_{*}\delta_{0}$ induce zero $BP_{*}$-homomorphism.

Write
$\delta^{\prime}=i^{\prime}_{s}j^{\prime}_{s}\in[\Sigma^{-sq-1}K_{s},K_{s}]$.
By Theorem 6.5.23 in chapter 6,
$\delta^{\prime}f^{\prime}-f^{\prime}\delta^{\prime}\in Mod_{*}$ and this
group is a commutative subring of $[\Sigma^{*}K_{s},K_{s}]$. Then we have
$f^{\prime}(\delta^{\prime}f^{\prime}-f^{\prime}\delta^{\prime})=(\delta^{\prime}f^{\prime}-f^{\prime}\delta^{\prime}f^{\prime})f^{\prime}$
or equivalently,
$(f^{\prime})^{2}\delta^{\prime}-\delta^{\prime}(f^{\prime})^{2}=2((f^{\prime})^{2}\delta^{\prime}-f^{\prime}\delta^{\prime}f^{\prime})$.
By induction we have
$(f^{\prime})^{s}\delta^{\prime}-\delta^{\prime}(f^{\prime})^{s}=s((f^{\prime})^{s}\delta^{\prime}-(f^{\prime})^{s-1}\delta^{\prime}f^{\prime})$
, $s\geq 1$. That is
(9.8.27) $\qquad
s\cdot(f^{\prime})^{s-1}\delta^{\prime}f^{\prime}=\delta^{\prime}(f^{\prime})^{s}+(s-1)(f^{\prime})^{s}\delta^{\prime}$,
$s\geq 1$

Let $\rho_{a,1}:K_{a}\rightarrow K_{1}$ be the projection in (9.8.4), then by
Theorem in chapter 6,
$\rho_{a,1}(f^{\prime})^{s}i^{\prime}_{a}i\in\pi_{*}K_{1}$ can be extended to
$k_{s}\in Mod_{*}\subset[\Sigma^{sp^{n}(p+1)q}K_{1},\\\ K_{1}]$ such that
$\rho_{a,1}(f^{\prime})^{s}i^{\prime}_{a}i=k_{s}i^{\prime}_{1}i$ and
$(k_{s})_{*}=v_{2}^{sp^{n}}$. Since
$j^{\prime}_{1}k_{s}i^{\prime}_{1}i=\alpha^{a-1}j^{\prime}_{a}(f^{\prime})^{s}i^{\prime}_{a}i$
and $(i^{\prime}_{1}j^{\prime}_{1}k_{s}-k_{s}i^{\prime}_{1}j^{\prime}_{1})\in
Mod_{*}$, then
$(i^{\prime}_{1}j^{\prime}_{1}k_{s}-k_{s}i^{\prime}_{1}j^{\prime}_{1})i^{\prime}_{1}i$
= 0 and so
$i^{\prime}_{1}j^{\prime}_{1}k_{s}=k_{s}i^{\prime}_{1}j^{\prime}_{1}$. By
applying the derivation $d$ to the equation
$\rho_{a,1}(f^{\prime})^{s}i^{\prime}_{a}\delta=k_{s}i^{\prime}_{1}\delta$
(where we write $\delta=ij$) we have
(9.8.28)
$\qquad\rho_{a,1}(f^{\prime})^{s}i^{\prime}_{a}=k_{s}i^{\prime}_{1}-d(\rho_{a,1})(f^{\prime})^{s}i^{\prime}_{a}\delta=k_{s}i^{\prime}_{1}-i^{\prime}_{1}\xi
j^{\prime}_{a}(f^{\prime})^{s}i^{\prime}_{a}\delta$

$\qquad=k_{s}i^{\prime}_{1}-i^{\prime}_{1}j^{\prime}_{a}(f^{\prime})^{s}i^{\prime}_{a}\xi\delta$,
$s\geq 1$,
where $\xi\in[\Sigma^{aq+1}M,M]\cap kerd$( cf. Prop. 9.8.26). Let $t\geq 2$ is
not divisible by $p$, then by (9.8.27)(9.8.28) we have
$i^{\prime}_{1}j^{\prime}_{a}(f^{\prime})^{t}i^{\prime}_{a}=\rho_{a,1}i^{\prime}_{a}j^{\prime}_{a}(f^{\prime})^{t}i^{\prime}_{a}=t\cdot\rho_{a,1}(f^{\prime})^{t-1}i^{\prime}_{a}j^{\prime}_{a}f^{\prime}i^{\prime}_{a}$
= $t\cdot k^{t-1}i^{\prime}_{1}j^{\prime}_{a}f^{\prime}i^{\prime}_{a}-t\cdot
i^{\prime}_{1}j^{\prime}_{a}(f^{\prime})^{t-1}i^{\prime}_{a}\xi\delta
j^{\prime}_{a}f^{\prime}i^{\prime}_{a}$ and
(9.8.29) $\qquad i^{\prime}_{1}j^{\prime}_{a}\phi=t\cdot
k^{t-1}i^{\prime}_{1}j^{\prime}_{a}f^{\prime}i^{\prime}_{a}$,
where we write
$\phi=(f^{\prime})^{t}i^{\prime}_{a}+t\cdot(f^{\prime})^{t-1}i^{\prime}_{a}\xi\delta
j^{\prime}_{a}f^{\prime}i^{\prime}_{a}$.

Let $X$ be the cofibre of
$i^{\prime}_{1}j^{\prime}_{a}f^{\prime}i^{\prime}_{a}:\Sigma^{p^{n+1}q+q-1}M\rightarrow
K_{1}$ given by the cofibration in the upper row of the following homnotopy
commutative diagram ($m=(t-1)p^{n}(p+1)q+(p^{n}-1)q$)

$\quad\Sigma^{-1}X\quad\stackrel{{\scriptstyle
u}}{{\longrightarrow}}\quad\Sigma^{p^{n+1}q+q-1}M\stackrel{{\scriptstyle
i^{\prime}_{1}j^{\prime}_{a}f^{\prime}i^{\prime}_{a}}}{{\longrightarrow}}\quad
K_{1}\qquad\stackrel{{\scriptstyle w}}{{\longrightarrow}}\qquad X$
(9.8.30)$\quad\Big{\downarrow}\bar{\phi}\quad\qquad\qquad\qquad\Big{\downarrow}\phi\qquad\qquad\qquad\quad\Big{\downarrow}t\cdot
k^{t-1}\qquad\quad\Big{\downarrow}\bar{\phi}$

$\quad\Sigma^{-m-1}K_{a+1}\stackrel{{\scriptstyle\rho_{a+1,a}}}{{\longrightarrow}}\Sigma^{-m-1}K_{a}\stackrel{{\scriptstyle
i^{\prime}_{1}j^{\prime}_{a}}}{{\longrightarrow}}\quad\Sigma^{-m+aq}K_{1}\stackrel{{\scriptstyle\psi_{1,a+1}}}{{\longrightarrow}}\Sigma^{-m}K_{a+1}$
Note that the above middle rectangle is homotopy commutative by (9.8.29), then
there exists $\bar{\phi}$ such that all the above rectangles commute up to
homotopy.

By $wi^{\prime}_{1}j^{\prime}_{a}f^{\prime}i^{\prime}_{a}$ = 0 we have
$wi^{\prime}_{1}j^{\prime}_{a}f^{\prime}=yj^{\prime}_{a}$ with
$y\in[\Sigma^{p^{n}(p+1)q}M,X]$ so that $uyj^{\prime}_{a}$ = 0 and
$uy=\lambda\cdot\alpha^{a}$ for some $\lambda\in[M,M]\cong Z_{p}\\{1_{M}\\}$,
that is we have
(9.8.31)$\qquad wi^{\prime}_{1}j^{\prime}_{a}f^{\prime}=yj^{\prime}_{a}$,
$uy=\lambda\cdot\alpha^{a}$ for some $\lambda\in Z_{p}$.
On the other hand,
$j^{\prime}_{a}(f^{\prime})^{s}\psi_{1,a}i^{\prime}_{1}=j^{\prime}_{a}(f^{\prime})^{s}i^{\prime}_{a}\alpha^{a-1}=\alpha^{a-1}j^{\prime}_{a}(f^{\prime})^{s}i^{\prime}_{a}=j^{\prime}_{1}\rho_{a,1}\\\
(f^{\prime})^{s}i^{\prime}_{a}=j^{\prime}_{1}k_{s}i^{\prime}_{1}$, then
$j^{\prime}_{a}f^{\prime}\psi_{1,a}=j^{\prime}_{1}k_{1}+\eta j^{\prime}_{1}$
with $\eta\in[\Sigma^{*}M,M]$ and so
$yj^{\prime}_{1}=wi^{\prime}_{1}j^{\prime}_{a}f^{\prime}\psi_{1,a}=wi^{\prime}_{1}j^{\prime}_{1}k_{1}+wi^{\prime}_{1}\eta
j^{\prime}_{1}=wk_{1}i^{\prime}_{1}j^{\prime}_{1}+wi^{\prime}_{1}\eta
j^{\prime}_{1}$ and we have
(9.8.32) $\qquad y=wk_{1}i^{\prime}_{1}+wi^{\prime}_{1}\eta+z\alpha$ with
$z\in[\Sigma^{*}M,X]$,

$\qquad\quad\bar{\phi}y=t\cdot\psi_{1,a+1}k_{t-1}k_{1}i^{\prime}_{1}+t\cdot\psi_{1,a+1}k_{t-1}i^{\prime}_{1}\eta+\bar{\phi}z\alpha$
which is obtained by (9.8.31).

We claim that
(9.8.33) $\bar{\phi}z\alpha i\in\pi_{tp^{n}(p+1)q+aq}K_{a+1}$ has $BP$
filtration $>0$
This will be proved in the last. Then
$\bar{\phi}yi=t\cdot\psi_{1,a+1}k_{t-1}k_{1}i^{\prime}_{1}i$ (modulo higher
filtration) is represented by $t\cdot v_{1}^{a}v_{2}^{tp^{n}}\in
Ext_{BP_{*}BP}^{0,*}(BP_{*},BP_{*}K_{a+1})$ in the Adams-Novikov spectral
sequence.

Hence , by (9.8.31)(9.8.30)(9.8.28)(9.8.27) we have
$\bar{\phi}yj^{\prime}_{a}=\bar{\phi}wi^{\prime}_{1}j^{\prime}_{a}f^{\prime}=t\cdot\psi_{1,a+1}k_{t-1}i^{\prime}_{1}j^{\prime}_{a}f^{\prime}=t\cdot\psi_{1,a+1}\rho_{a,1}(f^{\prime})^{t-1}i^{\prime}_{a}j^{\prime}_{a}f^{\prime}=\psi_{1,a+1}\rho_{a,1}(i^{\prime}_{a}j^{\prime}_{a}(f^{\prime})^{t}+(t-1)(f^{\prime})^{t}i^{\prime}_{a}j^{\prime}_{a})$
= $(t-1)\psi_{1,a+1}\rho_{a,1}(f^{\prime})^{t}i^{\prime}_{a}j^{\prime}_{a}$
and so
$\bar{\phi}y=(t-1)\psi_{1,a+1}\rho_{a,1}(f^{\prime})^{t}i^{\prime}_{a}+\tilde{f}\alpha^{a}$
with $\tilde{f}\in[\Sigma^{tp^{n}(p+1)q}M,K_{a+1}]$.

By the claim (9.8.33), $\bar{\phi}yi$ is represented by $t\cdot
v_{1}^{a}v_{2}^{tp^{n}}$ in the Adams-Novikov spectral sequence, then
$\tilde{f}\alpha^{a}i$ is represented by $v_{1}^{a}v_{2}^{tp^{n}}$ and so
$\tilde{f}i\in\pi_{tp^{n}(p+1)q}K_{a+1}$ is represented by
$v_{2}^{tp^{n}}+v_{1}x\in
Ext_{BP_{*}BP}^{0,tp^{n}(p+1)q}(BP_{*},BP_{*}K_{a+1})$, where
$v_{1}x=\Sigma_{r=1}^{[n/2]}\lambda_{r}v_{1}^{p^{n}-p^{n-2r}}\tilde{c}_{1}(a_{r}p^{n-2r})$
which is obtained by Lemma 9.8.25.

By [20], if $t\geq 2$ is not divisible by $p$ and $1\leq r\leq p$,
$v_{2}^{tp}\in Ext_{BP_{*}BP}^{0,*}(BP_{*},\\\ BP_{*}K_{r})$ is a permanent
cycle in the Adams-Novikov spectral sequence. Suppose inductively that
$v_{2}^{tp^{s}}\in Ext_{BP_{*}BP}^{0,*}(BP_{*},BP_{*}K_{r})$ are permanent
cycles for all $t\geq 2$ is not divisible by $p$, $1\leq r\leq p^{s}$ and
$s\leq n-1$. Then, it is easily seen that
$v_{1}^{p^{n}-p^{n-2r}}\tilde{c}_{1}(a_{r}p^{n-2r})$ is realizable in
$[\Sigma^{tp^{n}(p+1)q}K_{a+1},K_{a+1}]$ so that the above by the induction
hypothesis we know that $v_{1}x$ also is a permanent cycle. . So,
$v_{2}^{tp^{n}}\in Ext_{BP_{*}BP}^{0,*}(BP_{*},BP_{*}K_{a+1})$ is a permanent
cycle in the Adams-Novikov spectral sequence and there exists
$h\in\pi_{tp^{n}(p+1)q}K_{a+1}$ such that the induced $BP_{*}$-homomorphism
$h_{*}=v_{2}^{tp^{n}}$. Hence, for $1\leq r\leq a+1=p^{n}$,
$jj^{\prime}_{r}\rho_{a+1,r}h\in\pi_{tp^{n}(p+1)q-rq-2}S$ is just the
$\beta_{tp^{n}/s}$-element of the Theorem.

Now our remaining work is to prove the claim (9.8.33). Recall as known above
that $j^{\prime}_{a}f^{\prime}i^{\prime}_{a}i\in\pi_{*}M$ is represented by
$\beta^{\prime}_{p^{n}/p^{n}-1}\in Ext_{BP_{*}BP}^{1,*}(BP_{*},\\\ BP_{*}M)$
in the Adams-Novikov spectral sequence and
$\beta^{\prime}_{p^{n}/p^{n}-1}=v_{1}\beta^{\prime}_{p^{n}/p^{n}}$, then
$(i^{\prime}_{1})_{*}(\beta^{\prime}_{p^{n}/p^{n}-1})=0\in
Ext_{BP_{*}BP}^{1,*}(BP_{*},BP_{*}K_{1})$ and so
$i^{\prime}_{1}j^{\prime}_{a}f^{\prime}i^{\prime}_{a}i\in\pi_{*}K_{1}$ has
$BP$-filtration $\geq q+1$. Then, in the Adams-Novikov resolution of the
spectrum $K_{1}$ , $i^{\prime}_{1}j^{\prime}_{a}f^{\prime}i^{\prime}_{a}i$ can
be lifted to $\kappa\in\pi_{*}\widetilde{E}_{q+1}\wedge K_{1}$ such that
$(\tilde{a}_{0}\wedge 1_{K_{1}})\cdots(\tilde{a}_{q}\wedge
1_{K_{1}})\kappa=i^{\prime}_{1}j^{\prime}_{a}f^{\prime}i^{\prime}_{a}i$. Since
$K_{1}$ is an $M$-module spectrum , then $\kappa=\kappa^{\prime}\cdot i$ with
$\kappa^{\prime}\in[\Sigma^{*}M,\widetilde{E}_{q+1}\wedge K_{1}]$.
Consequently we have
$i^{\prime}_{1}j^{\prime}_{a}f^{\prime}i^{\prime}_{a}=(\tilde{a}_{0}\wedge
1_{K_{1}})\cdots(\tilde{a}_{q}\wedge 1_{K_{1}})\kappa^{\prime}+\sigma j$ with
$\sigma\in[\Sigma^{p^{n+1}q+q}S,K_{1}]$. Note that $(\tilde{b}_{0}\wedge
1_{K_{1}})\sigma\in\pi_{p^{n+1}q+q}BP\wedge K_{1}\cong
Hom_{BP_{*}BP}^{p^{n+1}q+q}(BP_{*},BP_{*}(BP\wedge K_{1}))$ is a $d_{1}$-cycle
in the Adams-Novikov resolution of $K_{1}$ and it represents an element in
$Ext_{BP_{*}BP}^{0,p^{n+1}q+q}(BP_{*},BP_{*}K_{1})$. However, this group is
zero by degree reason , this is because
$Ext_{BP_{*}BP}^{0,*}(BP_{*},BP_{*}K_{1})\cong Z_{p}[v_{2}]$). Then we have
$(\tilde{b}_{0}\wedge 1_{K_{1}})\sigma$ = 0 so that $\sigma$ can be lifted to
$\sigma^{\prime}\in\pi_{*}\widetilde{E}_{q+1}\wedge K_{1}$ such that
$(\tilde{a}_{0}\wedge 1_{K_{1}})\cdots(\tilde{a}_{q}\wedge
1_{K_{1}})\sigma^{\prime}=\sigma$. So we have
$i^{\prime}_{1}j^{\prime}_{a}f^{\prime}i^{\prime}_{a}=(\tilde{a}_{0}\wedge
1_{K_{1}})\cdots(\tilde{a}_{q}\wedge
1_{K_{1}})(\kappa^{\prime}+\sigma^{\prime}j)$. By this we know that the
following short exact sequence induced by the cofibration in the top row of
(9.8.30) is a split exact sequence of $BP_{*}BP$-comodule:

$\quad 0\rightarrow BP_{*}K_{1}\stackrel{{\scriptstyle
u_{*}}}{{\longrightarrow}}BP_{*}X\stackrel{{\scriptstyle
w_{*}}}{{\longrightarrow}}BP_{*}M\rightarrow 0$
where $\mid w_{*}\mid=-(p^{n+1}+1)q$
Moreover, this splitness also hold in the following
$Ext_{BP_{*}BP}^{0,*}$-stage :

$\quad 0\rightarrow Ext^{0}K_{1}\stackrel{{\scriptstyle
u_{*}}}{{\longrightarrow}}Ext^{0}X\stackrel{{\scriptstyle
w_{*}}}{{\longrightarrow}}Ext^{0}M\rightarrow 0$
That is to say, there is an invariant $BP_{*}$-homomorphism
$u^{\prime}:Ext^{0}X\rightarrow Ext^{0}K_{1}$ and
$w^{\prime}:Ext^{0}M\rightarrow Ext^{0}X$ such that
$u^{\prime}u_{*}=1_{Ext^{0}K_{1}},w_{*}w^{\prime}=1_{Ext^{0}M}$ and
$u_{*}u^{\prime}+w^{\prime}w_{*}=1_{Ext^{0}X}$, where we briefly write
$Ext_{BP_{*}BP}^{0,*}(BP_{*},\\\ BP_{*}X)$ as $Ext^{0}X$.

To prove the claim (9.8.33), suppose in contrast that $\bar{\phi}z\alpha
i\in\pi_{*}K_{a+1}$ has $BP$-filtration 0, then, by (9.8.32), it is
represented by $\lambda v_{1}^{a}v_{2}^{tp^{n}}\in Ext^{0}K_{a+1}$ in the
Adams-Novikov spectral sequence, where $\lambda\neq 0\in Z_{p}$. Then
$zi\in\pi_{tp^{n}(p+1)q+(a-1)q}X$ must have $BP$ filtration 0 and it is
represented by some $x\in Ext^{0,*}X$ and
$(\bar{\phi})_{*}(v_{1}x)=\lambda\cdot v_{1}^{a}v_{2}^{tp^{n}}$. However,

$\qquad x=u_{*}u^{\prime}(x)+w^{\prime}w_{*}(x)=w^{\prime}w_{*}(x)$
, this is because by degree reason we have $u^{\prime}(x)\in
Ext^{0,tp^{n}(p+1)q+(a-1)q}K_{1}$ = 0, Then

$x=\lambda^{\prime}w^{\prime}(v_{1}^{r})$, for some $\lambda^{\prime}\in
Z_{p},rq=tp^{n}(p+1)q+(a-1)q-p^{n+1}q-q$
since $w_{*}(x)\in Ext^{0,r}M\cong Z_{p}\\{v_{1}^{r}\\}$, Then

$\lambda
v_{1}^{a}v_{2}^{tp^{n}}=(\bar{\phi})_{*}(v_{1}x)=\lambda^{\prime}(\bar{\phi})_{*}w^{\prime}(v_{1}^{r+1})$

Moreover, since $w^{\prime}(v_{1}^{r+1})$ is belong to $Ext^{0,*}X$ the
summand which is isomorphic to $Ext^{0,*}M$ and $Ext^{0,*}M$ is a trivial
$Z_{p}[v_{2}]$-module, then we have

$\qquad
0=v_{2}^{p^{n}}\cdot(\bar{\phi})_{*}w^{\prime}(v_{1}^{r+1})=\lambda\cdot
v_{1}^{a}v_{2}^{(t+1)p^{n}}\in Ext^{0,*}K_{a+1}$
This is a contradiction and then shows the claim(9.8.33). Q.E.D.

After finishing the proof of Thoerem 9.8.1 on second periodicity elements in
the stable homotopy groups of spheres, we state the following Theorem on
further result on second periodicity families in the stable homotopy groups of
spheres without proof. The proof is done in base on the result of Theorem
9.8.1 and using some properties of the spectrum $M(p^{r},v_{1}^{ap^{s}})$
which is the geometric realization of $BP_{*}/(p^{r},v_{1}^{ap^{s}})$. The
details of the proof can be seen in [23] §3.

Theorem 9.8.34 Let $p\geq 5$. $j=cp^{i}\leq p^{n-i}-1$ if $t\geq 1$
($cp^{i}\leq p^{n-i}$ if $t\geq 2$), then the element

$\beta_{tp^{n}/j,i+1}\in Ext_{BP_{*}BP}^{2,*}(BP_{*},BP_{*})$

is a permanent cycle in the Adams-Novikov spectral sequence and it converges
to the corresponding homotopy element of order $p^{i+1}$ in $\pi_{*}S$.

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