Retraction notice to "Qualitative analysis of coupled system of sequential fractional integrodifferential equations" [AIMS Mathematics 7(5) (2022) 8012–8034]

1.   Introduction

It is well known that the Herbrand-Ribet theorem is about the relation between the $p$-th class group of cyclotomic field $\mathbb{Q}(\zeta_p)$ and the Bernoulli number.

We introduce some notations. Let $F = \mathbb{Q}(\zeta_p)$ be the cyclotomic field, and

be the Galois group, where $\sigma_a(\zeta_p) = \zeta_p^a$. Let $\omega$ be the Teichmuller character of group $(\mathbb{Z}/p)^\times$, that is, a character $\omega: (\mathbb{Z}/p)^\times \to \mathbb{Z}_p^\times$ such that for $a\in \mathbb{Z}, (a, p) = 1.$ Then $\omega(a)^{p-1} = 1$ and $\omega(a)\equiv a \mod p$. For the group ring $\mathbb{Z}_p[G]$, where $\mathbb{Z}_p$ is the $p$-adic integer ring, the idempotents are

Let $A$ be the $p$-part of $Cl(F)$, which is the class group of $F$. Then $A = \bigoplus_{i = 0}^{p-2}A_i$, where $A_i = \varepsilon_iA$.

The Herbrand theorem states that if $p$ divides the numerator of the Bernoulli number $B_{p-i}$, then $\varepsilon_iA\neq0.$ In 1976, Ribet ^[7] proved the converse of the Herbrand's theorem. So the Herbrand-Ribet theorem is as follow.

Theorem 1.1. Let i be an odd integer with $3\leq i \leq p-2$. If p divides the numerator of the Bernoulli number $B_{p-i}$, then $\varepsilon_iA\neq0.$

The Herbrand theorem is obtained by the properties of the Stickelberger element and the $p$-adic $L$-function. In ^[8], the Herbrand-Ribet theorem for function fields was obtained. In addition, Coats and Sinnott ^[2] proved an analogue of Stickelberger's theorem for the $K_2$ groups.

Throughout this paper, inspired by the above results, we obtain respectively the $K_2$ analogue of Herbrand-Ribet theorem and the $K_2$ analogue of the Vandiver conjecture.

2.   $K_2$ analogue of Herbrand-Ribet theorem

Let $S$ be a finite set of places of $F = \mathbb{Q}(\zeta_p)$ including the archimedean ones. Let $O_S$ denote the ring of $S$-integers in $F$, i.e., the ring of all $a \in F$ such that $v(a)\geq 0$ for each place $v\not\in S$. Then

By Quillen's localization sequence, we have the isomorphism $\ker d^S\simeq K_2(O_S)$, which is moreover a $G$-isomorphism if $S$ is stable under $G$ (see [9, P. 271]).

Let $K_2(\mathbb{Z}[\zeta_p])$ be the $K_2$ group of the ring of algebraic integers $\mathbb{Z}[\zeta_p]$, and let $C$ be the $p$-part of $K_2(\mathbb{Z}[\zeta_p])$. Then we have $C = \bigoplus_{i = 0}^{p-2}C_i$, $C_i = \varepsilon_iC$.

Lemma 2.1. There exist $G$-isomorphisms:

Proof. We note an isomorphism ^[4]

where $G$ acts on $\mu_p\otimes A$ by the formula

We claim that the above isomorphism is a $G$-isomorphism. Let $S$ be a set of the places of $\mathbb{Q}(\zeta_p)$ consisting of the archimedean ones and the finite ones above $p$. Let $S_c$ denote the set of complex places. Then there is a natural exact sequence (see [9, Theorem 6.2])

where $(\coprod\mu_p)_0$ denotes the subgroup of the direct sum consisting of the elements $z = (z_v)$ such that $\sum z_v = 0$. The map $h_1^S$ is that induced by the $l$-th power norm residue symbols for $v\in S-S_c$. Since $S$ is stable under $G$, the above exact sequence is sequence of $G$-modules with $G$-homomorphisms(see [9, P. 271]). By[11, Theorem 73], $C$ and the $p$-part of $K_2(\mathbb{Z}[\zeta_p, 1/p])$ are equal to $H_{\rmét}^2(\mathbb{Z}[\zeta_p, 1/p], \mathbb{Z}_p(2))$. Since $p\mathbb{Z}[\zeta_p] = (1-\zeta_p)^{p-1}$, the $p$-part of $Cl(O_S)$ is equal to $A$. Moreover, the fourth term in (2.2) is $0$ (see [11, Example 5]), we get that (2.1) is a $G$-isomorphism.

Then we consider the following homomorphism

Here, $\delta$ is not a homomorphism of $G$-modules. The kernel of $\delta$ is $pA$, so we get an isomorphism

Next we give the explicit description of $\delta$ under the Galois group action. For $z: = \zeta_{p^n}$, we have $\sigma_a(z) = z^{\omega(a)}$ (see [1, Lemma 3.3]), so there is

Therefore,

Hence

By (2.4), the action of idempotents $\varepsilon_j$ on (2.3) leads to

Since (2.1) is a $G$-isomorphism, combining with the above isomorphism, we obatin

as desired.

Next, we give the $K_2$ analogue of the Herbrand-Ribet theorem of the field $\mathbb{Q}(\zeta_p)$ as follow.

Theorem 2.1. Let $i$ be even, $4\leq i\leq p-3$. Then

Proof. It is clearly that

From Lemma 2.1, we have $\varepsilon_{i} C/p\simeq \varepsilon_{i-1} A/p.$ Utilizing Theorem 1.1, we get $C_i\neq0 \Leftrightarrow p|B_{p+1-i},$ as required.

However, the proof of $"\Rightarrow "$ can also be obtained by the properties of the Stickelberger element without using Theorem 1.1 and Lemma 2.1, We sketch the proof as follow.

Considering the Stickelberger element for the cyclotomic field $\mathbb{Q}(\zeta_p)$

where $\zeta(\sigma, s)$ is the partial zeta function, we can prove that $(c^2-\omega^i(c))B_{2, \omega^{-i}}$ annihilates $C_i,$ moreover, for $i = 4, 6, \cdots, p-3,$ $B_{2, \omega^{-i}}$ annihilates $C_i$.

We now suppose $C_i\neq0$. Then $B_{2, \omega^{-i}}\equiv0\ ({\rm{mod}}\ p)$. Since

we get

Therefore, $p|B_{p+1-i}$.

3.   $K_2$ analogue of Vandiver's conjecture

The Vandiver's conjecture states that $p$ does not divide the class number of $\mathbb{Q}(\zeta_p)^{+}$, where $\mathbb{Q}(\zeta_p)^{+}$ is the maximal real subfield of the cyclotomic field $\mathbb{Q}(\zeta_p)$. Equivalently, the Vandiver's conjecture says that all the even part $\varepsilon_iA$ are trivial.

Lemma 3.1. For any irregular prime $p$, $A_{2i} = 0,$ where $1\leq i \leq 14.$

Proof. From ^[10] (Tables §1 Bernoulli numbers), for $i = 1, 2, 3, 4, 5, 7,$ we have $p\nmid B_{2i}$. So from Theorem 1.1, we have $A_{p-2i} = 0, i = 1, 2, 3, 4, 5, 7.$ By the reflection theorem (see [10, Theorem 10.9])

we get $A_{2i} = 0$.

Let $P_n$ denote the maximal prime factor of $B_n$ if $B_n$ has a prime factor. For $i = 6, 8, 9, 10, 11, 12, 13, 14$, from ^[10] (Tables §1 Bernoulli numbers) we have

These primes are all less than 12,000,000. But it is well know that the Vandiver conjecture has been checked to be true for all irregular primes less than 12,000,000. So we get $A_{2i} = 0$ for $i = 6, 8, 9, 10, 11, 12, 13, 14.$

Now we can make a $K_2$-analogue of Vandiver's conjecture as follow.

Conjecture 3.1. For odd $i$, $\varepsilon_iC = 0$, where $C$ is the $p$-part of $K_2(\mathbb{Z}[\zeta_p])$.

It has been proved that $\varepsilon_{p-3}A$ always vanishes (see ^[5]) and that if the prime $p\equiv 3\ (mod \ 4)$, then $\varepsilon_{(p+1)/2}A$ is trivial (see ^[3,6]). Combining these results with Lemmas 2.1 and 3.1, we get the following result, which checks some cases of Conjecture 3.1.

Theorem 3.1. For any irregular prime $p$, $C_{2i+1} = 0 \ (1\leq i \leq 14),$ $C_{p-2} = 0$ and $C_{(p+3)/2} = 0$ if $p\equiv 3\ (mod \ 4)$.

4.   Conclusions

We gave the $K_2$ analogue of Herbrand-Ribet theorem and prove the case. The $K_2$ analogue of Vandiver's conjecture was also obtained, but this case is hard to prove. However, we just check some special circumstances of it.

Acknowledgments

The authors are thankful for the careful reviews of referees and the editor. The first author was supported by the National Natural Science Foundation of China (No. 11901079), China Postdoctoral Science Foundation (No. 2021M700751) and the Scientific and Technological Research Program Foundation of Jilin Province (No. JJKH20190690KJ; No. 20200401085GX; No. JJKH20220091KJ). The second author was supported by the National Natural Science Foundation of China (No. 11601211).

Conflict of interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.