Exercise 9.22

Proof. Let $\gamma \in D$ and suppose that $\gamma$ is not a cube in $D$. We will show that the equation $x^3\equiv \gamma [\pi ]$ is not solvable for infinitely primes $\pi \in D$.

By Exercise 9.2, we can write

where the ${\gamma }_i$ are distinct primary primes, not associate to $\lambda$. Let $v=3q+b,w=3q^{\prime }+c,n_i=3q_i+r_i$, with the remainders $b,c,r_i$ in $\{0,1,2\}$. Grouping the factors with null remainders, we obtain $\gamma ={\delta }^3{\gamma }^{\prime },{\gamma }^{\prime }={\omega }^b{\lambda }^c{\gamma }_1^{r_1}\cdots {\gamma }_l^{r_l}$, with $b,c,r_i$ in $\{1,2\},\delta \in D,l\ge 0$ ( $-1$ is a cube).

Moreover the equation $x^3\equiv \gamma [\pi ]$ is solvable iff the equation $x^3\equiv {\gamma }^{\prime }[\pi ]$ is solvable. So we may suppose that

without cubic factors.

$\text{∙}$

Case 1 : $l\ge 1$.

Let $A=\{{\lambda }_1,\dots ,{\lambda }_k\}$ a possibly empty set of distinct primary primes ${\lambda }_i$, distinct of the ${\gamma }_i$, not associate to $\lambda$, and such that the equation $x^3\equiv \gamma [{\lambda }_i]$ is not solvable. We will show that we can add a prime ${\lambda }_{k+1}$ with the same properties.

Suppose that $l\ge 1$. We have proved in Ex. 9.21 that there exists $\sigma \in D$ such that ${\chi }_{{\gamma }_l}(\sigma )=\omega$. Since $9,{\lambda }_i,{\gamma }_j$ are relatively prime, there exists $\beta \in D$ such that

$$\begin{array}{llll}\hfill \beta & \equiv -1[9]& \hfill & \\ \hfill \beta & \equiv 1[{\lambda }_i],1\le i\le k& \hfill & \\ \hfill \beta & \equiv 1[{\gamma }_i],1\le i\le l-1& \hfill & \\ \hfill \beta & \equiv \sigma [{\gamma }_l]& \hfill & \end{array}$$

$\beta \equiv -1[9]$, thus $\beta \equiv -1[3]$ : $\beta$ is primary, of the form $\beta =3M-1+3\mathit{N\omega }$.

$\beta =3M-1+3\mathit{N\omega }\equiv -1[9]$, so $3M+3\mathit{N\omega }\equiv 0[9]$, $M+\mathit{N\omega }\equiv 0[3]$, thus $3\mid M,3\mid N$.

By Exercise 9.18,

$$\begin{array}{llll}\hfill {\chi }_{\beta }(\omega )& ={\omega }^{M+N}=1& \hfill & \\ \hfill {\chi }_{\beta }(\lambda )& ={\omega }^{2M}=1& \hfill & \end{array}$$

As $\beta$ and ${\gamma }_i$ are primary, ${\chi }_{\beta }({\gamma }_i)={\chi }_{{\gamma }_i}(\beta )={\chi }_{{\gamma }_i}(1)=1(1\le i\le l-1)$.

${\chi }_{\beta }(\gamma )={\chi }_{\beta }{(\omega )}^b{\chi }_{\beta }{(\lambda )}^c{\chi }_{\beta }{({\gamma }_1)}^{r_1}\cdots {\chi }_{\beta }{({\gamma }_l)}^{r_l}={\chi }_{\beta }{({\gamma }_l)}^{r_l}={\chi }_{{\gamma }_l}{(\beta )}^{r_l}={\chi }_{{\gamma }_l}{(\sigma )}^{r_l}={\omega }^{r_l}\ne 1$, since $r_l\in \{1,2\}$.

$\beta =\pm {\beta }_1\cdots {\beta }_m$, with ${\beta }_i$ primary primes, therefore

$${\chi }_{\beta }(\gamma )=({\chi }_{{\beta }_1}\cdots {\chi }_{{\beta }_m})(\gamma )\ne 1.$$

Thus there exists a subscript $i$ such that ${\chi }_{{\beta }_i}(\gamma )\ne 1$, so $x^3\equiv \gamma [{\beta }_i]$ is not solvable. Moreover $\beta \equiv 1[{\gamma }_i]$, so ${\beta }_i$ is not associate to any ${\gamma }_j$. Similarly, ${\beta }_i$ is not associate to any ${\gamma }_j$, and $\beta \equiv -1[9]$, therefore ${\beta }_i$ is not associate to $\lambda$. So ${\lambda }_{k+1}={\beta }_i$ is convenient.

There exist infinitely many $\pi$ such that $x^3\equiv \gamma [\pi ]$ is not solvable.

$\text{∙}$

Case 2 : $l=0$, so $\gamma ={\omega }^b{\lambda }^c,1\le b\le 2,1\le c\le 2$.

${\pi }_0=2-3\omega$ is a primary prime ( $N({\pi }_0)=19$).

Let $A=\{{\lambda }_1,\dots ,{\lambda }_k\}$ a possibly empty set of distinct primary primes ${\lambda }_i\ne {\pi }_0$ such that the equation $x^3\equiv \gamma [{\lambda }_i]$ is not solvable. We will show that we can add a prime ${\lambda }_{k+1}$ with the same properties.

Let $\beta =9{(-1)}^{k-1}{\lambda }_1\cdots {\lambda }_k+2-3\omega$.

$\beta \equiv 2[3]$ : $\beta$ is primary.

Moreover ${(-1)}^{k-1}{\lambda }_1\cdots {\lambda }_k$ is primary, so

$${(-1)}^{k-1}{\lambda }_1\cdots {\lambda }_k=3m-1+3\mathit{n\omega },m\in \mathbb{Z},n\in \mathbb{Z}.$$

Then

$$\begin{array}{llll}\hfill \beta & =9(3m-1+3\mathit{n\omega })+2-3\omega & \hfill & \\ \hfill & =27m-7+(27n-3)\omega & \hfill & \\ \hfill & =3(9m-2)-1+3(9n-1)\omega & \hfill & \\ \hfill & =3M-1+3\mathit{N\omega },& \hfill & \end{array}$$

where $M=9m-2,N=9n-1$. Therefore

$$\begin{array}{llll}\hfill {\chi }_{\beta }(\omega )& ={\omega }^{M+N}={\omega }^{9m-2+9n-1}=1& \hfill & \\ \hfill {\chi }_{\beta }(\lambda )& ={\omega }^{2M}={\omega }^{2(9m-2)}={\omega }^2\ne 1& \hfill & \end{array}$$

$\beta =\pm {\beta }_1\cdots {\beta }_m$, where the ${\beta }_i$ are primary primes.

${\chi }_{\beta }(\gamma )={\chi }_{\beta }{(\omega )}^b{\chi }_{\beta }{(\lambda )}^c={\omega }^{2c}\ne 1$ since $c=1$ or $c=2$.

$${\chi }_{\beta }(\gamma )=({\chi }_{{\beta }_1}\cdots {\chi }_{{\beta }_m})(\gamma )\ne 1.$$

Thus there exists a subscript $i$ such that ${\chi }_{{\beta }_i}(\gamma )\ne 1$, so $x^3\equiv \gamma [{\beta }_i]$ is not solvable.

As ${\beta }_i\mid \beta =9{(-1)}^{k-1}{\lambda }_1\cdots {\lambda }_k+2-3\omega$, if ${\beta }_i={\lambda }_j$ for some subscript $j$, ${\lambda }_j\mid {\pi }_0=2-3\omega$, so ${\lambda }_j={\pi }_0$, which is a contradiction, thus ${\beta }_i\notin A$. Similarly, if ${\beta }_i={\pi }_0=2-3\omega$, then ${\pi }_0\mid 9{\lambda }_1\cdots {\lambda }_k$, and ${\pi }_0$ is relatively prime to $\lambda$, so ${\pi }_0={\lambda }_j$ for some subscript $j$ : this is a contradiction, thus ${\beta }_i\ne {\pi }_0$. ${\lambda }_{k+1}={\beta }_i$ is convenient.

So there exist infinitely many $\pi$ such that $x^3\equiv \gamma [\pi ]$ is not solvable.

$\text{∙}$ Conclusion :

if $\gamma$ is not a cube in $D$, there exist infinitely many primes $\pi$ such that $x^3\equiv \gamma [\pi ]$ is not sovable.

By contraposition, if the equation $x^3\equiv \gamma [\pi ]$ is solvable for every prime $\pi$, at the exception perhaps of the primes in a finite set, then $\gamma$ is a cube in $D$.