Exercise 9.21

Proof.

a)

As some primary elements of $D$ may be cubes, by example $53+36\omega ={(-1+3\omega )}^3$, we must of course suppose that $\gamma$ is not the cube of some element of $D$ (in the contrary case $x^3\equiv \gamma (\mathit{mod}\pi )$ is solvable for all prime $\pi$).

Note first that for all primes $\pi$ in $D$, there exists $\sigma \in D$ such that ${\chi }_{\pi }(\sigma )=\omega$. Indeed, there exist $(\mathit{N\pi }-1)∕3$ cubes in ${(D∕\mathit{\pi D})}^{\text{∗}}$, which has $\mathit{N\pi }-1$ elements, so there exists an element $\bar{\tau }\in {(D∕\mathit{\pi D})}^{\text{∗}}$ which is not a cube, therefore there exists $\tau \in D$ such that ${\chi }_{\pi }(\tau )\ne 1$. If ${\chi }_{\pi }(\tau )=\omega$, we put $\sigma =\tau$ and if ${\chi }_{\pi }(\tau )={\omega }^2$, we put $\sigma ={\tau }^2$. In the two cases, ${\chi }_{\pi }(\sigma )=\omega$.

Let $\gamma \in D$, where $\gamma$ is primary. Then $\gamma =\pm {\gamma }_1^{n_1}{\gamma }_2^{n_2}\cdots {\gamma }_p^{n_p}$, where the ${\gamma }_i$ are distinct primary primes. Write $n_i=3q_i+r_i,r_i\in \{0,1,2\}$. Then grouping in ${\gamma }^{\prime }$ the ${\gamma }^{r_i}$ such that $r_i\ne 0$, we can write $\gamma ={\delta }^3{\gamma }^{\prime },{\gamma }^{\prime }={\gamma }_1^{r_1}{\gamma }_2^{r_2}\cdots {\gamma }_l^{r_l},r_i\in \{1,2\},\delta =\pm {\gamma }_1^{q_1}\cdots {\gamma }_p^{q_p}\in D$ ( $-1$ is a cube). Since by hypothesis $\gamma$ is not a cube, $l\ge 1$. Moreover the equation $x^3\equiv \gamma (\mathit{mod}\pi )$ is solvable iff $x^3\equiv {\gamma }^{\prime }(\mathit{mod}\pi )$ is solvable. We may then suppose that

$$\gamma ={\gamma }_1^{r_1}{\gamma }_2^{r_2}\cdots {\gamma }_l^{r_l},1\le r_i\le 2,$$

without cubic factors.

Note that the ${\gamma }_i$ are not associate to $\lambda =1-\omega$ (see Ex. 9.17).

Let $A=\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_k\}$ a set (possibly empty) of distinct primary primes ${\lambda }_i$ (therefore they are not associate), and not associate neither to ${\gamma }_i,1\le i\le l$, nor to $\lambda =1-\omega$.

We will show that we can find a primary prime ${\lambda }_{k+1}$ distinct of the ${\lambda }_i$ with the same properties and such that the equation $x^3\equiv \lambda (\mathit{mod}{\lambda }_{k+1})$ is not solvable. This will prove the existence of infinitely many primes $\pi$ such that the equation $x^3\equiv \lambda (\mathit{mod}\pi )$ is not solvable.

Using the initial note, let $\sigma \in D$ such that ${\chi }_{{\gamma }_l}(\sigma )=\omega$. As $D$ is a principal ideal domain, the Chinese Remainder Theorem is valid. Since $3=\lambda \bar{\lambda }=-{\omega }^2{\lambda }^2$ is relatively prime to ${\gamma }_i,{\lambda }_i$, there exists $\beta \in D$ such that

$$\begin{array}{llll}\hfill \beta & \equiv 2[3],& \hfill & \\ \hfill \beta & \equiv 1[{\lambda }_i](1\le i\le k),& \hfill & \\ \hfill \beta & \equiv 1[{\gamma }_i](1\le i\le l-1),& \hfill & \\ \hfill \beta & \equiv \sigma [{\gamma }_l].& \hfill & \\ \hfill & & \hfill \end{array}$$

The first equation show that $\beta$ is primary, so $\beta ={(-1)}^{m-1}{\beta }_1\dots {\beta }_m$, where the ${\beta }_i$ are primary primes.

By Exercise 9.20,

$${\chi }_{\beta }(\gamma )={\chi }_{\beta }{({\gamma }_1)}^{r_1}\cdots {\chi }_{\beta }{({\gamma }_l)}^{r_l}={\chi }_{{\gamma }_1}{(\beta )}^{r_1}\cdots {\chi }_{{\gamma }_l}{(\beta )}^{r_l}.$$

As ${\chi }_{{\gamma }_i}(1)=1(1\le i\le l-1)$, and ${\chi }_{{\gamma }_l}(\beta )={\chi }_{{\gamma }_l}(\sigma )=\omega$, we obtain ${\chi }_{\beta }(\gamma )={\omega }^{r_l}\ne 1$, since $r_l=1$ or $r_l=2$.

By Exercise 9.18, ${\chi }_{\rho }(\alpha ){\chi }_{\gamma }(\alpha )={\chi }_{-\mathit{\rho \gamma }}(\alpha )$, with primary $\rho ,\gamma$, so by induction, as $\beta ={(-1)}^{m-1}{\beta }_1\cdots {\beta }_m$,

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

Thus there exists a subscript $j$ such that ${\chi }_{{\beta }_j}(\gamma )\ne 1$.

We can then take ${\lambda }_{k+1}={\beta }_j$. Indeed, since $\beta \equiv 1[{\lambda }_i]$ and $\beta \not\equiv 0[{\gamma }_i]$, ${\beta }_j$ is distinct of the ${\lambda }_i$ and ${\gamma }_i$, and ${\beta }_j$ is not associate to $\lambda$ since $\beta \equiv 2(\mathit{mod}3)$.

As ${\chi }_{{\lambda }_{k+1}}(\gamma )\ne 1$, the equation $x^3\equiv \gamma [{\lambda }_{k+1}]$ is not solvable, so ${\lambda }_{k+1}$ is convenient.

Conclusion : if $\gamma \in D$ is primary and is not a cube in $D$, there exist infinitely many primes $\pi \in D$ such that the equation $x^3\equiv \gamma [\pi ]$ is not solvable.

b)

We show that $x^3\equiv \omega [\pi ]$ has no solution for infinitely many primes $\pi$.

To initialize the induction, we display such a prime $\pi$, namely $\pi =2+3\omega$. Indeed, $N(\pi )=4+9-6=7$, 7 is a rational prime, so $\pi$ is a primary prime in $D$, of the form $\pi =3m-1+3\mathit{n\omega }$, with $n=m=1$, so ${\chi }_{\pi }(\omega )={\omega }^{m+n}={\omega }^2\ne 1$ : the equation $x^3\equiv \omega [\pi ]$ is not solvable. Moreover $\pi$ is not associate to $\lambda =1-\omega$.

Suppose now the existence of a set $A=\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_l\},l\ge 1$, of distinct primary primes ${\lambda }_i$, not associate to $\lambda$ and such the equation $x^3\equiv \omega [{\lambda }_i]$ is not solvable for each $i,1\le i\le l$. We will show that we can add a prime ${\lambda }_{l+1}$ to the set $A$ with the same properties.

Let

$$\beta =3{(-1)}^{l-1}{\lambda }_1\cdots {\lambda }_l-1.$$

${(-1)}^{l-1}{\lambda }_1\cdots {\lambda }_l$ is primary, so ${(-1)}^{l-1}{\lambda }_1\cdots {\lambda }_l=3m-1+3\mathit{n\omega },m,n\in \mathbb{Z}$.

$\beta =3(3m-1+3\mathit{n\omega })-1=3(3m-1)-1+9\mathit{n\omega }=3M-1+3\mathit{N\omega }$, where $M=3m-1,N=3n$. By Exercise 9.19,

$${\chi }_{\beta }(\omega )={\omega }^{M+N}={\omega }^{3m-1+3n}={\omega }^2\ne 1.$$

As $\beta =\pm {\beta }_1\cdots {\beta }_m$, where the ${\beta }_i$ are primary primes, ${\chi }_{\beta }(\omega )={\chi }_{{\beta }_1}(\omega )\cdots {\chi }_{{\beta }_m}(\omega )\ne 1$, so there exists a subscript $i$ such that ${\chi }_{{\beta }_i}(\omega )\ne 1$.

Since $\beta =3{(-1)}^{l-1}{\lambda }_1\cdots {\lambda }_l-1$, ${\beta }_i$ is associate neither to ${\lambda }_i$ nor to $\lambda$. Moreover ${\chi }_{{\beta }_i}(\omega )\ne 1$, thus the equation $x^3\equiv \omega [{\beta }_i]$ is not solvable: ${\lambda }_{l+1}={\beta }_i$ is convenient.

Conclusion: the equation $x^3\equiv \omega [\pi ]$ is not solvable for infinitely many primes $\pi$.

c)

We show that $x^3\equiv \lambda [\pi ]$ has no solution for infinitely many primes $\pi$.

To initialize the induction, we display such a prime $\pi$, namely $\pi =-4+3\omega$. Indeed, $N(\pi )=16+9+12=37$, 37 is a rational prime, so $\pi$ is a primary prime in $D$, of the form $\pi =3m-1+3\mathit{n\omega }$, with $m=-1,n=1$, so ${\chi }_{\pi }(\lambda )={\omega }^{2m}=\omega \ne 1$ : the equation $x^3\equiv \lambda [\pi ]$ is not solvable.

Suppose now the existence of a set $A=\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_l\},l\ge 1$, of distinct primary primes ${\lambda }_i$, not associate to $\lambda$ and such the equation $x^3\equiv \lambda [{\lambda }_i]$ is not solvable. We will show that we can add a prime ${\lambda }_{l+1}$ to the set $A$ with the same properties.

Let

$$\beta =3{(-1)}^{l-1}{\lambda }_1\cdots {\lambda }_l-1.$$

${(-1)}^{l-1}{\lambda }_1\cdots {\lambda }_l$ is primary, so ${(-1)}^{l-1}{\lambda }_1\cdots {\lambda }_l=3m-1+3\mathit{n\omega },m,n\in \mathbb{Z}$.

$\beta =3(3m-1+3\mathit{n\omega })-1=3(3m-1)-1+9\mathit{n\omega }=3M-1+3\mathit{N\omega }$, where $M=3m-1,N=3n$. By Exercise 9.19,

$${\chi }_{\beta }(\lambda )={\omega }^{2M}={\omega }^{2(3m-1)}=\omega \ne 1.$$

As $\beta =\pm {\beta }_1\cdots {\beta }_m$, where the ${\beta }_i$ are primary primes, ${\chi }_{\beta }(\omega )={\chi }_{{\beta }_1}(\omega )\cdots {\chi }_{{\beta }_m}(\omega )\ne 1$, so there exists a subscript $i$ such that ${\chi }_{{\beta }_i}(\lambda )\ne 1$.

Since $\beta =3{(-1)}^{l-1}{\lambda }_1\cdots {\lambda }_l-1$, ${\beta }_i$ is associate neither to ${\lambda }_i$ nor to $\lambda$. Moreover ${\chi }_{{\beta }_i}(\lambda )\ne 1$, thus the equation $x^3\equiv \lambda [{\beta }_i]$ is not solvable : ${\lambda }_{l+1}={\beta }_i$ is convenient.

Conclusion : the equation $x^3\equiv \lambda [\pi ]$ is not solvable for infinitely many primes $\pi$.