Monogenic pure cubics

The proof is a straightforward adaptation of steps in the proof of Theorem 1.1 given in [Mar18, p. 35–36] ∎

For (a), suppose $k^{2}m\not\equiv\pm 1$ mod $9$, let $\alpha=\sqrt[3]{k^{2}m}$, $K={\mathbb{Q}}(\alpha)$, and consider the integral basis $\{1,\alpha,\alpha^{2}/k\}$. To find $\theta\in\mathcal{O}_{K}$ such that $\mathcal{O}_{K}={\mathbb{Z}}[\theta]$, it suffices to consider $\theta$ of the form $\theta=u\alpha+v(\alpha^{2}/k)$ with $u,v\in{\mathbb{Z}}$. Then we have:

We represent $(1,\theta,\theta^{2})$ in terms of the given integral basis and the corresponding matrix has determinant $ku^{3}-mv^{2}$. Therefore $\mathcal{O}_{K}={\mathbb{Z}}[\theta]$ if and only if the equation $kX^{3}-mY^{3}=1$ has a solution $X,Y\in{\mathbb{Z}}$.

The proof of part (b) is similar with some tedious algebraic expressions as follows. Suppose $k^{2}m\equiv\pm 1$ mod $9$, let $\alpha=\sqrt[3]{k^{2}m}$, $K={\mathbb{Q}}(\alpha)$, and consider the integral basis $\{1,\alpha,\beta=(k^{2}\pm k^{2}\alpha+\alpha^{2})/(3k)\}$. As before, consider $\theta=u\alpha+v\beta$ with $u,v\in{\mathbb{Z}}$. Then depending on whether $k^{2}m\equiv\pm 1$ mod $9$, we have:

where $\displaystyle c_{3}=3k\left(u^{2}\pm\frac{2uvk}{3}+\frac{v^{2}k^{2}}{9}+\frac{2v^{2}}{9}\right)$, $\displaystyle c_{2}=\frac{2uvk}{3}+\frac{v^{2}m}{9}\mp k^{2}u^{2}-\frac{2uvk^{3}}{3}\mp\frac{v^{2}k^{4}}{9}$, and the precise value of $c_{1}$ is not needed for our purpose. We represent $(1,\theta,\theta^{2})$ in terms of the given integral basis and the corresponding matrix has determinant:

Therefore if $\mathcal{O}_{K}={\mathbb{Z}}[\theta]$ then the equation $kX^{3}-mY^{3}=9$ has a solution $X,Y\in{\mathbb{Z}}$. Conversely, if $(X_{0},Y_{0})$ is a solution, we can choose $v=Y_{0}$ and $u=(X_{0}\mp kY_{0})/3$ and we need to explain why $u\in{\mathbb{Z}}$. From $k^{2}m\equiv\pm 1$ mod $9$, we have $m\equiv\pm k^{4}$ mod $9$. Using this and the equation $kX_{0}^{3}-mY_{0}^{3}=9$, we have $X_{0}^{3}\equiv\pm k^{3}Y_{0}^{3}$ mod $9$. Hence $X_{0}\equiv\pm kY_{0}$ mod $3$. ∎

Let $L={\mathbb{Q}}(\sqrt[3]{b/a})$ and let $L^{\prime}$ be its Galois closure. Let $S$ be the set of primes $p\nmid 3ab$ such that $b/a$ is not a cube mod $p$. This means $p$ remains a prime in $L$ and $p\mathcal{O}_{L}$ splits completely in $L^{\prime}$; in other words the Frobenius of $p$ with respect to $L^{\prime}/{\mathbb{Q}}$ is the conjugacy class of the 2 elements of order 3. The Chebotarev density theorem gives that $S$ has Dirichlet as well as natural density $1/3$. Put $s(x)=|S\cap[1,x]|$ so that $s(x)=\displaystyle\frac{\pi(x)}{3}+o(\pi(x))$; put $r(x)=s(x)-\displaystyle\frac{\pi(x)}{3}$. Then partial summation gives:

thanks to the Prime Number Theorem and the fact that $r(t)=o(\pi(t))$. This implies

Now observe that if $m\in[1,N]$ is divisible by some $p\in S$ then the equation $aX^{3}-mY^{3}=b$ cannot have an integer equation since $b/a$ is not a cube mod $p$. By sieving [MV06, Chapter 3.2], the number of $m\in[1,N]$ such that $p\nmid m$ for all $p\in S$ is $O\left(\displaystyle\prod_{p\in S,p\leq N}\left(1-\frac{1}{p}\right)N\right)$ and we use (1) to finish the proof. ∎

The upper bound in Theorem 1.2 follows from Propositions 2.1 and 2.2. For the lower bound, first we consider $S_{k}$ and the equation $kX^{3}-mY^{3}=1$. We can always take $m=kX_{0}^{3}-1$ for $X_{0}\in[1,(N/k)^{1/3}]$ so that the above equation has a solution $(X_{0},1)$. We need that $k^{2}m\not\equiv\pm 1$ mod $9$ and $m$ is square-free for a positive proportion of such $X_{0}$. A direct calculation shows that regardless of the possibility of $k$ mod $9$, we can always find $r\in\{0,\ldots,8\}$ such that $k^{2}(kr^{3}-1)\not\equiv\pm 1$ mod $9$. We now choose $X_{0}$ of the form $X_{0}=9t+r$ for $t\in[1,cN^{1/3}]$ where $c$ is a positive constant depending only on $k$. By classical results of Hooley [Hoo67, Hoo68] (also see [Gra98] for a more general result assuming ABC), the irreducible cubic polynomial $f(t)=k(9t+r)^{3}-1\in{\mathbb{Z}}[t]$ admits square-free values for at least $c^{\prime}cN^{1/3}$ many $t$ where $c^{\prime}>0$ depends only on $k$ and $r$. The proof of $N^{1/3}\ll_{k}|\mathcal{T}_{k}\cap[1,N]|$ is completely similar. ∎

Let $\delta$ be a small positive number depending on $\epsilon$ that will be specified later. The implicit constants in this proof depends only on $a$, $b$, and $\delta$. Except for the finitely many $m$ for which $b/m$ is the cube of an integer, any $(m,X_{0},Y_{0})$ such that $aX_{0}^{3}-mY_{0}^{3}=b$, $|m|\leq N$, and $X_{0},Y_{0}\in{\mathbb{Z}}$ satisfies $mX_{0}Y_{0}\neq 0$. An immediate consequence of ABC gives:

From $aX_{0}^{3}-mY_{0}^{3}=b$, we get $|Y_{0}|\ll|m^{-1/3}X_{0}|$. Combining with the above, we get: $|X_{0}|^{3}\ll|m^{2/3}X_{0}^{2}|^{1+\delta}$. Put $\delta^{\prime}=\displaystyle\frac{2(1+\delta)}{3(1-2\delta)}-\frac{2}{3}$ so that we have:

Therefore, in order to estimate the number of $m$, we estimate the number of pairs $(X_{0},Y_{0})$ with $X_{0}=O(N^{(2/3)+\delta^{\prime}})$ and $Y_{0}=O(N^{(1/3)+\delta^{\prime}})$ such that $\displaystyle\frac{aX_{0}^{3}-b}{Y_{0}^{3}}$ is an integer in $[-N,N]$.

Fix such a $Y_{0}$, we have the obvious bound $|X_{0}|\ll N^{1/3}|Y_{0}|$ and we now study the congurence $aX_{0}^{3}\equiv b$ mod $Y_{0}^{3}$. Let $p$ be a prime divisor of $Y_{0}$ and let $d>0$ such that $p^{d}\parallel Y_{0}$. If $p\nmid ab$, the equation $aX^{3}\equiv b$ mod $p^{3d}$ has at most 3 solutions in ${\mathbb{Z}}/p^{3d}{\mathbb{Z}}$ thanks to the structure of $({\mathbb{Z}}/p^{3d}{\mathbb{Z}})^{*}$. If $p\mid ab$ and $3d>\max\{v_{p}(a),v_{p}(b)\}$, for the above congruence equation to have a solution, we must have that $v_{p}(b)-v_{p}(a)$ is a positive integer divisible by $3$ and any solution must have the form $p^{(v_{p}(b)-v_{p}(a))/3}x$ where $x$ satisfies $x^{3}\equiv u$ mod $p^{3d-(v_{p}(b)-v_{p}(a))}$ and $u$ is given by $\displaystyle\frac{b}{a}=p^{v_{p}(b)-v_{p}(a)}u$. Again, there are at most 3 solutions in this case. In conclusion, there are $O(3^{\omega(Y_{0})})$ many solutions in ${\mathbb{Z}}/Y_{0}^{3}{\mathbb{Z}}$ of the equation $aX^{3}\equiv b$ mod $Y_{0}^{3}$; here $\omega(n)$ denotes the number of distinct prime factors of $n$.

Overall, the number of pairs $(X_{0},Y_{0})$ is at most:

This is $O(N^{((1/3)+\delta^{\prime})(1+\delta^{\prime})})$ since $3^{\omega(Y_{0})}$ is dominated by $|Y_{0}|^{\delta^{\prime}}$. Now choosing $\delta$ sufficiently small so that $((1/3)+\delta^{\prime})(1+\delta^{\prime})<(1/3)+\epsilon$ and we get the desired conclusion. ∎

Write $g=g_{1}\cdots g_{u}$ and $h=h_{1}\cdots h_{v}$ where the $g_{i}$’s and $h_{j}$’s are irreducible over $F$. Let $n$ be the largest non-negative integer such that both $gX^{3}$ and $hY^{3}$ are $p^{n}$-th power of some element of $F[t]$. Write $gX^{3}=\tilde{X}^{p^{n}}$ and $hY^{3}=\tilde{Y}^{p^{n}}$, we have that $\tilde{X}-\tilde{Y}\in F^{*}$ and at least one of the derivatives $\tilde{X}^{\prime}$ and $\tilde{Y}^{\prime}$ is non-zero.

Since $p\neq 3$, from $gX^{3}=\tilde{X}^{p^{n}}$ and $hY^{3}=\tilde{Y}^{p^{n}}$ we can express $\tilde{X}$ and $\tilde{Y}$ as:

where the $b_{i}$’s and $c_{j}$’s are positive integer, $\gcd(X_{0},g_{1}\cdots g_{u})=\gcd(Y_{0},h_{1}\cdots h_{v})=1$, and $b_{i}p^{n}-1\equiv c_{j}p^{n}-1\equiv 0$ mod $3$ for $1\leq i\leq u$ and $1\leq j\leq v$.

Hence the $b_{i}$’s and $c_{j}$’s have the same non-zero congruence mod $3$. Depending on whether they are $1$ mod $3$ or respectively $2$ mod $3$, we can write

or respectively

We need to rule out the second possibility above. Indeed, suppose it happens then the Mason-Stothers theorem implies:

contradiction since the RHS is strictly smaller than the average of the 2 terms in the LHS. This finishes the proof. ∎

All the implicit constants in this proof depend only on $q$. For every positive interger $k$, let $d_{k}$ be the degree of the product of all monic irreducible polynomials of degree at most $k$. Since there are $(q^{n}+O(q^{n/2}))/n$ monic irreducible polynomials of degree $n$, we have

Now we choose the smallest $k$ such that $N\leq d_{k}$. This implies that $\omega(P)$ is at most the number of monic irreducible polynomials of degree at most $k$:

From the above formula for $d_{k}$ and the choice of $k$, we have that $q^{k}\ll N$ and $k\ll\log N$; this finishes the proof. ∎

Let $s_{N}$ be $\sum 3^{\omega(P)}$ where $P$ ranges over all monic polynomials of degree equal to $N$. It suffices to show $s_{N}=O(N^{2}q^{N})$. The generating series $\displaystyle\sum_{n}s_{n}T^{n}$ has the Euler product:

where $P$ ranges over all the monic irreducible polynomials over $F$. The denominator is simply the zeta-function $1/(1-qT)$ while the coefficients of the numerator are bounded above by the coefficients of $\displaystyle\prod_{P}(1+t^{\deg(P)}+t^{2\deg(P)}+\ldots)^{2}=\frac{1}{(1-qT)^{2}}$. Therefore the $s_{N}$’s are bounded above by the coefficients of $1/(1-qT)^{3}$ and this finishes the proof. ∎

For the lower bound, we simply study the equation $gX^{3}-hY^{3}=1$. Either by adapting the arguments in [Hoo67, Hoo68, Gra98] or using a general result of Poonen [Poo03, Theorem 3.4] which is valid for a multivariable polynomial, we have that for a positive proportion of the polynomials $X\in F[t]$ with $\deg(X)\leq(N-\deg(g))/3$, we have that $gX^{3}-1$ is squarefree; we now simply take $Y=1$ and $h=gX^{3}-1$ for those $X$’s. This proves the lower bound.

For the upper bound, we prove that for an arbitrary $\alpha\in F^{*}$, there are $O(q^{N/3})$ many $h$ of degree at most $N$ such that the equation $gX^{3}-hY^{3}=\alpha$ has a solution $X,Y\in F[t]$; since $\deg(g)>0$ we must have that $Y\neq 0$. By Lemma 3.3, we may assume that at least one of the derivatives $(gX^{3})^{\prime}$ and $(hY^{3})^{\prime}$ is non-zero. The Mason-Stothers theorem yields:

The first inequality gives $\deg(Y)\leq\deg(X)/2+O(1)$ then we use this and the second inequality to obtain $\deg(X)\leq\frac{2}{3}\deg(h)+O(1)$. Then it follows that $\deg(Y)\leq\frac{N}{3}+O(1)$.

We now count the number of pairs $(X,Y)$ with $Y\neq 0$ such that $\deg(Y)\leq\frac{N}{3}+O(1)$ and $\frac{gX^{3}-\alpha}{Y^{3}}$ is a polynomial in $F[t]$ of degree at most $N$. Hence $\deg(X)\leq\deg(Y)+(N/3)-\deg(g)$. Arguing as before, for each prime power factor $P^{n}$ of $Y$, the congruence equation $gX^{3}-\alpha=0$ mod $P^{3n}$ has at most 3 solutions mod $P^{3n}$. Therefore by the Chinese Remainder Theorem, the congruence equation $gX^{3}-\alpha=0$ mod $Y^{3}$ has at most $3^{\omega(Y)}$ solutions mod $Y^{3}$. Therefore once $Y$ is fixed, there are at most