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Shimrod
Shimrod
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I have a question which is fairly elementary, but first I must provide relevant context. Without it, my question would seem rather arbitrary and scarcely interesting. Note also that my question can be answered without reading what follows.

Let $j(\omega)$ be the modular invariant, $\Gamma$ be the modular group and $H$ the upper half-plane. Define $\gamma_2(\omega)=\sqrt[3]{j(\omega)}$ using the cube root which is real on the imaginary axis.

Let $\tau\in H$ satisfy \begin{equation} A\tau^2+B\tau+C=0,\ (A,B,C)=1. \end{equation}

By Class field theory the value $j(\tau)$ is an algebraic integer of degree exactly $h(D)$, where $D=B^2-4AC$. Since $\gamma_2(\tau)^3-j(\tau)=0$ one would expect the degree of $\gamma_2(\tau)$ to be $3h(D)$. But it turns out that, if $(A,3)=1$ and $3\mid B$, the degree of $\gamma_2(\tau)$ is also $h(D)$.

Now I indicate how this result can be proved.

Set $\varepsilon=e^{-\frac{2\pi i}{3}}$. It can be shown that for $\bigl( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \bigr)\in\Gamma$ we have

\begin{equation} \gamma_2{\left(\frac{a\omega+b}{c\omega+d}\right)}=\varepsilon^{ab-ac+cd-a^2cd}\gamma_2(\omega). \end{equation} This implies that $\gamma_2(\omega)=\gamma_2(\tau)$ if and only if $A\omega=\tau$, with $A=\bigl( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \bigr)\in\Gamma$ satisfying $ab-ac+cd-a^2cd \equiv 0 \ (\textrm{mod}\ 3)$, except when $\omega \equiv e^{-\frac{2\pi i}{3}} \ (\textrm{mod}\ \Gamma)$, in which case $\gamma_2(\omega)=0$.

Next define \begin{equation} S(n,3)=\bigg\lbrace\begin{pmatrix} r & s\\ 0 & t \end{pmatrix}:rt=n,r>0,(r,s,t)=1,3\mid s,0\leq s <3t \bigg\rbrace. \end{equation} Let $\omega$ be in the upper half-plane $H$. Consider the polynomial \begin{equation} \phi_{n,\omega}(x)=\prod_{M\in S(n,3)}\left(x-\gamma_2(M\omega)\gamma_2(\omega)^{-n} \right). \end{equation} It takes some work to show that the coefficients of this polynomial are invariant under $\Gamma$, and therefore rational functions in $j(\omega)$. After additional work we conclude that they are in fact polynomials with integral coefficients. In this way we get a polynomial $\Phi_n(x,y)\in \mathbb Z[x,y]$ such that $\Phi_n(x,j(\omega))=\phi_{n,\omega}(x)$. Applying a suitable substitution we obtain another polynomial $\Psi_n(x,y)\in \mathbb Z[x,y]$ satisfying for all $0\neq z \in \mathbb C$

  • $\Psi_n(z,z)=0$ is equivalent to $z=\gamma_2(\omega)=\gamma_2(M\omega)$ for some $M\in S(n,3)$ and $\omega\in H$.

Thus if $0\neq z \in \mathbb C$ is a root of $\Psi_n(x,x)=0$ then we must have for some $\omega$ and $M=\bigl(\begin{smallmatrix} r & s \\ 0 & t \end{smallmatrix} \bigr)\in S(n,3)$ \begin{equation} z=\gamma_2(\omega)=\gamma_2(M\omega) \end{equation} and consequently by the fourth paragraph \begin{equation} \frac{a\omega+b}{c\omega+d}=\frac{r\omega+s}{t} \end{equation} where $a,b,c,d$ are integers such that $ad-bc=1$ and $ab-ac+cd-a^2cd \equiv 0 \ (\textrm{mod}\ 3)$. Hence $\omega$ is a root of a quadratic equation with integral coefficients, and if $n$ is suitably chosen, then the middle coefficient will be divisible by $3$. Conversely if we can find $\bigl( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \bigr)$ and $\bigl(\begin{smallmatrix} r & s \\ 0 & t \end{smallmatrix} \bigr) $ with these properties, $\gamma_2(\omega)$ will be a root of $\Psi_n(x,x)=0$

On the other hand the roots of the equation \begin{equation} x^3-j(\omega)=0 \end{equation} are \begin{equation} \gamma_2(\omega),\ e^{-2\pi i/3}\gamma_2(\omega)=\gamma_2(\omega+1),\ e^{2\pi i/3}\gamma_2(\omega)=\gamma_2(\omega-1) \end{equation}

If we could guarantee that the only common root of $\Psi_n(x,x)$ and $x^3-j(\omega)$ is $\gamma_2(\omega)$, then we would know that $\gamma_2(\omega)$ and $j(\omega)$ had the same degree. But if $A$ is not divisible by $3$, $B$ is divisible by $3$ and \begin{equation} A\omega^2+B\omega+C=0,\ (A,B,C)=1. \end{equation} then it is easy to see that the numbers $\omega + 1$ and $\omega -1$ satisfy quadratic equations whose middle coefficients are not divisible by $3$. Therefore $\gamma_2(\omega)$ is the only common root and we are done.

Finally, my question is this: given $\omega$$\omega\in H$ such that $\omega^2+3\omega+\frac{9-D}{4}=0$ where $D\equiv 5 \ (\textrm{mod} \ 8)$ and $D$$D<0$ is not divisible by 3, how can I find $n\equiv -1 \ (\textrm{mod}\ 3)$ and $M\in S(n,3)$ such that $$A\omega=M\omega$$ holds for some $A=\bigl( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \bigr)\in\Gamma$ satisfying $ab-ac+cd-a^2cd \equiv 0 \ (\textrm{mod}\ 3)$?

I have a question which is fairly elementary, but first I must provide relevant context. Without it, my question would seem rather arbitrary and scarcely interesting. Note also that my question can be answered without reading what follows.

Let $j(\omega)$ be the modular invariant, $\Gamma$ be the modular group and $H$ the upper half-plane. Define $\gamma_2(\omega)=\sqrt[3]{j(\omega)}$ using the cube root which is real on the imaginary axis.

Let $\tau\in H$ satisfy \begin{equation} A\tau^2+B\tau+C=0,\ (A,B,C)=1. \end{equation}

By Class field theory the value $j(\tau)$ is an algebraic integer of degree exactly $h(D)$, where $D=B^2-4AC$. Since $\gamma_2(\tau)^3-j(\tau)=0$ one would expect the degree of $\gamma_2(\tau)$ to be $3h(D)$. But it turns out that, if $(A,3)=1$ and $3\mid B$, the degree of $\gamma_2(\tau)$ is also $h(D)$.

Now I indicate how this result can be proved.

Set $\varepsilon=e^{-\frac{2\pi i}{3}}$. It can be shown that for $\bigl( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \bigr)\in\Gamma$ we have

\begin{equation} \gamma_2{\left(\frac{a\omega+b}{c\omega+d}\right)}=\varepsilon^{ab-ac+cd-a^2cd}\gamma_2(\omega). \end{equation} This implies that $\gamma_2(\omega)=\gamma_2(\tau)$ if and only if $A\omega=\tau$, with $A=\bigl( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \bigr)\in\Gamma$ satisfying $ab-ac+cd-a^2cd \equiv 0 \ (\textrm{mod}\ 3)$, except when $\omega \equiv e^{-\frac{2\pi i}{3}} \ (\textrm{mod}\ \Gamma)$, in which case $\gamma_2(\omega)=0$.

Next define \begin{equation} S(n,3)=\bigg\lbrace\begin{pmatrix} r & s\\ 0 & t \end{pmatrix}:rt=n,r>0,(r,s,t)=1,3\mid s,0\leq s <3t \bigg\rbrace. \end{equation} Let $\omega$ be in the upper half-plane $H$. Consider the polynomial \begin{equation} \phi_{n,\omega}(x)=\prod_{M\in S(n,3)}\left(x-\gamma_2(M\omega)\gamma_2(\omega)^{-n} \right). \end{equation} It takes some work to show that the coefficients of this polynomial are invariant under $\Gamma$, and therefore rational functions in $j(\omega)$. After additional work we conclude that they are in fact polynomials with integral coefficients. In this way we get a polynomial $\Phi_n(x,y)\in \mathbb Z[x,y]$ such that $\Phi_n(x,j(\omega))=\phi_{n,\omega}(x)$. Applying a suitable substitution we obtain another polynomial $\Psi_n(x,y)\in \mathbb Z[x,y]$ satisfying for all $0\neq z \in \mathbb C$

  • $\Psi_n(z,z)=0$ is equivalent to $z=\gamma_2(\omega)=\gamma_2(M\omega)$ for some $M\in S(n,3)$ and $\omega\in H$.

Thus if $0\neq z \in \mathbb C$ is a root of $\Psi_n(x,x)=0$ then we must have for some $\omega$ and $M=\bigl(\begin{smallmatrix} r & s \\ 0 & t \end{smallmatrix} \bigr)\in S(n,3)$ \begin{equation} z=\gamma_2(\omega)=\gamma_2(M\omega) \end{equation} and consequently by the fourth paragraph \begin{equation} \frac{a\omega+b}{c\omega+d}=\frac{r\omega+s}{t} \end{equation} where $a,b,c,d$ are integers such that $ad-bc=1$ and $ab-ac+cd-a^2cd \equiv 0 \ (\textrm{mod}\ 3)$. Hence $\omega$ is a root of a quadratic equation with integral coefficients, and if $n$ is suitably chosen, then the middle coefficient will be divisible by $3$. Conversely if we can find $\bigl( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \bigr)$ and $\bigl(\begin{smallmatrix} r & s \\ 0 & t \end{smallmatrix} \bigr) $ with these properties, $\gamma_2(\omega)$ will be a root of $\Psi_n(x,x)=0$

On the other hand the roots of the equation \begin{equation} x^3-j(\omega)=0 \end{equation} are \begin{equation} \gamma_2(\omega),\ e^{-2\pi i/3}\gamma_2(\omega)=\gamma_2(\omega+1),\ e^{2\pi i/3}\gamma_2(\omega)=\gamma_2(\omega-1) \end{equation}

If we could guarantee that the only common root of $\Psi_n(x,x)$ and $x^3-j(\omega)$ is $\gamma_2(\omega)$, then we would know that $\gamma_2(\omega)$ and $j(\omega)$ had the same degree. But if $A$ is not divisible by $3$, $B$ is divisible by $3$ and \begin{equation} A\omega^2+B\omega+C=0,\ (A,B,C)=1. \end{equation} then it is easy to see that the numbers $\omega + 1$ and $\omega -1$ satisfy quadratic equations whose middle coefficients are not divisible by $3$. Therefore $\gamma_2(\omega)$ is the only common root and we are done.

Finally, my question is this: given $\omega$ such that $\omega^2+3\omega+\frac{9-D}{4}=0$ where $D\equiv 5 \ (\textrm{mod} \ 8)$ and $D$ is not divisible by 3, how can I find $n\equiv -1 \ (\textrm{mod}\ 3)$ and $M\in S(n,3)$ such that $$A\omega=M\omega$$ holds for some $A=\bigl( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \bigr)\in\Gamma$ satisfying $ab-ac+cd-a^2cd \equiv 0 \ (\textrm{mod}\ 3)$?

I have a question which is fairly elementary, but first I must provide relevant context. Without it, my question would seem rather arbitrary and scarcely interesting. Note also that my question can be answered without reading what follows.

Let $j(\omega)$ be the modular invariant, $\Gamma$ be the modular group and $H$ the upper half-plane. Define $\gamma_2(\omega)=\sqrt[3]{j(\omega)}$ using the cube root which is real on the imaginary axis.

Let $\tau\in H$ satisfy \begin{equation} A\tau^2+B\tau+C=0,\ (A,B,C)=1. \end{equation}

By Class field theory the value $j(\tau)$ is an algebraic integer of degree exactly $h(D)$, where $D=B^2-4AC$. Since $\gamma_2(\tau)^3-j(\tau)=0$ one would expect the degree of $\gamma_2(\tau)$ to be $3h(D)$. But it turns out that, if $(A,3)=1$ and $3\mid B$, the degree of $\gamma_2(\tau)$ is also $h(D)$.

Now I indicate how this result can be proved.

Set $\varepsilon=e^{-\frac{2\pi i}{3}}$. It can be shown that for $\bigl( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \bigr)\in\Gamma$ we have

\begin{equation} \gamma_2{\left(\frac{a\omega+b}{c\omega+d}\right)}=\varepsilon^{ab-ac+cd-a^2cd}\gamma_2(\omega). \end{equation} This implies that $\gamma_2(\omega)=\gamma_2(\tau)$ if and only if $A\omega=\tau$, with $A=\bigl( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \bigr)\in\Gamma$ satisfying $ab-ac+cd-a^2cd \equiv 0 \ (\textrm{mod}\ 3)$, except when $\omega \equiv e^{-\frac{2\pi i}{3}} \ (\textrm{mod}\ \Gamma)$, in which case $\gamma_2(\omega)=0$.

Next define \begin{equation} S(n,3)=\bigg\lbrace\begin{pmatrix} r & s\\ 0 & t \end{pmatrix}:rt=n,r>0,(r,s,t)=1,3\mid s,0\leq s <3t \bigg\rbrace. \end{equation} Let $\omega$ be in the upper half-plane $H$. Consider the polynomial \begin{equation} \phi_{n,\omega}(x)=\prod_{M\in S(n,3)}\left(x-\gamma_2(M\omega)\gamma_2(\omega)^{-n} \right). \end{equation} It takes some work to show that the coefficients of this polynomial are invariant under $\Gamma$, and therefore rational functions in $j(\omega)$. After additional work we conclude that they are in fact polynomials with integral coefficients. In this way we get a polynomial $\Phi_n(x,y)\in \mathbb Z[x,y]$ such that $\Phi_n(x,j(\omega))=\phi_{n,\omega}(x)$. Applying a suitable substitution we obtain another polynomial $\Psi_n(x,y)\in \mathbb Z[x,y]$ satisfying for all $0\neq z \in \mathbb C$

  • $\Psi_n(z,z)=0$ is equivalent to $z=\gamma_2(\omega)=\gamma_2(M\omega)$ for some $M\in S(n,3)$ and $\omega\in H$.

Thus if $0\neq z \in \mathbb C$ is a root of $\Psi_n(x,x)=0$ then we must have for some $\omega$ and $M=\bigl(\begin{smallmatrix} r & s \\ 0 & t \end{smallmatrix} \bigr)\in S(n,3)$ \begin{equation} z=\gamma_2(\omega)=\gamma_2(M\omega) \end{equation} and consequently by the fourth paragraph \begin{equation} \frac{a\omega+b}{c\omega+d}=\frac{r\omega+s}{t} \end{equation} where $a,b,c,d$ are integers such that $ad-bc=1$ and $ab-ac+cd-a^2cd \equiv 0 \ (\textrm{mod}\ 3)$. Hence $\omega$ is a root of a quadratic equation with integral coefficients, and if $n$ is suitably chosen, then the middle coefficient will be divisible by $3$. Conversely if we can find $\bigl( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \bigr)$ and $\bigl(\begin{smallmatrix} r & s \\ 0 & t \end{smallmatrix} \bigr) $ with these properties, $\gamma_2(\omega)$ will be a root of $\Psi_n(x,x)=0$

On the other hand the roots of the equation \begin{equation} x^3-j(\omega)=0 \end{equation} are \begin{equation} \gamma_2(\omega),\ e^{-2\pi i/3}\gamma_2(\omega)=\gamma_2(\omega+1),\ e^{2\pi i/3}\gamma_2(\omega)=\gamma_2(\omega-1) \end{equation}

If we could guarantee that the only common root of $\Psi_n(x,x)$ and $x^3-j(\omega)$ is $\gamma_2(\omega)$, then we would know that $\gamma_2(\omega)$ and $j(\omega)$ had the same degree. But if $A$ is not divisible by $3$, $B$ is divisible by $3$ and \begin{equation} A\omega^2+B\omega+C=0,\ (A,B,C)=1. \end{equation} then it is easy to see that the numbers $\omega + 1$ and $\omega -1$ satisfy quadratic equations whose middle coefficients are not divisible by $3$. Therefore $\gamma_2(\omega)$ is the only common root and we are done.

Finally, my question is this: given $\omega\in H$ such that $\omega^2+3\omega+\frac{9-D}{4}=0$ where $D\equiv 5 \ (\textrm{mod} \ 8)$ and $D<0$ is not divisible by 3, how can I find $n\equiv -1 \ (\textrm{mod}\ 3)$ and $M\in S(n,3)$ such that $$A\omega=M\omega$$ holds for some $A=\bigl( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \bigr)\in\Gamma$ satisfying $ab-ac+cd-a^2cd \equiv 0 \ (\textrm{mod}\ 3)$?

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Shimrod
Shimrod
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  • 22

The degree of the cube root of the $j$-invariant

I have a question which is fairly elementary, but first I must provide relevant context. Without it, my question would seem rather arbitrary and scarcely interesting. Note also that my question can be answered without reading what follows.

Let $j(\omega)$ be the modular invariant, $\Gamma$ be the modular group and $H$ the upper half-plane. Define $\gamma_2(\omega)=\sqrt[3]{j(\omega)}$ using the cube root which is real on the imaginary axis.

Let $\tau\in H$ satisfy \begin{equation} A\tau^2+B\tau+C=0,\ (A,B,C)=1. \end{equation}

By Class field theory the value $j(\tau)$ is an algebraic integer of degree exactly $h(D)$, where $D=B^2-4AC$. Since $\gamma_2(\tau)^3-j(\tau)=0$ one would expect the degree of $\gamma_2(\tau)$ to be $3h(D)$. But it turns out that, if $(A,3)=1$ and $3\mid B$, the degree of $\gamma_2(\tau)$ is also $h(D)$.

Now I indicate how this result can be proved.

Set $\varepsilon=e^{-\frac{2\pi i}{3}}$. It can be shown that for $\bigl( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \bigr)\in\Gamma$ we have

\begin{equation} \gamma_2{\left(\frac{a\omega+b}{c\omega+d}\right)}=\varepsilon^{ab-ac+cd-a^2cd}\gamma_2(\omega). \end{equation} This implies that $\gamma_2(\omega)=\gamma_2(\tau)$ if and only if $A\omega=\tau$, with $A=\bigl( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \bigr)\in\Gamma$ satisfying $ab-ac+cd-a^2cd \equiv 0 \ (\textrm{mod}\ 3)$, except when $\omega \equiv e^{-\frac{2\pi i}{3}} \ (\textrm{mod}\ \Gamma)$, in which case $\gamma_2(\omega)=0$.

Next define \begin{equation} S(n,3)=\bigg\lbrace\begin{pmatrix} r & s\\ 0 & t \end{pmatrix}:rt=n,r>0,(r,s,t)=1,3\mid s,0\leq s <3t \bigg\rbrace. \end{equation} Let $\omega$ be in the upper half-plane $H$. Consider the polynomial \begin{equation} \phi_{n,\omega}(x)=\prod_{M\in S(n,3)}\left(x-\gamma_2(M\omega)\gamma_2(\omega)^{-n} \right). \end{equation} It takes some work to show that the coefficients of this polynomial are invariant under $\Gamma$, and therefore rational functions in $j(\omega)$. After additional work we conclude that they are in fact polynomials with integral coefficients. In this way we get a polynomial $\Phi_n(x,y)\in \mathbb Z[x,y]$ such that $\Phi_n(x,j(\omega))=\phi_{n,\omega}(x)$. Applying a suitable substitution we obtain another polynomial $\Psi_n(x,y)\in \mathbb Z[x,y]$ satisfying for all $0\neq z \in \mathbb C$

  • $\Psi_n(z,z)=0$ is equivalent to $z=\gamma_2(\omega)=\gamma_2(M\omega)$ for some $M\in S(n,3)$ and $\omega\in H$.

Thus if $0\neq z \in \mathbb C$ is a root of $\Psi_n(x,x)=0$ then we must have for some $\omega$ and $M=\bigl(\begin{smallmatrix} r & s \\ 0 & t \end{smallmatrix} \bigr)\in S(n,3)$ \begin{equation} z=\gamma_2(\omega)=\gamma_2(M\omega) \end{equation} and consequently by the fourth paragraph \begin{equation} \frac{a\omega+b}{c\omega+d}=\frac{r\omega+s}{t} \end{equation} where $a,b,c,d$ are integers such that $ad-bc=1$ and $ab-ac+cd-a^2cd \equiv 0 \ (\textrm{mod}\ 3)$. Hence $\omega$ is a root of a quadratic equation with integral coefficients, and if $n$ is suitably chosen, then the middle coefficient will be divisible by $3$. Conversely if we can find $\bigl( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \bigr)$ and $\bigl(\begin{smallmatrix} r & s \\ 0 & t \end{smallmatrix} \bigr) $ with these properties, $\gamma_2(\omega)$ will be a root of $\Psi_n(x,x)=0$

On the other hand the roots of the equation \begin{equation} x^3-j(\omega)=0 \end{equation} are \begin{equation} \gamma_2(\omega),\ e^{-2\pi i/3}\gamma_2(\omega)=\gamma_2(\omega+1),\ e^{2\pi i/3}\gamma_2(\omega)=\gamma_2(\omega-1) \end{equation}

If we could guarantee that the only common root of $\Psi_n(x,x)$ and $x^3-j(\omega)$ is $\gamma_2(\omega)$, then we would know that $\gamma_2(\omega)$ and $j(\omega)$ had the same degree. But if $A$ is not divisible by $3$, $B$ is divisible by $3$ and \begin{equation} A\omega^2+B\omega+C=0,\ (A,B,C)=1. \end{equation} then it is easy to see that the numbers $\omega + 1$ and $\omega -1$ satisfy quadratic equations whose middle coefficients are not divisible by $3$. Therefore $\gamma_2(\omega)$ is the only common root and we are done.

Finally, my question is this: given $\omega$ such that $\omega^2+3\omega+\frac{9-D}{4}=0$ where $D\equiv 5 \ (\textrm{mod} \ 8)$ and $D$ is not divisible by 3, how can I find $n\equiv -1 \ (\textrm{mod}\ 3)$ and $M\in S(n,3)$ such that $$A\omega=M\omega$$ holds for some $A=\bigl( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \bigr)\in\Gamma$ satisfying $ab-ac+cd-a^2cd \equiv 0 \ (\textrm{mod}\ 3)$?