Finding an elliptic curve which has CM by $\mathbb{Q}(\sqrt{-143})$.

Question

Let $k = \mathbb{Q}(\sqrt{-143})$ and $\Lambda \subseteq \mathbb{C}$ the lattice with $\mathbb{Z}$-basis $\{\omega_1, \omega_2\}$, where $\omega_1 = 1$ and $\omega_2 = (1 + \sqrt{-143})/2$. I know that the elliptic curve $E = \mathbb{C}/\Lambda$ defined over $k(j)$, ($j$ is the $j$-invariant of $E$) has CM by $\mathbb{Q}(\sqrt{-143})$, but I would like to write $E$ in Weirstrass form. We have $E: y^2 = 4x^3 -g_2x -g_3$, where $g_2$ and $g_3$ come from the weights $4$ and $6$ Eisenstien series: \begin{align*} g_2 = 60 \sum_{\omega \in \Lambda \atop \omega\neq 0} \frac{1}{\omega^4} \hskip10mm \text{and} \hskip10mm g_3 = 140 \sum_{\omega \in \Lambda \atop \omega\neq 0} \frac{1}{\omega^6} \end{align*} Is there any way to compute these values in this example, or is this just to difficult to compute? I know that some of these values are known like when $k = \mathbb{Q}(i)$ and $\mathbb{Q}(\sqrt{-3})$. I've also looked for this example in LMFDB and didn't come up with anything.

Answer 1

First of all, one can obtain by "numerical guess" the value of the $j$-invariant of the given elliptic curve with associated lattice $$ \Lambda_2=\Bbb Z\cdot \omega_1+\Bbb Z\cdot \omega_2 =\Bbb Z\cdot 1+\Bbb Z\cdot \frac 12(-1+\sqrt{-143})\ . $$ A sign in $\omega_2$ was changed, this is not changing the story. Let $\tau_0$ be $\omega_2$ in alternative notation used below, since most modular functions below want a $\tau$-argument with a corresponding $q$-nome.

Then the computation of $j(\Lambda_0)=j(\tau_0)=:j_0$ already gives valuable information connecting $g_2$ and $g_3$. (The quotient $g_2^3/g_3^2$ , i.e. up to a known factor the quotient $G_2^3/G_3^2$ , i.e. up to a known factor the quotient $E_2^3/E_3^2$ is determined by $j_0$.)

It turns out that $j_0=\approx -20680840776711129.69108257742558129\dots$ is an algebraic integer, even more, $j_1:=j_0/5^3$ is also one. Instead of typing below the minimal polynomial for $j_0$, let us show the "shorter" minimal polynomial $f$ of $j_1$: To get $f$ in sage, the following "guess" code was launched:

sage: tau0 = (-1 + sqrt(-143))/2
sage: j0_approx = elliptic_j(tau0, 100000)
sage: f = (j0_approx / 5^3).algdep(10)
sage: latex(f)

(The guess can be converted into a proof.) The latex form of $f$ is: $$ \begin{aligned} f &= x^{10} + 165446726211309x^{9} \\ &\qquad- 393768093803579723x^{8} \\ &\qquad+ 218981818859893019825993647x^{7} \\ &\qquad- 518576281498995946961126719091x^{6} \\ &\qquad+ 1138907111817275820098533879450173x^{5} \\ &\qquad- 175531791674450070874162808746877854x^{4} \\ &\qquad+ 12666460576084039651631279180004665607x^{3} \\ &\qquad- 133940745852901279669264803596050247482x^{2} \\ &\qquad+ 23211436505681484601511877941186160239x \\ &\qquad- 157724749072656668777243335931768363351 \ . \end{aligned} $$ The roots of $f$, multiplied by $5^3$ are numerically...

sage: [ 5^3*r for r in f.roots(ring=QQbar, multiplicities=False)]
[-2.068084077671113?e16,
 1562.181458481542?,
 3.854446832015781? - 136.0937184913752?*I,
 3.854446832015781? + 136.0937184913752?*I,
 743.999999998961? - 1.438083473805263?e8*I,
 743.999999998961? + 1.438083473805263?e8*I,
 9235.24174842493? - 8467.87383023411?*I,
 9235.24174842493? + 8467.87383023411?*I,
 137988.1586167921? - 237712.6134039348?*I,
 137988.1586167921? + 237712.6134039348?*I]

and $j_0=j(\tau_0)$ is the first entry in the list. We can check for instance:

sage: j0 = 5^3*f.roots(ring=QQbar, multiplicities=False)[0]
....: abs( j0_approx - j0 ) < 10e-30000
True

(So the distance between the numerically computed $j0$, and $5^3$ times the algebraic root of the monic $f$ is smaller than $10^{-30\,000}$.)

Does $j_0$ live in the Hilbert class field $H$ of $K:=\Bbb Q(\sqrt{-143})$? Yes. We build $H$, and ask for the roots of $f$ in $H$. Then we count them, all ten are in this field:

sage: K.<a> = QuadraticField(-143)
sage: H.<b> = K.hilbert_class_field()
sage: len(f.roots(ring=H, multiplicities=False))
10
sage: H.defining_polynomial()
x^10 - 6*x^9 + 12*x^8 - 13*x^7 + 9*x^6 - 3*x^5 - 3*x^4 + 6*x^3 - 6*x^2 + 3*x - 1

One can associate now for the "guessed" algebraic $j_0\in\overline{\Bbb Q}$ the elliptic curve $$ E'\ :\qquad Y^2 + XY = X^3 - \frac{36}{j_0-1728}X - \frac 1{j_0-1728}\ , $$ which has CM by $K=\Bbb Q(\sqrt{-143})$, so $E$ and $E'$ are isomorphic over $\Bbb C$. It remains to compute the needed scaling between $\Lambda_0=\Lambda(E)$ and $\Lambda(E')$. Some checks so far are as follows:

K.<j1> = NumberField(f, embedding=-2e16 / 5^3)
j0 = 5^3 * j1
EE = EllipticCurve(K, [1, 0, 0, -36/(j0 - 1728), -1/(j0 - 1728)])
print(f'Is the j-invariant of EE equal to j0? {bool(EE.j_invariant() == j0)}')

With the above we get...

sage: EE.has_cm()
True
sage: EE.cm_discriminant()
-143
sage: Lambda = EE.period_lattice( K.complex_embeddings()[0] )
sage: Lambda.basis(prec=100)
(75.135968200478499095024772208,
 37.567984100239249547512386104 + 3.1415926535898497483734228045*I)

Then understanding (periods for) $E'$ (in code EE) leads to understanding (periods of) $E$.

---

The above is just a guess, and we have only a check for $j_0\in H$. To work in a more rigurous manner, let us recall the structure, collect some formulas, also introducing notations. I am using

Noriko Yui, Don Zagier, On the singular values of Weber modular functions, Mathematics of computation, 66 (1997)

as a reference. We collect some data: $$ \begin{aligned} d &= -143 \\ \tau_0 &=\omega_2=\frac 12(-1+\sqrt{-143}) &\tau_0 &\text{ is a root of } &Q_0 &=[1,1,36] \\ q_0 &=e^{2\pi i\;\tau_0}\\ q &= q(\tau) =e^{2\pi i\;\tau}\text{ (nome)}\\ \Lambda(\tau) &=\Bbb Z\cdot 1+\Bbb Z\cdot \tau\\[2mm] % g_2(\tau) &= 60 G_4(\tau) = 60\sum_{\substack{\omega\in\Lambda(\tau)\\\omega\ne 0}}\frac 1{\omega^4} & G_4 &=\frac 1{45}\pi^4\; E_4 & E_4(\tau) &= 1+240\sum_{n\ge 1}\frac{n^3q^n}{1-q^n}\\ \\ g_3(\tau) &= 140 G_4(\tau) = 140\sum_{\substack{\omega\in\Lambda(\tau)\\\omega\ne 0}}\frac 1{\omega^6} & G_6 &=\frac 1{945}\pi^6\; E_6 & E_6(\tau) &= 1-504\sum_{n\ge 1}\frac{n^5q^n}{1-q^n}\\ \\[2mm] E(\tau)\ &:\ y^2 =4x^3 -g_2(\tau)x-g_3(\tau)\\ j(\tau) &= j(E(\tau))=\frac {1728}{\displaystyle 1-\frac{E_6(\tau)^2}{E_4(\tau)^3}} \\[2mm] \eta &=\text{ Dedekind eta function} \\[2mm] \mathfrak f(\tau) &= q^{-\frac 1{48}}\prod_{n\ge 0}\left( 1+q^{n-\frac 12}\right) & \mathfrak f(\tau) &= \frac{\eta\left( \frac {\tau+1}2 \right)}{\zeta_{48}\;\eta(\tau)} \\ \mathfrak f_1(\tau) &= q^{-\frac 1{48}}\prod_{n\ge 0}\left( 1-q^{n-\frac 12}\right) & \mathfrak f_1(\tau) &= \frac{\eta\left( \frac \tau2 \right)}{\eta(\tau)} \\ \mathfrak f_2(\tau) &= \sqrt 2\; q^{\frac 2{48}}\prod_{n\ge 0}\left( 1+q^n\right) & \mathfrak f_2(\tau) &= \sqrt 2\frac{\eta(2\tau)}{\eta(\tau)} \\ E_4&=\underbrace{(\ \mathfrak f^8\overbrace{(\mathfrak f_1^8 +\mathfrak f_2^8)}^{=\mathfrak f^8} -\mathfrak f_1^8\mathfrak f_2^8\ )}_{\gamma_2}\; \eta^8 \\ E_6&=\underbrace{(\mathfrak f^8 + \mathfrak f_1^8)(\mathfrak f^8 +\mathfrak f_2^8)(\mathfrak f_1^8-\mathfrak f_2^8)}_{\gamma_3}\; \eta^{12} \\[3mm] f(Q_0) &= \zeta_{48}\mathfrak f_2(\tau_0') \\ W_d &=\text{ minimal pol of }f(Q_0)\ . \end{aligned} $$ Here the function $f$ is as in the Proposition from loc. cit. page 1647. Then for $Q_0=[a,b,c]=[1,1,36]$ we are in the third case, $2\not|a$ and $2|c$, the factor $\varepsilon_d$ is $+1$, so $$ f(\tau_0) = f(Q_0) = \zeta_{48}\;\mathfrak f_2(\tau_0) \ , $$ and the table from loc. cit. gives for $d=-143$ the line: $$ \begin{array}{|c|c|c|c|c|} \hline d & h(d) & W_d(x) & \operatorname{Disc}(W_d) \\\hline -143 & 10 & 1,-3,6,-6,3,3,-9,13,-12,6,-1 & 5^2\cdot 11^4\cdot 13^5\\\hline \end{array} $$ We obtain the specific value of $\mathfrak f_2(\tau_0)$ as $$ \begin{aligned} \mathfrak f_2(\tau_0) &=\frac 1{\zeta_{48}}\cdot\Xi\ ,\qquad\text{ where} \\ \Xi &=\operatorname{Root}\Big( x^{10} - 3x^9 + 6x^8 - 6x^7 + 3x^6 + 3x^5 - 9x^4 + 13x^3 - 12x^2 + 6x - 1 \Big)\ , \\ &\Xi\approx0.29559675118017300\dots \end{aligned} $$ From this information and from $$ \begin{aligned} \mathfrak f\; \mathfrak f_1 \;\mathfrak f_2 &= \sqrt 2\ ,\\ \mathfrak f_1^8 + \mathfrak f_2^8 &= \mathfrak f^8\ ,\\ \end{aligned} $$ we can determine all needed values that show up in $\gamma_2$ and $\gamma_3$, for instance: $$ \begin{aligned} 2^4 &= \mathfrak f^8 \;\mathfrak f_1^8 \;\mathfrak f_2^8 \\ 0&= \mathfrak f^8 - \mathfrak f_2^8 -\mathfrak f_1^8 & 0 &=\mathfrak f_1^8 + \mathfrak f_2^8 -\mathfrak f^8 \\ 0&= \mathfrak f^8 - \mathfrak f_2^8 -\frac {16}{\mathfrak f^8\mathfrak f_2^8} & 0 &=\mathfrak f_1^8 + \mathfrak f_2^8 - \frac {16}{\mathfrak f_1^8\mathfrak f_2^8} \\ 0&= \mathfrak f^{16} - \mathfrak f_2^8\; \mathfrak f^8 -\frac {16}{\mathfrak f_2^8} & 0&= \mathfrak f_1^{16} + \mathfrak f_2^8\; \mathfrak f_1^8 -\frac {16}{\mathfrak f_2^8} \\ \mathfrak f^8 &=\frac 12\left[ + \mathfrak f_2^8 \pm\sqrt{\mathfrak f_2^{16} +\frac{64}{\mathfrak f_2^8}} \right] & \mathfrak f_1^8 &=\frac 12\left[ - \mathfrak f_2^8 \pm\sqrt{\mathfrak f_2^{16} +\frac{64}{\mathfrak f_2^8}}\right] \end{aligned} $$ and now $$ \begin{aligned} \gamma_2 &= \mathfrak f^8\underbrace{(\mathfrak f_1^8 +\mathfrak f_2^8)}_{=\mathfrak f^8} - \mathfrak f_1^8 \;\mathfrak f_2^8 = \mathfrak f^{16} - \mathfrak f_1^8 \;\mathfrak f_2^8 \\ &= \mathfrak f^8 \;\mathfrak f_2^8 +\frac {16}{\mathfrak f_2^8} - \mathfrak f_1^8 \;\mathfrak f_2^8 = (\mathfrak f^8 - \mathfrak f_1^8)\;\mathfrak f_2^8 +\frac {16}{\mathfrak f_2^8} \\ &= \mathfrak f_2^{16} +\frac {16}{\mathfrak f_2^8} = \frac 1{\mathfrak f_2^8}(\mathfrak f_2^{24} + 16) =\frac {\zeta_{48}^8}{\Xi^8}(-\Xi^{24} + 16) \\ &= \zeta_6\left(\frac {16}{\Xi^8} - \Xi^{16}\right)\ . \\[3mm] &\qquad\text{ And in a similar manner:} \\[3mm] \gamma_3 &= (\mathfrak f^8 + \mathfrak f_1^8)(\mathfrak f^8 + \mathfrak f_2^8)(\mathfrak f_1^8 - \mathfrak f_2^8) \\ &=\pm\sqrt{\mathfrak f_2^{16} +\frac {64}{\mathfrak f_2^8}} \cdot\frac 12\left(+3\mathfrak f_2^8 \pm\sqrt{\mathfrak f_2^{16} +\frac {64}{\mathfrak f_2^8}}\right) \cdot\frac 12\left(-3\mathfrak f_2^8 \pm\sqrt{\mathfrak f_2^{16} +\frac {64}{\mathfrak f_2^8}}\right) \\ &=\pm\sqrt{\mathfrak f_2^{16} +\frac {64}{\mathfrak f_2^8}} \cdot\frac 12\cdot\frac 12\cdot 8\left(\mathfrak f_2^{16} -\frac 8{\mathfrak f_2^8}\right) \\ &=-\frac i{\Xi^{12}}(8-\Xi^{24})\sqrt{64-\Xi^{24}}\ . \end{aligned} $$

---

The above formulas for $\gamma_2$, $\gamma_3$ give explicit description of the explicit algebraic integer numbers appearing in the formulas for $E_4$, $E_6$ in front of the $\eta$-factors.

It remains to compute these eta-factors, expected to be some algebraic numbers times gamma factors.

At this point, taking a look at Theorem 9.3 from

Values of the Dedekind Eta Function, Alfred van der Poorten, Kenneth S. Williams

we are expecting a formula of the shape: $$ \eta\left(\frac {-1+\sqrt{-143}}2\right) = e^{2\pi i/48}(2\pi\cdot 143)^{-1/4} \left(\prod_{1\le m\le 143}\gamma\left(\frac m{143}\right)^{\left(\frac dm\right)}\right)^{\displaystyle\frac 2{8\cdot 10}}\cdot \alpha\ , $$ where $\alpha$ is an algebraic number. There is an explicit formula for $\alpha$ in loc. cit. but making it explicit as a number would cost me further hours. Numerical experiments are supporting the following simple value for $\alpha$: $$ \begin{aligned} \alpha^{20} &=\operatorname{Root}\Big( x^{10} + 7 \, x^{9} - 869 \, x^{8} - 5747 \, x^{7} + 260467 \, x^{6} + 3436949 \, x^{5} \\ &\qquad\qquad + 13451413 \, x^{4} + 35857588 \, x^{3} - 13067736 \, x^{2} + 1161137 \, x - 1 \Big)\ ,\\ \alpha &\approx 0.497457562194359381428542\dots \end{aligned} $$ As a further pro argument for the above claim, note that $\alpha^{20}$ is an element in the Hilbert class field $H$ of $K=\Bbb Q(\sqrt{-143})$, with $\alpha$ defined as above.

---

---

It remains to put all together, for $g_2, g_3$ pass to $G_4,G_6$, then to $E_4,E_6$, which are $\gamma_2\eta^8$, $\gamma_3\eta^{12}$, then use the algebraic values for $\gamma_2$, $\gamma_3$ computed above, and the powers of the above $\eta$-value, which in essence involves powers of gamma factors multiplied with algebraic integers related to the Hilbert class field of $K$.