Mixing
How to know the j-invariant of the modular elliptic curve from the modular form?
2
$begingroup$
How do people compute the $j$-invariant of an elliptic curve $E$ over $mathbb Q$ from the associated modular form $f=sum_{n=1}^{infty} a_nq^n$? In other words, how to compute (at least giving some estimates) $j(E)$ using $a_n$?
number-theory elliptic-curves modular-forms
asked Jan 17 at 6:34
zzyzzy
2,6161420
$endgroup$
add a comment |
2
$begingroup$
How do people compute the $j$-invariant of an elliptic curve $E$ over $mathbb Q$ from the associated modular form $f=sum_{n=1}^{infty} a_nq^n$? In other words, how to compute (at least giving some estimates) $j(E)$ using $a_n$?
number-theory elliptic-curves modular-forms
asked Jan 17 at 6:34
zzyzzy
2,6161420
$endgroup$
1
$begingroup$
Silverman AEC.VIII.8.3 explains the minimal equation, this is an arithmetic choice of curve $E$. The $g_2,g_3$ of the lattice of the newform gives an analytic canonical choice $E'$. There is a 3rd one : from the maps $X_0(N)$ to $X_0(N)_{alg-mathbb{Q}}$ to $J_0(N)_{alg-mathbb{Q}}$ containing the algebraic canonical curve $E''$. I think the point is they know the latter Jacobian has good reduction for $p nmid N$, they transfer it to $E''$, relating $E,E''$ outside of $p | N$. Then they probably make $X_0(N) to J_0(N)_{alg-mathbb{Q}}$ explicit.
$endgroup$
– reuns
Jan 17 at 23:59
add a comment |
2
2
2
1
$begingroup$
How do people compute the $j$-invariant of an elliptic curve $E$ over $mathbb Q$ from the associated modular form $f=sum_{n=1}^{infty} a_nq^n$? In other words, how to compute (at least giving some estimates) $j(E)$ using $a_n$?
number-theory elliptic-curves modular-forms
asked Jan 17 at 6:34
zzyzzy
2,6161420
$endgroup$
How do people compute the $j$-invariant of an elliptic curve $E$ over $mathbb Q$ from the associated modular form $f=sum_{n=1}^{infty} a_nq^n$? In other words, how to compute (at least giving some estimates) $j(E)$ using $a_n$?
number-theory elliptic-curves modular-forms
number-theory elliptic-curves modular-forms
asked Jan 17 at 6:34
zzyzzy
2,6161420
asked Jan 17 at 6:34
zzyzzy
2,6161420
asked Jan 17 at 6:34
zzyzzy
2,6161420
asked Jan 17 at 6:34
zzyzzy
2,6161420
asked Jan 17 at 6:34
zzyzzy
2,6161420
2,6161420
1
$begingroup$
Silverman AEC.VIII.8.3 explains the minimal equation, this is an arithmetic choice of curve $E$. The $g_2,g_3$ of the lattice of the newform gives an analytic canonical choice $E'$. There is a 3rd one : from the maps $X_0(N)$ to $X_0(N)_{alg-mathbb{Q}}$ to $J_0(N)_{alg-mathbb{Q}}$ containing the algebraic canonical curve $E''$. I think the point is they know the latter Jacobian has good reduction for $p nmid N$, they transfer it to $E''$, relating $E,E''$ outside of $p | N$. Then they probably make $X_0(N) to J_0(N)_{alg-mathbb{Q}}$ explicit.
$endgroup$
– reuns
Jan 17 at 23:59
add a comment |
1
$begingroup$
Silverman AEC.VIII.8.3 explains the minimal equation, this is an arithmetic choice of curve $E$. The $g_2,g_3$ of the lattice of the newform gives an analytic canonical choice $E'$. There is a 3rd one : from the maps $X_0(N)$ to $X_0(N)_{alg-mathbb{Q}}$ to $J_0(N)_{alg-mathbb{Q}}$ containing the algebraic canonical curve $E''$. I think the point is they know the latter Jacobian has good reduction for $p nmid N$, they transfer it to $E''$, relating $E,E''$ outside of $p | N$. Then they probably make $X_0(N) to J_0(N)_{alg-mathbb{Q}}$ explicit.
$endgroup$
– reuns
Jan 17 at 23:59
1
1
$begingroup$
Silverman AEC.VIII.8.3 explains the minimal equation, this is an arithmetic choice of curve $E$. The $g_2,g_3$ of the lattice of the newform gives an analytic canonical choice $E'$. There is a 3rd one : from the maps $X_0(N)$ to $X_0(N)_{alg-mathbb{Q}}$ to $J_0(N)_{alg-mathbb{Q}}$ containing the algebraic canonical curve $E''$. I think the point is they know the latter Jacobian has good reduction for $p nmid N$, they transfer it to $E''$, relating $E,E''$ outside of $p | N$. Then they probably make $X_0(N) to J_0(N)_{alg-mathbb{Q}}$ explicit.
$endgroup$
– reuns
Jan 17 at 23:59
$begingroup$
Silverman AEC.VIII.8.3 explains the minimal equation, this is an arithmetic choice of curve $E$. The $g_2,g_3$ of the lattice of the newform gives an analytic canonical choice $E'$. There is a 3rd one : from the maps $X_0(N)$ to $X_0(N)_{alg-mathbb{Q}}$ to $J_0(N)_{alg-mathbb{Q}}$ containing the algebraic canonical curve $E''$. I think the point is they know the latter Jacobian has good reduction for $p nmid N$, they transfer it to $E''$, relating $E,E''$ outside of $p | N$. Then they probably make $X_0(N) to J_0(N)_{alg-mathbb{Q}}$ explicit.
$endgroup$
– reuns
Jan 17 at 23:59
add a comment |
1 Answer
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oldest
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3
$begingroup$
This is not a well-defined question, because the modular form $f$ corresponds to an isogeny class of elliptic curves, not a single elliptic curve; and the elliptic curves in the isogeny class can have different $j$-invariants.
However, if you ask for the $j$-invariant of some elliptic curve in the isogeny class, then this is a very well-studied problem, and there is an effective way of doing so using the period lattice. This is all described comprehensively and beautifully in John Cremona's book Algorithms for modular elliptic curves (available for free online here); the algorithm you're after is described in section 2.14.
answered Jan 17 at 10:41
David LoefflerDavid Loeffler
6,917923
$endgroup$
$begingroup$
If $f in S_2(Gamma_0(n))$ newform with rational coefficient and $F(p) = int_{p_0}^p f(z)dz$ then $gamma mapsto F(gamma(p_0))$ is an homomorphism $Gamma_0(n) to mathbb{C}$ and its image is a lattice $ Lambda=omega_1mathbb{Z}+omega_2 mathbb{Z}$. Is the section 2.10-14 taking some generators $gamma_1,ldots,gamma_m$ of $Gamma_0(n)/[Gamma_0(n),Gamma_0(n)]$ to find $omega_1,omega_2$ as a $mathbb{Z}$-linear combination of the $F(gamma_l(p_0))$ ? Then that $g_2(Lambda) ,g_3(Lambda) $ are (almost) integers is something about minimal models given $j(Lambda) in mathbb{Q}$ ?
$endgroup$
– reuns
Jan 17 at 11:46
$begingroup$
Thank you very much! So one can only get an abelian variety up to isogeny? But it seems in chapter 10 section 2 at wstein.org/edu/Fall2003/252/lectures/all/252.pdf, one can multiply positive integer $n$ to get a real abelian variety $A_f$
$endgroup$
– zzy
Jan 17 at 15:48
$begingroup$
@zzy $E,E', u:E to E',u^*u = m$ elliptic curves and pair of dual isogeny everything defined over a same number field. For almost all primes : everything has good reduction so the curves $tilde{E},tilde{E}'$ are isogeneous and their Frobenius $phi,phi'$ satisfy $u^* phi' = phi u^*, phi'u =uphi,u^*phi u = mphi$ so $phi'^2-t'phi'+q=0$ means $0=u^*phi'^2u-t'u^*phi'u+u^*qu =phi^2 m - t' m phi+qm = 0$ so $t' = t$. Whence $L(s,E) = L(s,E')$ up to finitely many primes, and the corresponding modular forms have the same newform representative.
$endgroup$
– reuns
Jan 17 at 21:59
$begingroup$
@reuns Thank you! But I think somehow there is a canonical choice for the elliptic curve?
$endgroup$
– zzy
Jan 17 at 22:57
add a comment |
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$begingroup$
This is not a well-defined question, because the modular form $f$ corresponds to an isogeny class of elliptic curves, not a single elliptic curve; and the elliptic curves in the isogeny class can have different $j$-invariants.
However, if you ask for the $j$-invariant of some elliptic curve in the isogeny class, then this is a very well-studied problem, and there is an effective way of doing so using the period lattice. This is all described comprehensively and beautifully in John Cremona's book Algorithms for modular elliptic curves (available for free online here); the algorithm you're after is described in section 2.14.
answered Jan 17 at 10:41
David LoefflerDavid Loeffler
6,917923
$endgroup$
$begingroup$
If $f in S_2(Gamma_0(n))$ newform with rational coefficient and $F(p) = int_{p_0}^p f(z)dz$ then $gamma mapsto F(gamma(p_0))$ is an homomorphism $Gamma_0(n) to mathbb{C}$ and its image is a lattice $ Lambda=omega_1mathbb{Z}+omega_2 mathbb{Z}$. Is the section 2.10-14 taking some generators $gamma_1,ldots,gamma_m$ of $Gamma_0(n)/[Gamma_0(n),Gamma_0(n)]$ to find $omega_1,omega_2$ as a $mathbb{Z}$-linear combination of the $F(gamma_l(p_0))$ ? Then that $g_2(Lambda) ,g_3(Lambda) $ are (almost) integers is something about minimal models given $j(Lambda) in mathbb{Q}$ ?
$endgroup$
– reuns
Jan 17 at 11:46
$begingroup$
Thank you very much! So one can only get an abelian variety up to isogeny? But it seems in chapter 10 section 2 at wstein.org/edu/Fall2003/252/lectures/all/252.pdf, one can multiply positive integer $n$ to get a real abelian variety $A_f$
$endgroup$
– zzy
Jan 17 at 15:48
$begingroup$
@zzy $E,E', u:E to E',u^*u = m$ elliptic curves and pair of dual isogeny everything defined over a same number field. For almost all primes : everything has good reduction so the curves $tilde{E},tilde{E}'$ are isogeneous and their Frobenius $phi,phi'$ satisfy $u^* phi' = phi u^*, phi'u =uphi,u^*phi u = mphi$ so $phi'^2-t'phi'+q=0$ means $0=u^*phi'^2u-t'u^*phi'u+u^*qu =phi^2 m - t' m phi+qm = 0$ so $t' = t$. Whence $L(s,E) = L(s,E')$ up to finitely many primes, and the corresponding modular forms have the same newform representative.
$endgroup$
– reuns
Jan 17 at 21:59
$begingroup$
@reuns Thank you! But I think somehow there is a canonical choice for the elliptic curve?
$endgroup$
– zzy
Jan 17 at 22:57
add a comment |
3
$begingroup$
This is not a well-defined question, because the modular form $f$ corresponds to an isogeny class of elliptic curves, not a single elliptic curve; and the elliptic curves in the isogeny class can have different $j$-invariants.
However, if you ask for the $j$-invariant of some elliptic curve in the isogeny class, then this is a very well-studied problem, and there is an effective way of doing so using the period lattice. This is all described comprehensively and beautifully in John Cremona's book Algorithms for modular elliptic curves (available for free online here); the algorithm you're after is described in section 2.14.
answered Jan 17 at 10:41
David LoefflerDavid Loeffler
6,917923
$endgroup$
$begingroup$
If $f in S_2(Gamma_0(n))$ newform with rational coefficient and $F(p) = int_{p_0}^p f(z)dz$ then $gamma mapsto F(gamma(p_0))$ is an homomorphism $Gamma_0(n) to mathbb{C}$ and its image is a lattice $ Lambda=omega_1mathbb{Z}+omega_2 mathbb{Z}$. Is the section 2.10-14 taking some generators $gamma_1,ldots,gamma_m$ of $Gamma_0(n)/[Gamma_0(n),Gamma_0(n)]$ to find $omega_1,omega_2$ as a $mathbb{Z}$-linear combination of the $F(gamma_l(p_0))$ ? Then that $g_2(Lambda) ,g_3(Lambda) $ are (almost) integers is something about minimal models given $j(Lambda) in mathbb{Q}$ ?
$endgroup$
– reuns
Jan 17 at 11:46
$begingroup$
Thank you very much! So one can only get an abelian variety up to isogeny? But it seems in chapter 10 section 2 at wstein.org/edu/Fall2003/252/lectures/all/252.pdf, one can multiply positive integer $n$ to get a real abelian variety $A_f$
$endgroup$
– zzy
Jan 17 at 15:48
$begingroup$
@zzy $E,E', u:E to E',u^*u = m$ elliptic curves and pair of dual isogeny everything defined over a same number field. For almost all primes : everything has good reduction so the curves $tilde{E},tilde{E}'$ are isogeneous and their Frobenius $phi,phi'$ satisfy $u^* phi' = phi u^*, phi'u =uphi,u^*phi u = mphi$ so $phi'^2-t'phi'+q=0$ means $0=u^*phi'^2u-t'u^*phi'u+u^*qu =phi^2 m - t' m phi+qm = 0$ so $t' = t$. Whence $L(s,E) = L(s,E')$ up to finitely many primes, and the corresponding modular forms have the same newform representative.
$endgroup$
– reuns
Jan 17 at 21:59
$begingroup$
@reuns Thank you! But I think somehow there is a canonical choice for the elliptic curve?
$endgroup$
– zzy
Jan 17 at 22:57
add a comment |
3
3
3
$begingroup$
This is not a well-defined question, because the modular form $f$ corresponds to an isogeny class of elliptic curves, not a single elliptic curve; and the elliptic curves in the isogeny class can have different $j$-invariants.
However, if you ask for the $j$-invariant of some elliptic curve in the isogeny class, then this is a very well-studied problem, and there is an effective way of doing so using the period lattice. This is all described comprehensively and beautifully in John Cremona's book Algorithms for modular elliptic curves (available for free online here); the algorithm you're after is described in section 2.14.
answered Jan 17 at 10:41
David LoefflerDavid Loeffler
6,917923
$endgroup$
This is not a well-defined question, because the modular form $f$ corresponds to an isogeny class of elliptic curves, not a single elliptic curve; and the elliptic curves in the isogeny class can have different $j$-invariants.
However, if you ask for the $j$-invariant of some elliptic curve in the isogeny class, then this is a very well-studied problem, and there is an effective way of doing so using the period lattice. This is all described comprehensively and beautifully in John Cremona's book Algorithms for modular elliptic curves (available for free online here); the algorithm you're after is described in section 2.14.
answered Jan 17 at 10:41
David LoefflerDavid Loeffler
6,917923
answered Jan 17 at 10:41
David LoefflerDavid Loeffler
6,917923
answered Jan 17 at 10:41
David LoefflerDavid Loeffler
6,917923
answered Jan 17 at 10:41
David LoefflerDavid Loeffler
6,917923
6,917923
$begingroup$
If $f in S_2(Gamma_0(n))$ newform with rational coefficient and $F(p) = int_{p_0}^p f(z)dz$ then $gamma mapsto F(gamma(p_0))$ is an homomorphism $Gamma_0(n) to mathbb{C}$ and its image is a lattice $ Lambda=omega_1mathbb{Z}+omega_2 mathbb{Z}$. Is the section 2.10-14 taking some generators $gamma_1,ldots,gamma_m$ of $Gamma_0(n)/[Gamma_0(n),Gamma_0(n)]$ to find $omega_1,omega_2$ as a $mathbb{Z}$-linear combination of the $F(gamma_l(p_0))$ ? Then that $g_2(Lambda) ,g_3(Lambda) $ are (almost) integers is something about minimal models given $j(Lambda) in mathbb{Q}$ ?
$endgroup$
– reuns
Jan 17 at 11:46
$begingroup$
Thank you very much! So one can only get an abelian variety up to isogeny? But it seems in chapter 10 section 2 at wstein.org/edu/Fall2003/252/lectures/all/252.pdf, one can multiply positive integer $n$ to get a real abelian variety $A_f$
$endgroup$
– zzy
Jan 17 at 15:48
$begingroup$
@zzy $E,E', u:E to E',u^*u = m$ elliptic curves and pair of dual isogeny everything defined over a same number field. For almost all primes : everything has good reduction so the curves $tilde{E},tilde{E}'$ are isogeneous and their Frobenius $phi,phi'$ satisfy $u^* phi' = phi u^*, phi'u =uphi,u^*phi u = mphi$ so $phi'^2-t'phi'+q=0$ means $0=u^*phi'^2u-t'u^*phi'u+u^*qu =phi^2 m - t' m phi+qm = 0$ so $t' = t$. Whence $L(s,E) = L(s,E')$ up to finitely many primes, and the corresponding modular forms have the same newform representative.
$endgroup$
– reuns
Jan 17 at 21:59
$begingroup$
@reuns Thank you! But I think somehow there is a canonical choice for the elliptic curve?
$endgroup$
– zzy
Jan 17 at 22:57
add a comment |
$begingroup$
If $f in S_2(Gamma_0(n))$ newform with rational coefficient and $F(p) = int_{p_0}^p f(z)dz$ then $gamma mapsto F(gamma(p_0))$ is an homomorphism $Gamma_0(n) to mathbb{C}$ and its image is a lattice $ Lambda=omega_1mathbb{Z}+omega_2 mathbb{Z}$. Is the section 2.10-14 taking some generators $gamma_1,ldots,gamma_m$ of $Gamma_0(n)/[Gamma_0(n),Gamma_0(n)]$ to find $omega_1,omega_2$ as a $mathbb{Z}$-linear combination of the $F(gamma_l(p_0))$ ? Then that $g_2(Lambda) ,g_3(Lambda) $ are (almost) integers is something about minimal models given $j(Lambda) in mathbb{Q}$ ?
$endgroup$
– reuns
Jan 17 at 11:46
$begingroup$
Thank you very much! So one can only get an abelian variety up to isogeny? But it seems in chapter 10 section 2 at wstein.org/edu/Fall2003/252/lectures/all/252.pdf, one can multiply positive integer $n$ to get a real abelian variety $A_f$
$endgroup$
– zzy
Jan 17 at 15:48
$begingroup$
@zzy $E,E', u:E to E',u^*u = m$ elliptic curves and pair of dual isogeny everything defined over a same number field. For almost all primes : everything has good reduction so the curves $tilde{E},tilde{E}'$ are isogeneous and their Frobenius $phi,phi'$ satisfy $u^* phi' = phi u^*, phi'u =uphi,u^*phi u = mphi$ so $phi'^2-t'phi'+q=0$ means $0=u^*phi'^2u-t'u^*phi'u+u^*qu =phi^2 m - t' m phi+qm = 0$ so $t' = t$. Whence $L(s,E) = L(s,E')$ up to finitely many primes, and the corresponding modular forms have the same newform representative.
$endgroup$
– reuns
Jan 17 at 21:59
$begingroup$
@reuns Thank you! But I think somehow there is a canonical choice for the elliptic curve?
$endgroup$
– zzy
Jan 17 at 22:57
$begingroup$
If $f in S_2(Gamma_0(n))$ newform with rational coefficient and $F(p) = int_{p_0}^p f(z)dz$ then $gamma mapsto F(gamma(p_0))$ is an homomorphism $Gamma_0(n) to mathbb{C}$ and its image is a lattice $ Lambda=omega_1mathbb{Z}+omega_2 mathbb{Z}$. Is the section 2.10-14 taking some generators $gamma_1,ldots,gamma_m$ of $Gamma_0(n)/[Gamma_0(n),Gamma_0(n)]$ to find $omega_1,omega_2$ as a $mathbb{Z}$-linear combination of the $F(gamma_l(p_0))$ ? Then that $g_2(Lambda) ,g_3(Lambda) $ are (almost) integers is something about minimal models given $j(Lambda) in mathbb{Q}$ ?
$endgroup$
– reuns
Jan 17 at 11:46
$begingroup$
If $f in S_2(Gamma_0(n))$ newform with rational coefficient and $F(p) = int_{p_0}^p f(z)dz$ then $gamma mapsto F(gamma(p_0))$ is an homomorphism $Gamma_0(n) to mathbb{C}$ and its image is a lattice $ Lambda=omega_1mathbb{Z}+omega_2 mathbb{Z}$. Is the section 2.10-14 taking some generators $gamma_1,ldots,gamma_m$ of $Gamma_0(n)/[Gamma_0(n),Gamma_0(n)]$ to find $omega_1,omega_2$ as a $mathbb{Z}$-linear combination of the $F(gamma_l(p_0))$ ? Then that $g_2(Lambda) ,g_3(Lambda) $ are (almost) integers is something about minimal models given $j(Lambda) in mathbb{Q}$ ?
$endgroup$
– reuns
Jan 17 at 11:46
$begingroup$
Thank you very much! So one can only get an abelian variety up to isogeny? But it seems in chapter 10 section 2 at wstein.org/edu/Fall2003/252/lectures/all/252.pdf, one can multiply positive integer $n$ to get a real abelian variety $A_f$
$endgroup$
– zzy
Jan 17 at 15:48
$begingroup$
Thank you very much! So one can only get an abelian variety up to isogeny? But it seems in chapter 10 section 2 at wstein.org/edu/Fall2003/252/lectures/all/252.pdf, one can multiply positive integer $n$ to get a real abelian variety $A_f$
$endgroup$
– zzy
Jan 17 at 15:48
$begingroup$
@zzy $E,E', u:E to E',u^*u = m$ elliptic curves and pair of dual isogeny everything defined over a same number field. For almost all primes : everything has good reduction so the curves $tilde{E},tilde{E}'$ are isogeneous and their Frobenius $phi,phi'$ satisfy $u^* phi' = phi u^*, phi'u =uphi,u^*phi u = mphi$ so $phi'^2-t'phi'+q=0$ means $0=u^*phi'^2u-t'u^*phi'u+u^*qu =phi^2 m - t' m phi+qm = 0$ so $t' = t$. Whence $L(s,E) = L(s,E')$ up to finitely many primes, and the corresponding modular forms have the same newform representative.
$endgroup$
– reuns
Jan 17 at 21:59
$begingroup$
@zzy $E,E', u:E to E',u^*u = m$ elliptic curves and pair of dual isogeny everything defined over a same number field. For almost all primes : everything has good reduction so the curves $tilde{E},tilde{E}'$ are isogeneous and their Frobenius $phi,phi'$ satisfy $u^* phi' = phi u^*, phi'u =uphi,u^*phi u = mphi$ so $phi'^2-t'phi'+q=0$ means $0=u^*phi'^2u-t'u^*phi'u+u^*qu =phi^2 m - t' m phi+qm = 0$ so $t' = t$. Whence $L(s,E) = L(s,E')$ up to finitely many primes, and the corresponding modular forms have the same newform representative.
$endgroup$
– reuns
Jan 17 at 21:59
$begingroup$
@reuns Thank you! But I think somehow there is a canonical choice for the elliptic curve?
$endgroup$
– zzy
Jan 17 at 22:57
$begingroup$
@reuns Thank you! But I think somehow there is a canonical choice for the elliptic curve?
$endgroup$
– zzy
Jan 17 at 22:57
add a comment |
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1
$begingroup$
Silverman AEC.VIII.8.3 explains the minimal equation, this is an arithmetic choice of curve $E$. The $g_2,g_3$ of the lattice of the newform gives an analytic canonical choice $E'$. There is a 3rd one : from the maps $X_0(N)$ to $X_0(N)_{alg-mathbb{Q}}$ to $J_0(N)_{alg-mathbb{Q}}$ containing the algebraic canonical curve $E''$. I think the point is they know the latter Jacobian has good reduction for $p nmid N$, they transfer it to $E''$, relating $E,E''$ outside of $p | N$. Then they probably make $X_0(N) to J_0(N)_{alg-mathbb{Q}}$ explicit.
$endgroup$
– reuns
Jan 17 at 23:59