Assume $p\neq 3$ is a prime. Let $f$ be a modular function on $X_0(3p)$ whose divisor is support on the upper half plane (i.e. $f$ has no zeros and poles at the cusps). If the divisor $D(f)=3D$ where $D$ is another degree zero divisor on $X_0(3p)$. Can we conclude that the cubic root $f^{1/3}(x)$ of $f$ is also a modular function on $X_0(3p)$ or some other congruence subgroup ?
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$\begingroup$ Not necessarily. If $f^{1/3}$ is a rational function, its divisor $D$ is linearly equivalent to 0. But the class of $D$ could be a nonzero (3-torsion) point in the Jacobian. $\endgroup$abx– abx2025-04-14 06:55:46 +00:00Commented Apr 14 at 6:55
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$\begingroup$ But $f^{1/3}$ maybe modular function on $X_0(9p)$ or some other index $3$ congruence subgroup of $\Gamma_0(3p)$, right ? @abx $\endgroup$yhb– yhb2025-04-14 07:08:31 +00:00Commented Apr 14 at 7:08
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$\begingroup$ Or maybe we can ask whether all the index 3 subgroups of $\Gamma_0(3p)$ are congruence subgroups ? $\endgroup$yhb– yhb2025-04-14 07:17:47 +00:00Commented Apr 14 at 7:17
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$\begingroup$ Yes, $f^{1/3}$ will be modular on some subgroup (I think you might need index 9). Whether it is a congruence subgroup I don't know. $\endgroup$abx– abx2025-04-14 13:37:32 +00:00Commented Apr 14 at 13:37
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$\begingroup$ Why index 9 ? The function field $\mathbb{C}(X_0(3p))$ will get a cubic extension after adding $f^{1/3}$ if $f^{1/3}\notin \mathbb{C}(X_0(3p))$. @abx $\endgroup$yhb– yhb2025-04-14 14:11:47 +00:00Commented Apr 14 at 14:11
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