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Franz Lemmermeyer
Franz Lemmermeyer
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Let N be a prime integer. We know that the element c=(0)-(\infty)$c=(0)-(\infty)$ generates the torsion subgroup of J_0(N)$J_0(N)$ and it has order Num( (N-1)/12). Now, there is a natural map $\pi^*:J_0(N) \rightarrow J_1(N)$, coming from the covering map $\pi:X_1(N) \rightarrow X_0(N)$. My question is what is the image of c under this map? Specifically, is it possible for $\pi^*(c)=0$?

Let N be a prime integer. We know that the element c=(0)-(\infty) generates the torsion subgroup of J_0(N) and it has order Num( (N-1)/12). Now, there is a natural map $\pi^*:J_0(N) \rightarrow J_1(N)$, coming from the covering map $\pi:X_1(N) \rightarrow X_0(N)$. My question is what is the image of c under this map? Specifically, is it possible for $\pi^*(c)=0$?

Let N be a prime integer. We know that the element $c=(0)-(\infty)$ generates the torsion subgroup of $J_0(N)$ and it has order Num( (N-1)/12). Now, there is a natural map $\pi^*:J_0(N) \rightarrow J_1(N)$, coming from the covering map $\pi:X_1(N) \rightarrow X_0(N)$. My question is what is the image of c under this map? Specifically, is it possible for $\pi^*(c)=0$?

there were incorrect claims in my original question. I removed those claims.
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Soroosh
Soroosh
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Let N be a prime integer. We know that the element c=(0)-(\infty) generates the torsion subgroup of J_0(N) and it has order Num( (N-1)/12). Now, there is a natural map $\pi^*:J_0(N) \rightarrow J_1(N)$, coming from the covering map $\pi:X_1(N) \rightarrow X_0(N)$. When $N=11$, we can verify explicitly that $\pi^* (c)=0$. We also know that degree of $\pi$ is equal to the order of $c$. My question is thenwhat is the image of c under this map? Specifically, is it true that $\pi^*(c)=0$possible for all $N$? How would one prove that$\pi^*(c)=0$?

Let N be a prime integer. We know that the element c=(0)-(\infty) generates the torsion subgroup of J_0(N) and it has order Num( (N-1)/12). Now, there is a natural map $\pi^*:J_0(N) \rightarrow J_1(N)$, coming from the covering map $\pi:X_1(N) \rightarrow X_0(N)$. When $N=11$, we can verify explicitly that $\pi^* (c)=0$. We also know that degree of $\pi$ is equal to the order of $c$. My question is then, is it true that $\pi^*(c)=0$ for all $N$? How would one prove that?

Let N be a prime integer. We know that the element c=(0)-(\infty) generates the torsion subgroup of J_0(N) and it has order Num( (N-1)/12). Now, there is a natural map $\pi^*:J_0(N) \rightarrow J_1(N)$, coming from the covering map $\pi:X_1(N) \rightarrow X_0(N)$. My question is what is the image of c under this map? Specifically, is it possible for $\pi^*(c)=0$?

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Soroosh
Soroosh
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Image of the cuspidal subgroup of J_0(N) in J_1(N)

Let N be a prime integer. We know that the element c=(0)-(\infty) generates the torsion subgroup of J_0(N) and it has order Num( (N-1)/12). Now, there is a natural map $\pi^*:J_0(N) \rightarrow J_1(N)$, coming from the covering map $\pi:X_1(N) \rightarrow X_0(N)$. When $N=11$, we can verify explicitly that $\pi^* (c)=0$. We also know that degree of $\pi$ is equal to the order of $c$. My question is then, is it true that $\pi^*(c)=0$ for all $N$? How would one prove that?