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Crystallineperiodic
Crystallineperiodic
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Let $k$ be an algebraically closed field of characteristic $p>0.$ How can I construct two projective curves $C_1,C_2$ of genus $ g \geq 2$ so that the abelianizations $\pi_1(C_i)^{ab},i=1,2$ are isomorphic, but $\pi_1(C_1) \not \equiv \pi_1(C_2)?$

One could try to construct curves $C_1$ and $C_2$ and try to arrange it so that a degree $n$ cover has different abelianization, but I failed to write something concrete.

Let $k$ be an algebraically closed field of characteristic $p>0.$ How can I construct two curves $C_1,C_2$ of genus $ g \geq 2$ so that the abelianizations $\pi_1(C_i)^{ab},i=1,2$ are isomorphic, but $\pi_1(C_1) \not \equiv \pi_1(C_2)?$

One could try to construct curves $C_1$ and $C_2$ and try to arrange it so that a degree $n$ cover has different abelianization, but I failed to write something concrete.

Let $k$ be an algebraically closed field of characteristic $p>0.$ How can I construct two projective curves $C_1,C_2$ of genus $ g \geq 2$ so that the abelianizations $\pi_1(C_i)^{ab},i=1,2$ are isomorphic, but $\pi_1(C_1) \not \equiv \pi_1(C_2)?$

One could try to construct curves $C_1$ and $C_2$ and try to arrange it so that a degree $n$ cover has different abelianization, but I failed to write something concrete.

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Crystallineperiodic
Crystallineperiodic
  • 303
  • 1
  • 6

Two curves of genus $g \geq 2$ in characteristic $p >0 $ with isomorphic abelianizations

Let $k$ be an algebraically closed field of characteristic $p>0.$ How can I construct two curves $C_1,C_2$ of genus $ g \geq 2$ so that the abelianizations $\pi_1(C_i)^{ab},i=1,2$ are isomorphic, but $\pi_1(C_1) \not \equiv \pi_1(C_2)?$

One could try to construct curves $C_1$ and $C_2$ and try to arrange it so that a degree $n$ cover has different abelianization, but I failed to write something concrete.