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.



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


*Nakajima, Shoichi*, On generalized Hasse-Witt invariants of an algebraic curve, Galois groups and their representations, Proc. Symp., Nagoya/Jap. 1981, Adv. Stud. Pure Math. 2, 69-88 (1983). [ZBL0529.14016](https://zbmath.org/?q=an:0529.14016).


The classical Hasse-Witt invariant, $\gamma$, is the rank of the $p$-torsion in the Jacobian of $C$, it is an integer between $0$ and $g$. We always have
$$\pi\_1(C)^{\text{ab}} \cong \prod\_{\ell \neq p} \mathbb{Z}\_{\ell}^{2g} \times \mathbb{Z}\_p^{\gamma}.$$


So two curves (both over an algebraically closed field of characteristic $p$) have isomorphic $\pi\_1(C)^{\text{ab}}$ if and only if they have the same genus and same Hasse-Witt invariant.


The "generalized Hasse-Witt" invariants of this paper control covers of the curve with Galois group $C\_n \ltimes C\_p^m$ where $n|p^m-1$. (Here $C\_n$ is the cyclic group of order $n$.)


In particular, in Section 6, the author considers the genus $2$ curve
$$y^2+y = Ax + \frac{B}{x} + \frac{C}{x+1} \qquad ABC \neq 0$$
over a field of characteristic $2$. This curve always has $g=\gamma = 2$, but the author shows that the $C\_3 \ltimes C\_2^2$ covers (in other words, $A\_4$ covers) can be either $40$, $39$ or $38$, according to the values of $(A,B,C)$.




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I discovered this paper while researching how to compute a similar answer I thought of; I'll record how that answer would work. Let $p \neq 2$ and let $X$ be a curve of genus $g$. Then $X$ has a unique $C\_2^{2g}$ cover, call that cover $Y$. Then $X$ will have a cover with Galois group $C\_2^{2g} \ltimes C\_p^r$ iff $r$ is less than or equal to the Hasse invariant of $Y$. (How $C\_2^{2g}$ acts on $C\_p^r$ will be determined by how $C\_2^{2g}$ acts on the $p$-torsion in $J(Y)$, but all I'll need is the rank.) I therefore set out to find two curves $X\_1$ and $X\_2$ with the same genus and Hasse-Witt invariant, but where the corresponding covers $Y\_1$ and $Y\_2$ have different Hasse-Witt invariant.


Suppose that $X$ is hyperelliptic of the form $y^2 = \prod\_{i=1}^{2g+2} (x-\alpha\_i)$. For $S$ any subset of $\{ \alpha\_1, \alpha\_2, \ldots, \alpha\_{2g+2} \}$ of cardinality $2k+2$, let $C\_S$ be the genus $k$ hyperelliptic curve $y^2 = \prod\_{\alpha\_i \in S} (x-\alpha\_i)$. I get that $J(Y)$ is isogenous to $\prod\_S J(C\_S)$.


In particular, I took $p=11$ and the two genus $2$ curves
$$X\_1 := \{y^2 = (x-1)(x-2)(x-3)(x-4)(x-5)(x-6)\}$$
and
$$X\_2 := \{ y^2 = (x-1)(x-2)(x-3)(x-4)(x-5)(x-7) \}.$$


I compute that both of these are ordinary (Hasse-Witt invariant $2$). However, of the $\binom{6}{4}$ genus $1$ curves of the form $C\_S$ in the product above, I compute that $11$ of them are ordinary in the $X\_1$ case and only $10$ are for $X\_2$.


Thus, $X\_1$ should have a $C\_2^{4} \ltimes C\_{11}^{11+2}$ cover, but $X\_2$ should not.


