# 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 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.

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.

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 $$(A,B,C)$$.

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.