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Given a smooth projective curve $X/\mathbb{Q}_{p}$ of genus $g \ge 2$, there is an induced monodromy action of the Galois group on the $\ell$-adic cohomology of

$$X \otimes \bar{\mathbb{Q_{p}}}.$$

What is an explicit example where the action of wild inertia is non-trivial?

By ``explicit," I meant I would like an example that is not, say, a modular curve. Ideally, $X$ would be, say, presented as an effective divisor on smooth surface. Similarly, I would like to see the monodromy described without using automorphic methods.

(Here $\ell \ne p$ is a prime.)

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    $\begingroup$ Just take an elliptic curve with very bad reduction at $p$ e.g. $y^2=x^3-x$ with $p=2$; this is an effective divisor in projective 2-space, and I think it might also be a modular curve but hopefully you won't notice this. $\endgroup$
    – Kevin Buzzard
    Commented Mar 27, 2012 at 23:23
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    $\begingroup$ [Yes, it's modular: the 2-isogenous curve $y^2 = x^3 + 4x$ which is $X_0(32)$; the isogeny is the quotient by the subgroup consisting of the identity and the 2-torsion point $(0,0)$; and translation by $(0,0)$ comes from an element of the normalizer of $\Gamma_0(32)$ in ${\rm SL}_2({\bf R})$. But yes, one can compute directly with the explicit equation, so modularity should notbe counted against this example.] $\endgroup$
    – Noam D. Elkies
    Commented Mar 28, 2012 at 3:11
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    $\begingroup$ One way of seeing that monodromy will be wild would just be to set $\ell=3$ and compute the 3-torsion e.g. by looking up the formula for multiplication by 3 on an elliptic curve and then figuring out the number field in the kernel of the action on the 3-torsion. This number field will already be wildly ramified at 2 I should think. I guess one could also use local-global and an analysis of the automorphic form attached to the ell curve at 2, but somehow I get the impression that you don't want this kind of answer because it easily generalises to the illegal answer "look at modular curves". $\endgroup$
    – Kevin Buzzard
    Commented Mar 28, 2012 at 7:18
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    $\begingroup$ Expanding on Kevin's comment, if you take an elliptic curve $E/\mathbf{Q}$, then wild inertia at $p$ acts non trivially on $H^1_\ell(E)$ if and only if the exponent of the conductor of $E$ at $p$ is $>2$, which also amounts to say that $\mathbf{Q}(E[\ell])$ is wildly ramified at $p$. This can only happen if $p \in \{2,3\}$. Explicit examples (taken from Silverman's Advanced topics in the arithmetic of elliptic curves p. 387) are $y^2=x^3+3$ at $p=3$ and $y^2+2xy=x^3-x^2+2x$ at $2$. To determine the exponent, you can use Ogg's formula or Tate's algorithm (loc. cit.). $\endgroup$
    – François Brunault
    Commented Mar 28, 2012 at 7:56
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    $\begingroup$ To get an example in genus $>1$, start with a genus-1 curve $X_0$ and consider any cover $X$ of $X_0$ (since $H^1(X_0)$ then injects into $H^1(X)$). For example, the $p=2$ and let $X$ be the Fermat quartic $X^4+Y^4=Z^4$, which (as Fermat in effect knew) dominates $y^2 = x^3 - x$. $\endgroup$
    – Noam D. Elkies
    Commented Mar 28, 2012 at 21:53

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