## 64.24 The Legendre family

Let $k$ be a finite field of odd characteristic, $X = \mathop{\mathrm{Spec}}(k[\lambda , \frac{1}{\lambda (\lambda - 1)}])$, and consider the family of elliptic curves $f : E \to X$ on $\mathbf{P}^2_ X$ whose affine equation is $y^2 = x(x - 1)(x - \lambda )$. We set $\mathcal{F} = Rf_*^1\mathbf{Q}_\ell = \left\{ R^1f_*\mathbf{Z}/\ell ^ n\mathbf{Z}\right\} _{n\geq 1} \otimes \mathbf{Q}_\ell$. In this situation, the following is true

• for each $n \geq 1$, the sheaf $R^1f_*(\mathbf{Z}/\ell ^ n\mathbf{Z})$ is finite locally constant – in fact, it is free of rank 2 over $\mathbf{Z}/\ell ^ n\mathbf{Z}$,

• the system $\{ R^1f_*\mathbf{Z}/\ell ^ n\mathbf{Z}\} _{n\geq 1}$ is a lisse $\ell$-adic sheaf, and

• for all $x\in |X|$, $\det (1 - \pi _ x\ T^{\deg x} |_{\mathcal{F}_{\bar x}}) = (1 - \alpha _ x T^{\deg x})(1 - \beta _ x T^{\deg x})$ where $\alpha _ x, \beta _ x$ are the eigenvalues of the geometric frobenius of $E_ x$ acting on $H^1(E_{\bar x}, \mathbf{Q}_\ell )$.

Note that $E_ x$ is only defined over $\kappa (x)$ and not over $k$. The proof of these facts uses the proper base change theorem and the local acyclicity of smooth morphisms. For details, see [SGA4.5]. It follows that

$L(E/X) := L(X, \mathcal{F}) = \prod _{x\in |X|} \frac{1}{(1-\alpha _ xT^{\deg x})(1-\beta _ xT^{\deg x })} .$

Applying Theorem 64.20.2 we get

$L(E/X) = \prod _{i = 0}^2 \det \left(1 - \pi _ X^*T |_{H_ c^ i(X_{\bar k}, \mathcal{F})}\right)^{(-1)^{i+1}},$

and we see in particular that this is a rational function. Furthermore, it is relatively easy to show that $H_ c^0(X_{\bar k}, \mathcal{F}) = H_ c^2(X_{\bar k}, \mathcal{F}) = 0$, so we merely have

$L(E/X) = \det (1 - \pi _ X^*T |_{H_ c^1(X, \mathcal{F})}).$

To compute this determinant explicitly, consider the Leray spectral sequence for the proper morphism $f : E \to X$ over $\mathbf{Q}_\ell$, namely

$H_ c^ i(X_{\bar k}, R^ jf_*\mathbf{Q}_\ell ) \Rightarrow H_ c^{i+j}(E_{\bar k}, \mathbf{Q}_\ell )$

which degenerates. We have $f_*\mathbf{Q}_\ell = \mathbf{Q}_\ell$ and $R^1f_*\mathbf{Q}_\ell = \mathcal{F}$. The sheaf $R^2f_*\mathbf{Q}_\ell = \mathbf{Q}_\ell (-1)$ is the Tate twist of $\mathbf{Q}_\ell$, i.e., it is the sheaf $\mathbf{Q}_\ell$ where the Galois action is given by multiplication by $\# \kappa (x)$ on the stalk at $\bar x$. It follows that, for all $n\geq 1$,

\begin{align*} \# E(k_ n) & = \sum \nolimits _ i (-1)^ i \text{Tr}({\pi _ E^ n}^* |_{H_ c^ i(E_{\bar k}, \mathbf{Q}_\ell )}) \\ & = \sum \nolimits _{i, j} (-1)^{i+j} \text{Tr}({\pi ^ n_ X}^* |_{H_ c^ i(X_{\bar k}, R^ jf_*\mathbf{Q}_\ell )}) \\ & = (q^ n - 2) + \text{Tr}({\pi _ X^ n}^* |_{H_ c^1(X_{\bar k}, \mathcal{F})}) + q^ n(q^ n - 2) \\ & = q^{2n} - q^ n - 2 + \text{Tr}({\pi _ X^ n}^* |_{H_ c^1(X_{\bar k}, \mathcal{F})}) \end{align*}

where the first equality follows from Theorem 64.20.5, the second one from the Leray spectral sequence and the third one by writing down the higher direct images of $\mathbf{Q}_\ell$ under $f$. Alternatively, we could write

$\# E(k_ n) = \sum _{x \in X(k_ n)} \# E_ x(k_ n)$

and use the trace formula for each curve. We can also find the number of $k_ n$-rational points simply by counting. The zero section contributes $q^ n -2$ points (we omit the points where $\lambda = 0, 1$) hence

$\# E(k_ n) = q^ n - 2 + \# \{ y^2 = x(x - 1)(x - \lambda ), \lambda \neq 0, 1\} .$

Now we have

$\begin{matrix} \# \{ y^2 = x(x - 1)(x - \lambda ),\ \lambda \neq 0, 1\} \\ \quad = \# \{ y^2 = x(x - 1)(x - \lambda )\text{ in }\mathbf{A}^3\} - \# \{ y^2 = x^2(x - 1)\} - \# \{ y^2 = x(x - 1)^2\} \\ \quad = \# \{ \lambda = \frac{-y^2}{x(x - 1)} + x,\ x\neq 0, 1\} + \# \{ y^2 = x(x - 1)(x - \lambda ), x = 0, 1\} - 2(q^ n - \varepsilon _ n) \\ \quad = q^ n(q^ n - 2)+2q^ n - 2(q^ n - \varepsilon _ n) \\ \quad = q^{2n}-2q^ n+2\varepsilon _ n \end{matrix}$

where $\varepsilon _ n = 1$ if $-1$ is a square in $k_ n$, 0 otherwise, i.e.,

$\varepsilon _ n = \frac{1}{2}\left(1+\left(\frac{-1}{k_ n}\right)\right) = \frac{1}{2}\left(1+(-1)^{\frac{q^ n - 1}{2}}\right).$

Thus $\# E(k_ n) = q^{2n} - q^ n - 2+ 2\varepsilon _ n$. Comparing with the previous formula, we find

$\text{Tr}({\pi _ X^ n}^* |_{H_ c^1(X_{\bar k}, \mathcal{F})}) = 2 \varepsilon _ n = 1 + (-1)^{\frac{q^ n - 1}{2}},$

which implies, by elementary algebra of complex numbers, that if $-1$ is a square in $k_ n^*$, then $\dim H_ c^1(X_{\bar k}, \mathcal{F}) = 2$ and the eigenvalues are $1$ and $1$. Therefore, in that case we have

$L(E/X) = (1 - T)^2.$

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